 Thank you very much for the introduction and for inviting me to speak here. I'm very pleased to be able to do it. So the planning of a talk is fairly conventional. So first, there'll be an introduction and a state new result for you. And then quite a lot of the talk, I want to spend discussing the proof. So most of the discussion will be pretty general. And I hope quite accessible, more or less will tell you all backgrounds. But there is a more technical bit as well, which if I have time, I'll say something about the end. And then I'll finish off with a couple of remarks. So to motivate this, I want to start off by talking about the Mobius function, which is something that many people know a lot about, but maybe not everybody. So the Mobius function is a function of natural number n. It's really important in analytic number theory. And this is the definition. So if n is divisible by any squares of primes, then mu of n is just 0. And if not, so if n doesn't have any repeated prime factors, then mu of n is minus 1 or 1, depending on the parity of the number of prime factors. So for example, mu of 1 and mu of 6 are both 1, because 1 has 0 prime factors and 6 has 2 prime factors, which are both distinct. Mu of 2 and mu of 3 and mu of 5, and more generally, mu of any prime is minus 1. And mu of 4 is 0, because 4 is divisible by 2 square. And a key property of this function, which is kind of resonant with the title of my talk, and this is the most important thing for today, is that the Mobius function is multiplicative. And that means that mu of n times m is mu of n times mu of m, provided nm are co-prime. And if you want to check this, the co-primality of nm just means that the no repeated prime factors condition splits up into a separate condition for n and for m, because the prime factors of nm don't sort of interfere with each other. And then if you have the square freeing us condition, then go repeated prime factors condition, then the parity just splits up as the parity of n and the parity of m. Okay, and why is this so important? Well, here is a little hint about why. So if the real part of S is bigger than one, if S is some complex number, you can form this Dirichlet series who's coefficient to the Mobius function. And this converges absolutely and defines a homomorphic function on this range. And if you multiply this by the Riemann's zeta function, zeta of S and see what you get. On this range of S, the zeta function is given by an absolutely convergent product and all the product, which is this. So product of all primes p, one minus one over p to the s to the power minus one. And on this range, this Dirichlet series with coefficients, Mobius is also given by where the product, where the term is one minus one over p to the s. And if you want, if you're already bored with what I'm saying, you can try and prove this by just formally expanding out the products and collecting the terms. And the minus one here corresponds to the minus one to the homomorphic in the definition of you. And if you look at this, clearly every term in this product cancels with the corresponding term in that one. So this is all just one. But at least when the real part of S is bigger than one, zeta times this Dirichlet series is always one. So that says that small values or zeros of zeta functions sort of correspond to large values or to poles of this Dirichlet series with coefficient mu. So that's a bit of a hint of why this Mobius function is interesting. And I'll come back to that a bit later. Okay, so in the forties, Winkner, who is maybe the first person ever to look at random multiplicative functions, what he wants to do is to construct a random model for this function, the Mobius function. So he wanted to find a random real sequence or a random real valued function that mimics some of the properties of the Mobius function. So I want to explain what his construction was. So he took on the primes, just an independent sequence of coin tosses. So an independent sequence of plus or minus ones probability half each. And these are called Radimack and random variables. Sometimes these kinds of plus or minus one random variables. And then he extended this function to a functional all the natural numbers in the following way. So if n is not square free, if n has any repeated prime factors, he just set f of n to be zero. And if not, if n is square free, so all the prime factors are very different, then you set f of n just to be the product of f of p for all primes p dividing it. And this random object is called a Radimacker random multiplicative function because it's constructed from these underlying Radimack and random variables. And if you think about how this behaves, but we do have multiplicativity, we have f of n m is f of n f of m provided n m a co-prime, just like for the Mobius function. Because once again, the square free-ness condition, if n m a co-prime just splits up as a separate square free-ness condition for n and for m. And if they are co-prime, then the prime factorizations just concatenate here. So you get this support. Clearly, f of n and mu of n both to zero, if n has non-trivial square factors, just by definition. And on the square free-n, f of n takes values plus or minus one and so does mu. So f mimics these properties of this, okay? You can wonder whether this is a really good model for the Mobius function or not. In my opinion, it's not such a great model for certain questions, but it is the model that Winkler wanted to answer that. I want to mention a variant