 So, today let us look at some properties of functions of a process which is Brownian motion or a little more generally. Let us see what happens if we have a general Langevin type equation which I will now call by a slightly different name in a minute and the corresponding Fokker Planck equation and from these 2 let us try and see if we can deduce the equation satisfied by functions of the variable random variables concerned. So, to refresh your memory we have been talking about a Langevin equation of the form x dot equal to some f of x plus a g of x times white noise this is Gaussian white noise Delta correlated in the usual fashion. But as I said the rigorous way to do this is not to deal in terms of a singular object like white noise but rather it is integral namely the Wiener process. So, the right way the mathematically better way of writing this is to write this as dx equal to f of x dt plus g of x dw of t. So, this thing here stands for what is loosely written as eta of t dt because remember that W of t was the integral of white noise. So, eta of t dt is essentially the increment in the Wiener process this process itself this Wiener process this had a property W of t average was equal to 0 and W of t W of t prime the average value is the lesser of t and t prime a non-stationary process okay. So, this equation in general is called the Eto equation Eto equation and it is a starting point of the so called Eto calculus about which I will make just a few preliminary remarks okay. It is a slightly different calculus than the one you are used to because this process here this dw is a very weird one as we saw when I explained a little bit about Brownian trails and so on. So, we will see what it implies what sort of rule one has when you have when you are dealing with the Wiener process everything as as non-standard as the Wiener process okay. Now corresponding to this we know there is a Fokker-Planck equation satisfied by the conditional density of this variable X of a random variable X. So, that equation was and I wrote this down without proof this quantity P of X t for some initial condition X naught say this satisfies the Fokker-Planck equation it is the master equation and that equation was delta P over delta t equal to minus delta over delta X F of X P plus one half g square of X times P. So, this is the correspondence between the stochastic differential equation and the conditional density for the corresponding mark conditional density of the Markov process okay. We are going to exploit this we did not prove this but this can be regressively established from ito calculus that the moment you have this equation with this kind of noise you have this for the conditional density the result is a Markov process X of t the driven process and its conditional density then satisfies this equation Fokker-Planck equation okay. So, we will exploit this fact and back and forth okay. So, the right equation the right ito equation or Langevin equation in various cases will be obtained by taking recourse to this Fokker-Planck equation in the cases we can actually solve for it for instance we will be able to find out what related quantities what kind of equation related quantity satisfy okay. Now the matter is quite trivial when g is a constant because then you have pure additive noise and then these niceties can be sort of slurred over but when you when it is a multiplicative noise one has to be cautious about the ito equation okay. So, the first example we look at is what happens to diffusion in d dimensions d spatial dimensions remember that we have an equation for the conditional density of the position p of r t given that it starts at some point say the origin for instance this quantity satisfies the equation delta p over delta t equal to d d2p over del squared sorry del squared p of r and t where del squared is a Laplacian in d Euclidean dimensions okay. Now we are interested in the solution which corresponds to p of r and 0 equal to delta d dimensional delta function at the origin. So, the particle starts the diffusing diffusion starts from the origin and then it spreads out in d dimensional Euclidean space okay. Now this quantity is spherically symmetric and the equation is spherically symmetric as well and we are looking for a solution which is got which satisfies natural boundary conditions namely p of r and t dies to 0 as r tends to infinity along any direction in space okay. So, the boundary condition is spherically symmetric the equation is spherically symmetric and the initial condition is spherically symmetric therefore the solution is spherically symmetric right. So, we are really looking for a solution that is spherically symmetrical and we know what the answer is you can separate this in various dimensionalities can separate this for instance into Cartesian coordinates solve each one dimensional equation etc or you can do this in polar coordinates spherical polar coordinates. But we know the solution and we will exploit this and work backwards from the solution is the solution is p of r, t equal to 1 over 4 pi dt to the power d over 2 in d dimensions e to the minus r squared over 4 dt this is the fundamental Gaussian solution this r squared stands for x 1 squared plus x 2 squared up to x d squared square of the radial coordinate. But what is the actual equation satisfied by this quantity here and just to remind ourselves that we are looking at a spherically symmetric solution let me in an abuse of notation write this p of r vector t as p of radial coordinate r and t although we must remember that this is the probability density of the position of the displacement vector and not of the radial coordinate alone okay. So, what is the equation