of this as well. So as well as the Mobius function, there are other very interesting number of theoretic functions in a multiplicative. So their functions like n to the minus i t is some fixed t, mu is a function of n. All of the context through which they can describe n for people who know what they are. So these are functions that are very important for studying the distribution of things in arithmetic progression. And to model things like this, we use Steinhaus random multiplicative functions, which are a variant of the previous definition. So this is how it works. So now on the primes, you take the values f of p to be independent, but uniform on the unit circle in the context, instead of being plus or minus one. And this kind of sequence is called a Steinhaus sequence, sometimes Steinhaus random variables. And then we make these into a multiplicative function just by setting f of n to be the product of f of p to the power a, where p to the a is the highest power of p, which divides that. So if n has prime factorization, p to the a one, p one to the a one times p two to the a two through to p w to the a w, then f of n is f of p one to the power a one times f of p two to the power a two, and so on. So here we don't worry about the square freedoms. And you can check that in this case, f of m n is equal to f of m f of n, actually for all m again, without the co-primality description, just like for n minus i to the power of k. Now the key thing to note about both of these, especially if you're from maybe a more probabilistic background and not so interested in the number theory motivation, is that these values f of n, they're not independent random variables. So some of them are on the primes, they are by construction. But if you look at the whole sequence f of n, they're definitely not. For example, f of six just by construction in both of these cases is f of two times f of three. So if you know f of two and you know f of three, then you completely know f of six. So they can't possibly be independent. And this means that the classical probabilistic techniques don't transfer at least directly to study these random multiplicative functions f of n. And that's what makes them challenging and interesting to study. So maybe this would be a good point to ask. If anybody has any questions about the definitions, please ask now because you won't understand much else before I say if this isn't clear. Okay, so let me continue. So for the merbiast function, it's conjecture that you should have a lot of cancellation in the partial sums. So this sum here could in theory have size order x because mu takes values plus or minus one, zero. But it's conjecture that it should be a lot smaller than and it's conjecture that this should be bounded by a constant times x to the half plus epsilon for any fixed. And this is actually equivalent to the real hypothesis. So this kind of manifests what I hinted at earlier that if you know that sums of merbiast or the corresponding Dirichlet series don't get too big, that tells you that you can't have zeros of a z to function in bad cases. So we expect a lot of cancellation in this sum, but we also expect it should be very hard to prove because it's equivalent to our rate. So what Wiener was interested in doing is proving bands for his random model object, his random multiplicative function to try to get an idea of what might be going on for merbiast. And the kind of bands he wanted were almost sure bands. So that means bands that hold a probability one. And let me just emphasize, so something that holds a probability one, it doesn't mean it necessarily always occurs. So there are events that can happen, but they happen with probability zero. But something that holds the probability one, that would seem to give a reasonable kind of indication of what might be going on. And what he particularly had in mind was this classical result, the Lore-Veterated Algorithm. So this says if you have a sequence of plus or minus one point-osers, which are just completely independent, so a very classical probabilistic set up, and you look at the partial sums, they're almost surely, so with probability one over all these kinds of sequences, the Limb-Soup will look like square root two x local x when x gets big, and the Limb-Inf will look like minus square root two x local x. So almost surely these partial sums will get as big as root two x local x, infinitely often, and they'll get to minus root two x local x in the negative direction, but they won't get bigger than that. So this is a very precise statement about the almost sure fluctuations of these sums. And what he was interested in is having something like this in this more arithmetic situation in this random order of model. And let me just emphasize here. So if I fix an x, and I look at the sum of the x long n by the central midterm, this has a roughly Gaussian distribution. So if I fixed x, this is roughly Gaussian with mean zero and variance x, which is the number of times. And we expect something like that to be of size like root x roughly, like the standard deviation. But the growth you get here is bigger. This is root x local x. So when you take the Limb-Sook, when you look at all the values of x, simultaneously, you get larger fluctuations than if you just fix one point. And that makes sense, but that's kind of the heart of the result to understand the effect of looking at all these x simultaneously. So as I said, what he wanted is something like this in the random multiple effects. So let me give a kind of a potted history of what's known about this. So these are all in the in the Radimacker case, the first case I explained, which you can think of throughout the talk if you want. And these are all almost sure-banned. So these hold with probability of one. So Winter himself proved that the sum is bigger of x to the half plus epsilon, and is not bigger of x to the half minus epsilon. So it is bounded by x to the half plus epsilon, almost surely. And it does fluctuate at least as big as x to the half minus epsilon, if you're from. So this is like the equivalent of the Riemann hypothesis for this random. This was improved by Erdos and some unpublished work. So he got to Rutex times a positive power of log for the upper bound, and Rutex times a negative power of log for the lower bound. And then there's a really beautiful seminal paper of Holash in this area from the 80s where he improved this again. So he got to something roughly like Rutex e to the root of log x in the upper bound, this extra factor that I won't quite say, because I'll make a moment full of it. And Rutex e to the minus root of log x in the lower bound. So just to be very clear here, so the log x is e to the log of log x. So the improvement here is this is e to the root of log x, so the improvement is the square root in the exponent here. And there are lots of nice new ideas in this paper. So if the upper bound, he uses conditioning and he applies moment bounds just to the randomness from the large primes, and this connects with some nice issues in harmonic analysis actually with some hypercontractive qualities. And for lower bound, he makes a connection with large values of the other products. So he can produce large values of these sums if he can produce large values of the corresponding random or the product. And then he creates just by hand, the very clever argument for doing that. And that's where this comes from. And then most recently, the upper bound was improved to big O Rutex log log x squared roughly by Pascal and Lao Teng of Balmang Wu, not really independently. They sort of think you will, each other we're doing. And this is by improving the treatment of the moment in whole ashes upper bound. So they introduce a splitting device that makes the moment calculations more efficient. And that's where this improvement comes from. And I prove that the sum is almost surely not the goal of Rutex divided by log x to five over two. And this is by improving the estimation this random or the product in whole ashes lower bound. So instead of using the very clever ad hoc argument that he did, I introduced some tools from a Gaussian process theory to do this. It gives a sharper result. Now the thing to notice about all these bands is that lower bands are all smaller than Rutex. Okay, none of them grow fast in the Rutex or even as fast as Rutex. And that seems like quite a weird situation to a number theorist, I would say. So we don't usually expect be seeing consistently smaller than square root point directions. But it's not clear that it might not be happening in this case. So Hollash asked whether it might be true that the sums of bigger Rutex almost surely. And on the other hand, Erdisch conjectured that that shouldn't happen. Conjectured almost surely the sum should not be bigger Rutex. I've just mentioned that this was in the Steinhaus case and this was in the Rademacher case to be fair to what they actually said. But I don't think we expect really any difference between these. So there's no reason to think that looking at Steinhaus or Rademacher should change the behavior in this respect. And Erdisch also said in this same paper that, as far as he knew, he had a plausible guess for what the real size of the issue. So the results I want to talk about today resolves this issue to an extent at least. So this is the new theorem. So for both Steinhaus or Rademacher, random multiple functions, if V of X is anything that tends to infinity, however, slowly, then almost surely, so with probability one, there'll be large values of X where the sums of bigger than Rutex times log log X to the one quarter divided by V of X. So in particular, you could take V of X here to be log log X to the power 0.001, let's say. And then the right-hand side, you'd be growing, I appreciate the faster than square Rutex. And the theorem says that almost surely, you'll get fluctuations that big infinitely often. So this shows that the asked Holash's question is no, and Erdisch's conjecture, it's true. And regarding this bit, nobody has a plausible guess. Well, I still wouldn't claim to have a plausible guess for the exact order of the magnitude, but it is plausible that the exponent one quarter in this band might be sharp, and I'll explain at the end of the talk, why that might be. So what I want to do now, unless there are any questions at this point, is start talking about the proof, okay? So what happens here is that we sort of give up on the basic strategy of proof that Holash introduced in his paper, and we go back to try to produce these large fluctuations to a more classical probabilistic strategy like you might apply to the classical or the iterative algorithm, which I'll remind you of here. So what I want to do is try to explain in reasonable amount of detail how you prove the existence of these large fluctuations in this or the iterative algorithm. And I'll do that in a way that certainly isn't new, but if you've seen just kind of one proof of the or the iterative algorithm before or no proofs, then