satisfied by this this becomes equal to d times divided by r to the d minus 1 delta over delta r r to the d minus 1 p sorry that is the Laplacian written out in spherical polar coordinates in d dimensions and retaining only the radial part the angular parts are not the p does not depend on the angular parts and therefore those do not appear in the del squared at all right. Of course when you go to d equal to 3 you get the familiar 1 over r squared d over dr r squared d over dr and let us write that out so this is equal to d times d minus 1 r to the d minus 2 over d minus 1 so this over r delta p over delta r plus d times so you have a term which is not just the second derivative but also first derivative term sitting here. Now let us say we are interested in the probability density function of the radial distance from the origin that is the modulus of r and we need to know what its equation is what its solution what equation its distribution satisfies and what is the actual corresponding stochastic differential equation. So we are trying to work backwards and therefore let us call rho of r, t the radial distribution function of this fellow r so r equal to mod r and we want the distribution function of this now what is the customary we are doing it is to say look I take this fellow the actual solution for the pdf of the vector r and then I integrate over all angles and what is left is the radial distribution in 3 dimensions it will be 4 pi r squared e to the minus r squared over dt and so on. What is it in d dimensions you need the analog of this 4 pi in d dimensions all solid angles if you like the surface of a unit sphere in d dimensions is what you need in 2 dimensions it is 2 pi because the radial circumference is 2 pi and in 4 dimensions it is 4 in 3 dimensions it is 4 pi in 4 it is 2 pi squared and so on. The general formula is 2 pi to the d over 2 over gamma d over 2 multiplied of course by this factor r to the d minus 1 and then a p of r and t. So if I multiply this by this factor here and then put in this volume element or factor in the volume element I get the radial distribution you can check I mean when d equal to 2 for instance it is going to be equal to 2 pi this is 1 2 pi r times p and when d is equal to 3 then you get 2 pi to the 3 halves that is pi root pi over gamma 3 halves which is half root pi the root pi cancels the half goes up there and it gives you 4 pi r square and so okay. So this is what rho is whatever it is this constant factor is actually irrelevant because this whole thing is homogeneous in p and therefore it is going to cancel out what is relevant is that you have an r to the d minus 1 times p. So all you have to do is to remember that rho is r to the d minus 1 times p apart from a constant and therefore p is rho divided by r to the d minus 1. So this immediately says delta rho over delta t equal to d times d minus 1 over r delta over delta r rho divided by r to the d minus 1 plus d you are guaranteed that rho is going to satisfy this differential equation right and now the task is to simplify it right out this second derivative term and so on and you get an equation which is going to tell you what so this will lead to implies a Fokker-Planck equation for rho of r that is trivial to write down you write down what this equation is and then you work backwards you work this correspondence backwards to find out what is the stochastic differential equation satisfied by r the process r okay. So this is not a very trivial process in that sense because writing down the equation for each Cartesian component of this position vector is one matter you have uncorrelated noise in the different Cartesian components but now you are saying what is the stochastic equation satisfied by the square root of x 1 squared plus x 2 squared up to x d squared okay and to navigate between the Fokker-Planck equation and the stochastic differential equation you got to be very careful as you will see in a minute if you did that finally the stochastic differential equation or the Ito equation for r of t for this process itself turns out to be the following it turns out to be r dot or dr equal to minus d into d minus 1 over r dt plus root 2d times dw or if you like the equation satisfied by rho will be delta rho over delta t equal to d sorry d here plus here minus d delta over delta r minus delta over delta r d times d minus 1 over r rho plus d d 2 rho over dr. So I am skipping the intermediate steps what you have to do is to write this out as in equation for rho by taking out these factors recombining them and so on and eventually you end up with a thing like this which of course will immediately imply a stochastic differential equation of this kind here. So notice that there is an extra term that is appeared here what is happened is that this guy gives you a drift this looks like the F of x comma t this is like a drift here and the drift says that you are getting pushed away from the origin okay and as I explained very briefly last time a little earlier this is a real effect because what is really happening is that the chances of any fluctuation increasing r a greater than the ones that decrease r and this tendencies enhanced as you go towards r equal to 0 because if you are at the origin itself the moment there is a fluctuation you have increased r definitely. So this drift term is a real effect it is really there okay even though the noise in different Cartesian components is completely uncorrelated the moment you combine these fellows together into the square root of x 1 squared up to x d squared you get this systematic drift term here even though there is no external potential there is no force of any kind you still get in the radial variable a drift term okay. Now of course we already know the solution to the corresponding Fokker Planck equation because we work backwards it is right here the normalized solution is all you have to do is to substitute this Gaussian form here and simplify it a little bit and that is the solution but our interest here is in seeing what kind of itto equation you get for this process okay. What about the square of this variable what about r squared what happens if you deal with the random variable let us say r squared equal to x 1 squared sorry r equal to r squared which is x 1 squared plus x d squared what sort of equations that going to satisfy what kind of Fokker Planck equation or whatever you get for that well one possibility is to say the following is to say look this implies that 2 that dr equal to 2 r dr just differentiating and then I put that in here I multiply both sides by 2 r right and then I end up with dr equal to twice d into d minus times d minus 1 the r cancels out and then there is a dt plus twice root 2d times little r but that is a square root of capital R so d so the question is this the itto equation for dr looks very straight forward all I have to do is to multiply this by 2 r and that is the equation that is wrong this is wrong it is not possible to do it this way because the itto calculus stops you from doing this it is incorrect to do this I will mention a little bit about the itto calculus and then you will see what goes wrong in this because this quantity here in a sense the fact that w of t w of t prime is minimum dt prime shows that dw is of order square root of dt so that is got to be kept in mind when you play around with these differentials here the right way to do this is to say alright one way to do this is to say the following is to say okay I have this r equal to r squared I have a solution for this little row in fact I have the Fokker Planck's satisfaction equation satisfied by it so what I should do is to see how the distribution of this capital R is related to the distribution of little r okay and use the relation capital R is little r squared right then if I call the density function of this fellow let us say it is pdf equal to some pi of r, t say I want of a better notation then I know that pi of r t dr equal to and that is a monotonically increasing function of little r so I do not have to worry about the sign of the Jacobian this is equal to r of r dr so this will tell me that that will immediately tell me that row of little r and t equal to pi of r and t dr over dr that is equal to pi of r, t and this dr over dr of course is to r that is certainly true so this is equal to twice square root of r but I have a Fokker Planck equation for row so I put this into that equation for row and convert all delta over delta little r in delta over delta capital R and then delta r over delta little r okay which is twice square root of r so I should be careful to do that and once I do that I have a Fokker Planck equation for this pi in which the independent variable the random variable is the variable of which I am interested in is capital R so the derivatives will be with respect to capital R on the right hand side and from that Fokker Planck equation I can go back to the stochastic differential equation satisfied by this capital R okay now that is a completely correct method because we have not done anything playing around with the Eto equation directly what we have done is to go to the Fokker Planck equation and say that instead of little r I use capital R as a random variable and then it turns out that the correct answer is almost this it turns out and I leave this as an exercise to you is 2d that is the correct equation not the one you would nively get by multiplying this by 2r so it is the logic clear we are doing some we are taking a shortcut to this whole business I started with one solution to the diffusion equation the spherically symmetric solution that is the Gaussian solution in d dimensions right and then when I want functions of the various coordinates which are sufficiently symmetric namely spherically symmetric then what I do is to start with that original diffusion equation write it in radial coordinates in polar coordinates for the radial variable and then make changes of variables every time in the Fokker Planck equation to go from one independent variable to another and that tells me what the correct Fokker Planck equation is for corresponding variable and then the density and from there I go back to the stochastic differential equation okay and that will give me the right answer always in every case so notice that again for this capital R there is a drift here and so on okay. Now what does this thing actually look like what is this distribution actually looked like for the radial variable remember that for the radial variable this row so by the way this process r of t it is called a Bessel process this is square root of x 1 squared up to x d squared square root of d squares of d Brownian motions in the case of d equal to 2 the distribution is very nice and easy to write down particularly easy to write down it was r to the d minus 1 e to the minus r squared whatever it is and in two dimensions it is proportional to r e to the minus r squared over 4 d some constant times this is what row of r equal to this in 2d this is called a Rayleigh distribution the special name for it just trying to check whether yes it is called a Rayleigh distribution what does it look like for small r linearly increases and then of course it dies down very fast exponentially in this form it is particularly convenient it is applied in various other places we will not talk about that right now but it is got a name because it is