the proof I explained probably won't be the one you've seen before. And once I've explained that, we'll try and see how this can maybe transfer over to the random multiple. Okay, so I claim that to prove the existence of large fluctuations in the lower band part, if you like, the classical or the iterative algorithm, it will suffice to prove that the maximum of these sums when I renormalize by one over root x is bigger than about root two local x with probability tending to one, as x goes to infinity. So let me just remind you, the epsilon ms here are just independent, completely independent plus or minus one random signs. And obviously don't worry about exactly what this interval is, just some reasonably big interval. So why will it suffice to do this? Well, if this holds, then the probability the maximum is smaller than about root two local x will be little over one, would be just the complement of this. And if that's true, then I can select a sequence of x values as sufficiently sparse sequence of x values so that these little low ones look like, let's say the reciprocal of the squares along the sequence of x. And then if I add up these probabilities, these probabilities of the opposite of this event, so these failure probabilities, it will converge along this special sparse sequence of x values, where the little or one terms are looking like the reciprocal of the squares. And then it's a kind of basic and actually quite easy to prove probabilistic fact, the first Royal Canterre lemma, but if you have a sequence of events whose probabilities converge, then with probability one, only finitely many of those events can occur. And if only finitely many of these failure events occur, that the thing is less than or equal to root two local x, that means that from some point onwards, this event is always occurring. So with probability one, at some point onwards along our sparse sequence of x, this maximum will be bigger than this. And that says that one of these sums is bigger than root two x local x, if you move the root x to the other side. And that's what we want. So this is different from the proof you might have seen of the Lore-Beat rates algorithm because here I'm only using the first Royal Canterre lemma. So the usual proof uses the second Royal Canterre lemma and that has a good thing about it, which is that you don't need bands here that hold with probability close to one. You just need bands that hold with sufficiently positive probability, let me say it like that. But the bad thing about the second Royal Canterre lemma is that it requires independence of the events that you're working with. And that will not transfer at all well to the random multiplicative points. So what we've done here is we've moved to some of the difficulty about, so we're only using the first Royal Canterre lemma which doesn't require any independence assumptions. But we now need to get this kind of band with probability very close to one. And what I want to explain is how we can do that. First of all, in this classical setting and then how it transfers over to the multiplicative. Okay, so I already remarked that these partial sums have roughly Gaussian distributions by the central limit theorem. So when I've renormalized by one over root x, this will now have a roughly standard normal distribution. So a Gaussian would mean zero and variance one. And we understand very well how a single standard normal distribution behaves. And what we need to understand to prove logarithmic rate of logarithm is the interaction between all these sums for different x because let me just go back. So here we have a maximum over x. So we need to understand the effect of taking this maximum on the behavior of these ones. And whenever I give one of these probabilistic number three type of talks, what I try to emphasize about Gaussians is that Gaussians are great because to understand their dependence, you just need to compute their correlations or their covariances, which isn't true for general randomness. So we know that the means are zero, the variance is one, what's the covariance which is the same as the correlation if the means are zero and the variance is a one. So the covariance is just the expectation of the product of this thing at two different points, x and y. You can easily compute what this is. So the square root x and the square root y come out. They are. And then when you take the expectation of the two sums because of the independence of the epsilon n's, the only terms that survive are the ones up to the minimum of x and y. So this is exactly what the covariance is. And you can see that this is bounded by one ways, which is good because that's the variance. And if we look at this for the special points, x and y being powers of some lambda, then if x and y are different, the biggest of x and y must be at least lambda times as big as the smallest. So that means that the denominator here, the xy is at least lambda times as big as numerator. So when you take the square root, you can only do at least square root lambda times as big. So if you look at this special kind of sequence of points, if x and y are different, you can say that this covariance is bounded by one over root lambda. And so if the covariance is close to one, if it were exactly equal to one, that'd be saying that these random variables were very dependent, essentially the same random variable. But if the covariance is getting smaller, that's saying that they're becoming more and more independent. The Gaussians, this