got some special significance this process here is useful elsewhere as well not just in the context of Brownian motion okay. Now this formula here with this value little d if this were d minus 1 it would vanish this term would vanish in d equal to 1 right but you can independently do the following thing I can start with the one dimensional diffusion Brownian motion and look at the square of that variable look at x squared right and ask what is the probability density function of x squared of course x squared will have a sample space running from 0 to infinity but you can ask what is the distribution of this x squared and that is not very hard to write down and then from there from that solution you can now go back find out what is the Fokker Planck equation corresponding to it it is easy to write down and then go back to what the Ito equation is and it will turn out to be exactly this equation with little d set equal to 1 so if you look at ordinary Brownian motion in one dimension so d equal to 1 now you have an equation the stochastic equation is dx equal to square root of 2d dw of t that is it because I had written this earlier as x dot as square root of 2d times 8 of t now I would need a little more rigorously if you like in terms of as an Ito equation now what is the Fokker Planck equation corresponding to it it is delta p over delta t equal to d over dx 2 of course that is the usual diffusion equation right and the solution the fundamental solution p of x and t is 1 over root 4 pi dt e to the minus x squared over 4d and now I can ask what about the distribution of the random variable r equal to x squared it would be tempting to multiply this by 2x but that is wrong that is wrong because there is no drift term here at all it is not correct what you have to do is to write this density so let us call the density of this pi so you have pi of r and t dr equal to p of x and t dx actually what you have is dr over dx you have this in this form and then compute this guy paying attention to the fact in this case what attention have I not paid I have not paid attention to one simple fact it looks looks very right right when this thing looks absolutely right what what is missing here it is not monotonic it is not monotonic because remember minus x and plus x give you the same capital R right so what I have forgotten to do is the conversion which I did between the radial coordinate and this are I it involved an angular integration surface of the unit sphere in d dimensions it was 2 pi in d equal to 2 it was 4 pi in d equal to 3 2 pi squared in d equal to 4 what is it in d equal to 1 2 because how do you define a unit sphere centered at the origin the locus of all points unit distance away from this point right so in the case of one dimension there are two points so you have 0 and then your plus 1 minus 1 those two points so there is a factor 2 this surface so you really got to multiply this by 2 this guy gets multiplied by 2 you did this earlier in the quiz we had you had to find the distribution of the probability that the modulus of the difference between two Poisson variables was equal to 3 so what you did such a few has got the right answer what you did was to say that this difference could be either plus 3 or minus 3 so you just add up the probabilities in that case it was not just multiplying by 2 because the distribution was not symmetric but in this case it is because this Gaussian is symmetric under x goes to minus x so this pi has a factor 2 extra factor 2 multiplying when you put that in you get in a solution for this pi out here and then you can work backwards and ask what is the corresponding Fokker-Plank equation and then what is the actual diffusion what is the Eto equation for the variable x and it will turn out to be this with d equal to 1 so it turns out in this case here this variable here obeys it d r equal to 2 d d t plus 2 2 d r notice that there is multiplicative noise this implies multiplicative noise because if you write the Fokker-Plank equation down for it what would that be this would say delta pi over delta t equal to minus delta over delta r 2 d times pi out here plus 1 half the square of this guy the 2 goes away so there is a 4 d delta 2 over delta r 2 r squared p sorry r times pi so this is the Fokker-Plank equation satisfied by the square of Brownian motion one-dimensional Brownian motion by the density probability density for the square of Brownian one-dimensional Brownian motion okay there is this term here which you would not get if you did not take the correct Eto equation okay what about the nth power of ordinary Brownian motion what about this guy x to the power n so I will call this some other variables i equal to x to the power n where x satisfies this Eto equation so what you do is again say let me find first the density function of this guy here or at least the equation satisfied by the density of this variable by starting with the density of x itself paying attention to whether it is monotonic etc etc and then once I have that Fokker-Plank equation I will go backwards and write the Eto equation for psi okay so I leave you to prove that this thing here also satisfies an Eto equation which has got a drift term and a diffusion term and there is multiplicative noise in this case so the nth power of Brownian motion or a diffusion process is also a diffusion process okay what about the exponential of Brownian motion what happens if I exponentiate it let us see where that gets us so let us put psi equal to e to the power x actually I should put e to the x I should be careful about dimensions but let us assume I measure x and some units appropriate units I put an L for a scale factor if you like but let us assume this