is true. This certainly isn't true for general random variable. Okay, we need to fit all these points into this interval that we had here to start with. So how many of those points can we get in this interval? Well, basically it's the log of this end point divided by log lambda minus the log of the other end point divided by log lambda. So it's some constant multiple of log x roughly divided by log lambda. Okay, and then for Gaussians, as I said, everything works great because you can understand all kinds of statistics of Gaussians if you understand their variances. So if I have some standard normal random variables, g1 up to gn, and I know that their variances, their correlations are bounded by rho, there's some fixed rho between zero and one, I'd say. Then with probability very close to one, the maximum would be bigger than about square root two one minus rho log n. So if rho was zero, this would correspond to these Gaussians being independent and it's quite well known that the maximum of n independent Gaussians is roughly square root two log n with high probability. And this is the generalization of that to the case where there are some non-trivial correlations. And this is actually very easy, relatively easy, let me say to prove. People will know about this. This follows very easily from Slepian's lemma, whatever that is. Okay, so if we go back to the setting here, so we have a bound one over root lambda for the non-trivial covariances. So rho is going to be one over root lambda. And this is our number of Gaussians, so here's n. So if you plug this in up here, you get a lower bound with high probability, one plus the low one square root two, one minus one over root lambda. And then log of this is still roughly log log n. There's a little bit of dependence on lambda from the denominator, which I've proved the low one. Okay, and lambda here can be anything arbitrarily large but fixed. So if I take lambda big enough, I can get this to be as close to square root two log log x as I want. And that proves the lower bound bit of Kynchian's law of iterative law. So let me emphasize that I've really shown you basically a complete proof here. I didn't give a precise statement of the form of the central limit theorem that you'd want to apply here. I didn't prove this for you, but you could certainly do both of those things rigorously. So this is really a full proof on these three slides. And what was absolutely crucial here is that we're working with Gaussian random variables. So that meant that to understand the size of the maximum, we just need to understand these correlation. Okay, so can this be made to work then for random multiplicative? Well, there aren't any prizes for guessing that just by the anthropic principle, the answer is yes, because if the answer would know, I wouldn't have bothered to spend some time explaining that. But there are some significant problems to address. And the main one is that, unlike in the classical case, the sums where f of n is a random multiplicative function do not have Gaussian distributions. So that's not obvious. This is quite open for a while, whether that would be true or not. But we now know that they definitely don't. And actually, if you renormalize the sum in this way by root x, then this will converge to zero in probability. So I proved a few years ago, but published earlier this year that actually this sum renormalized like this usually has size like one of the log log x to one quarter. So it's getting small when x is five, certainly not roughly standard. But nevertheless, let's just proceed and see what we can make it work. So just like before, it will suffice to prove that these maxima of these sums are bigger than log log x to one quarter of a v of x with probability tending to one. So now we don't have root two log log x here because we're not shooting for that. We have log log x to one quarter of a v of x like in the statement of the theorem for the random multiplicative functions. And I claim that it will actually suffice to prove something like this just for a bit of the sum. So just for the part of the sum where the numbers have a pretty big prime factor, a prime factor bigger than capital X. So the real part here you can forget about really. So if we're in the Radimack case, everything's real anyway. So this real part doesn't make any difference. In the Steinheiser case, taking the real parts just makes it simple, but you don't have to worry about that for the purpose of, okay. So why will it suffice to handle just these subsums? Well, if this bit of the sum is big, if we have a band like this for this part, the only way that the full sum won't also be big is if the other part of the sum, the part where the numbers only have smaller prime factors, were also very big at the same point little x and sort of canceled out with this bit of the sum. And we expect that for any given little x, it should be quite rare to have a large value of either piece of the sum. So to have a large value of both of them simultaneously should be even rarer, should be really rare. So if we can produce a large value of this, it should be very difficult for the other bit of the sum to cancel that. So that's a heuristic reason why it's enough to handle this and actually the way we set this up, it's quite easy to show that proving this is enough. And what's good about this bit of the sum is that these numbers, so they have a prime factor, which is bigger than capital X, but they can't have two because they're not large enough to have two prime factors that big. So these numbers, they look like some big prime between capital X and capital X, the fourth first, prime sum, a smaller number m. And then since f is multiplicative, f of n, which is f of m times p, just splits up as f of p times f of m. So this is a rewriting of the sum, which is there. Okay, and so what was the point of doing this? Well, now, so if I condition on these inner sums, so that means if I kind of freeze the randomness of these inner sums, so because p is big, x of a p is small, so these only depend on the f of p for the small primes p. So if I freeze these, then this sum, it looks like some kind of coefficient, which is now fixed, it's frozen, it's conditioned on, times these independent random variables f of p. And the sum of independent random variables f of p times some coefficients is the kind of thing that we might expect to have Gaussian. So what I want to really emphasize here is that this sum, or indeed this sum, does not have Gaussian behavior. It's not close to being Gaussian. The tails are completely different. So if you think of it as being Gaussian, you're making a big mistake. That's not the situation. But if you do a lot of conditioning, if you freeze and you kind of extract out to the outside all the randomness coming from these bits, then there is a little bit of behavior left, which is like Gaussian behavior, additional Gaussian behavior. And actually you can make this precise using a multivariate form of the central limit. So if you have some probabilistic background, I hope that makes sense to you. If you don't, it might not make very much sense. But just think that there's some procedure that you can do that makes these things look a bit ocean in behavior. And that allows us to make some connection with the presentation I gave before from the classical case. So now to understand this, we need to compute the means and the variances and the co-variances just like before. But now they're the conditional means and the conditional variances and the conditional covariance. The means though are still very easy. So whatever these things are that we've frozen, conditioned on, the FFPs are independent and they have mean zero. So the conditional means are still always zero. So this is easy. So how about the variances and the co-variances? Well, the variance calculation, the conditional variance calculation is very similar to one I already did in my previous work where I showed this band. So I won't say very much about that. It turns out that the conditional variance here is roughly like this kind of integral where F is the product. So F of S is the product of all the primes up to capital X of one minus F of P over P to the S to the minus one. I'd say it's the random or the product corresponding to that. And you can show that with probability tending to one, these integrals will be greater than greater than roughly one over root local X, one over V of X squared root local X where V is this function that's slowly tending to infinity. So let me say here, if you want a probability that tends to one, which we do so that we can apply the first real can tell you lemma, then you need this V of X here. If you just had a lower band like one over root local X, you couldn't say this happened with probability tending to one. So you need to make the band a bit smaller by this V of X factor to have this going. Also, so this is a band for the variance. So if you take the square root, and this becomes a band for the standard deviation, if you take the square root of this, you get something roughly like one over local X to the one quarter. Okay, and that explains the one over local X to the one quarter here. So the reason that this usually has size one over local X to the one quarter is because conditionally, the standard deviation is roughly like one over local X to the one quarter. And for the expert bit of the audience, if there are any such people who know a bit more about this, this is coming from the multiplicative chaos connection that exists in this problem. That's one over root local X factor. So what I want to say a bit more about is the covariance calculation, which is the kind of new thing in this problem. So it turns out that the conditional covariance is the same kind of integral, very roughly speaking, but with this factor X over Y to the minus IT. If you look at the covariance for the sums up to X over Y. So if X is equal to Y, this is just one, you get the same integral as up there like you'd expect. And if X and Y are very close, then this will be close to one and this one to the minus IT, this thing close to one to the minus IT won't change very much over this whole range of integration. So this will still be pretty close to that as you'd expect if X and Y are close together. So the sums are very highly connected. If X and Y get further apart, then this will start to oscillate over this range of integration. You could hope for some cancellation from this factor. And you can show, and I'll say a few words about this. This is gonna be the more technical bit of a book that with probability tending to one, over all the realizations of this sort of product, this will be a lot smaller than this band we have for the variance for a lot of pairs X and Y. So that says that for a lot of pairs X and Y, the conditional covariance is a lot smaller than the conditional variance. So they're behaving like conditionally independent. And that's enough just like in the classical LIL to get the lower bands out for maximum. So I want to explain roughly how you treat this, a little bit of how you treat this. So I'll talk for about five minutes about this. This will be more technical, I think. And then I'll finish with a few remarks which should be more accessible, maybe. Okay, so I want to restrict to looking at point X and Y would look like capital X to the power eight seventh. So just this end point multiplied by E to the power two pi R where R is some integer. So this is very like the situation I had before where I had my powers of lambda. Basically, I'm taking lambda to be E to the power two pi. And I need to fit all these points into this interval so I can take R up to some constant multiple of log X and still have all these points in it. So then if you look at these integrals for this choice of points X and Y, then the capital X to the eight sevenths cancel and you just get here E to the two pi some integer to the minus IT. So you're getting expression like this. And if you want to know that for a lot of integers R this thing can't be too big. The standard number theoretic thing to do is you form a sum like this. So on the outside I sum over all the R then I take some power of this integral, the two k power. I want to show this is a lot smaller than it would be if all of these integrals were this big. Okay, so I want to get some band for this that's a lot smaller than this thing to the power two k times log X, the number of R bands. And a technical point but an important point here is that to make this work we're going to have to take K big. K is going to have to grow with capital X. Actually K will end up being something roughly like log log X inside. And the reason for that is that it's quite well understood in these kind of arguments that to get a good bound you need some kind of balance between the size of the thing that you're taking the power of and the number of terms in the outer sum. And the number of terms here is roughly like log X. And we want to show that it's really a small proportion where this could be bad. It's a very small proportion of those log X terms where this could be big. And by a small proportion I mean some power of log X more than the first power. So we want it to be really almost all the R values where we have a good bound for this sum. So to do that we need the kind of size estimates rough sizes that we have for this to the power two K will be sort of matching the number of terms in this outer sum, matching with the log X terms here. And the rough kind of size of this is going to be like one over some power of log log. So we need to take K big enough that one over some power of log log to the power K sort of roughly compensates for the number of terms in the outer sum. And that's where this choice of K comes from. So as I warned you at the start this is the more technical bit but maybe if you do know about this kind of arguments that makes some sense to you. So in particular taking Ks be one or two or something certainly will not suffice. It's enough for good values of R. Okay, so if you believe me about that then once you have something like this the obvious thing to do is to just expand the two K power and then move the sum over R inside and try and perform the summation. So if you do that you'll end up summing up things like this E to the two pi R and then a sum and some difference of the TIs which come from expanding the two K power of this. And this was the reason for putting the two pi there. So this is a kind of nice sum for which we have some easy bands. So you get a band like this. It's at most the minimum of a number of terms which is roughly log X or something that depends on the distance from this thing in the bracket to the nearest integer. So if this thing isn't very close to an integer so if this distance to the nearest integer is at least one over log S to the power two thirds let's say then we get a band here which will be at most log X to the power two thirds if you just plug this band in there and that's very good. That's a saving by a power of log X. And what you might think now is well the part of the region of integration where this doesn't hold so where this is smaller than one over log two thirds X that's a very little piece of the whole range of integration. So we might expect that that should give a very little contribution but that's not true. So it turns out it does give a small contribution and that's why the argument works but it doesn't give a very small contribution. And the reason for that is that there are quite big contributions to these kind of integrals even from very small intervals of T. So we expect there usually to be points T where the product has size roughly log X very roughly speaking. So the square of it will have size like log squared X and if you look on a little T interval around there I think one of log is you get a contribution like log which will cancel with this one of a log and give you quite a substantial piece of this whole interval. So there are very little bits of these intervals that contribute quite a lot not everything but quite a lot. So that says that successfully handling the complementary bit will be quite delicate. It's not just a question of saying the complementary bit is just a little piece of the whole integral so it must contribute just a small proportion of the whole integral that's not the case. Okay, so let me very quickly say something about how you do this and then I'll go on to my closing remarks. So what you want to do is to impose in the integral some constraints on the size not just of the whole or the product but also of the partial or the products. So of the partial products over the primes on certain ranges. So I won't really get into this because I think it's too technical but for anybody who really knows about this just know that the band we're imposing here has this factor local gets to 1000 in the normally. So this is quite big this fact. So I'm posing a more stringent condition on the growth of the partial products than you usually do in these kind of products. But it turns out you can do this in the acceptable average contribution to the whole interval. And once you have these bands on the growth then you just distinguish two cases. So if in your tuple T1 up to T2K a lot of the points are quite close together. So let's say all the K of them are quite close together then you use these bounds and you get a big saving for these one of the local gets to 1000 savings. You get a saving like this to the power K roughly. You get a saving like this for all of the points that are very close together. And that's a big win. So that's enough. If these points aren't all very close together then so we sort of know if the points are well spaced then you have some kind of rough independence between the other products. So then you can perform expectation calculations reasonably successfully and then you can employ this condition and get a big saving. So I definitely don't expect anybody who doesn't already know about this to really follow what was on the last slide. But that's the experts. If anybody does have questions I'm happy to answer them again. So let me finish with something sort of gentler with some closing remarks. So I said a long time ago that maybe this exponent one-quarter in the theorem is sharp. Why is that? Well, in the classical or reiterated logarithm you get a bound like root X times square root log X roughly and that's because at a single point X the typical size of the sums is like root X like standard deviation of the Gaussian and then looking at the limb so it increases that by this factor root log log X. Now, simply here the typical size of the point is roughly root X divided by log log X to one-quarter that's the result I mentioned that I proved a few years ago. And again, the band we have for the large fluctuations is this multiplied by root log X. So it's root X times log log X to one-quarter. So the band we have for the large fluctuations in the random multiplicative case is like the one in the classical or reiterated logarithm just adjusted for this difference in the typical size. So that's some reason to think this might be roughly as far from certain for other ideas but if I had to bet a small amount of money might say what it is. Okay, let me finish with something that doesn't have any randomness in it in case you don't like randomness and haven't left already. So a number theoretic analog of the problem I've been talking about I said a while ago that I don't think these random multiplicative functions really are such a good model for the merbiest function even though that's what they were originally introduced for. But what they are a good model for I would say is for Dirichlet characters, a randomly chosen Dirichlet characters well modeled by a Steinhaus random multiplicative function. So understanding the behemoth of the maximum of the sum of F of n over root X with high probability that's kind of like asking for the behavior of the maximum of the character sum the most Dirichlet characters. So asking for something with high probabilities like asking for something to hold for 99% or asymptotically 100% of Dirichlet characters. So the analog of theorem one would be asking how this maximum behaves for most Dirichlet characters model. And what I would expect is that if capital X isn't too big compared with the conductor R so for example capital X at most R to the three quarters so that capital X to the fourth third is most R then I would expect this to be well modeled by the random problem by theorem one. So I expect that for most characters mod R this maximum should be bigger than about local X to the one quarter. And so the questions are what can you prove rigorously in the direction of this? And also what happens for larger capital X? So when capital X gets closer to R this can't continue to hold. There are lots of reasons for knowing that. So there should be some transition that happens when capital X gets close to R. So the question is what is this transition? And this question, so a lot is known a lot of work has been done on large values of character sums but almost all of this work is about what happens for kind of peculiar characters. So can you construct weird characters that make the character sum unusually big by kind of finding characters that behave strangely on the small primes, let's say. And what I'm asking here is a different question. So I don't want these to be weird characters some special subsets of the characters. I want to know that for most characters so for a typical character you can get large values just by moving the end point of the range of summation. So as far as I know, essentially nothing on trivial is known about this problem. So I would conjecture that this should be the answer for this range of capital X at least. And if anybody could prove anything as strong as this that would be great. But proving anything more trivial I think would be something that's not known. Okay, that's all I want to say. I think that's fine. So that's it. Thank you very much.