is dimensionless now and then ask what is the equation satisfied by this exponential here okay now this is of course a monotonically increasing function of x starts at 0 when x is minus infinity goes to 1 and goes off to plus infinity as x tends to infinity right so there is no hassle about folding and doubling and so on and so forth then if I call the density of this row of psi t d psi equal to p of x t dx so d psi over dx d psi over dx is e to the x which is I itself right so immediately I get p equal to x row and then I go back and say delta over delta t sorry psi times row there is a row here and there is a psi outside because that is not getting differentiated it is an independent variable this is equal to on the right hand side I have d times d 2 over dx 2 times p that is I row so let us write this as delta over delta x psi row but this is delta over delta psi times delta psi over delta x which is I itself and this delta over delta x let us write it as delta over delta psi and multiply by a psi but that removes this and that is it I simplify this guy and whatever I get is the stochastic is the Fokker Planck equation for room okay from that I can get the equation satisfied by psi itself so let us do that what is this imply delta row over delta t equal to d times delta over delta psi times that is the first term when I differentiate this let us do it slowly I psi row plus psi squared delta row over delta psi this equal to d times delta over delta psi is a row not so thrilled by this plus d psi delta row over delta psi not sure if I made a mistake somewhere okay let us see what happens this is not the way to write it is not the best way to write it 3 there this is not the best way to write it because I need to write it as a drift term which is a delta over delta psi with a linear term row on the right hand side bracket some function f of psi times row plus d 2 over dx d psi 2 g squared times row so I need to manipulate this a little bit well not thinking too clearly but anyway so you got to write this back as minus delta over delta psi f of psi row plus one half d 2 over delta psi 2 d squared of psi times row so you have to bring it to that form yeah yeah zeta square row minus row times 2 psi then you will get so this term I write it as equal to d delta over delta psi psi row is already there plus delta over delta psi of psi squared row mine minus 2 psi 2 psi row that is right thank you yeah perfect so this will give us equal to minus d delta over delta psi of psi times row this is a minus sign plus d d 2 over d psi 2 psi square very good that is it so that is the usual Fokker Planck equation so what will that imply now for the stochastic differential equation this will immediately imply that d psi is equal to d psi dt plus square root of 2d times psi times dw of t so when I take a half the square of it I get a d and then inside gets a xi squared okay. So notice in this case that both the drift and the diffusion terms in the Ito equation are linear and xi the standard one for x itself this is absent of course and this guy here is a constant but now you got a xi here the honestly no one back had this as a constant and this linear but this is in between very strange combination here okay and this is called geometric Brownian motion in quotation marks there is a small generalization of this and that is the model that is used in economics when they want to model stock market fluctuations and so on the model used is d psi equal to some alpha xi dt plus beta xi dw of t and this is called the black Scholes model the process itself the xi process itself if you use the Ito calculus turns out to be a functional of v of inner process it turns out to be xi has the same distribution as e to the power alpha times Brownian motion plus alpha minus half beta square times t next time I will say a little more about the Ito calculus we will write down the rules the integration rule and the differentiation rule for the Ito calculus then this will become clear immediately. So the way to handle this strange object this dw is to change the rules of the calculus somewhat and we will do that in an intuitive way we will try to justify it in an intuitive way it is after all a rule there are other rules other ways of arriving at stochastic differential equations when you have multiplicative noise this is the problem the correct stochastic differential equation to lead to the mastery equation you need to write the correct stochastic differential equation and I will explain this two different conventions there is the Ito convention there is the Stratonovitch convention there are other conventions as well and you will see what the differences are between these it has to do with how you write an integral as a sum over increments and depending on what convention you use you get different equations but the physics cannot be different finally the probability distribution of the variables you are talking about uniquely specified in any case. So the way to put it very roughly you start with an equation for average values or moments in engineering equation and then you make a model to a stochastic differential equation by adding noise on it and from there you go to a Fokker Planck equation for the density function probability density the initial point and the final point have to be unique they are measurable they are testable in between but the point is what is the intermediate stochastic differential equation you are going to write down and what is the prescription to go from that stochastic differential equation to the mastery equation and you can have two different roots in each of these cases. So the equations could differ in between but the root from the equation to the Fokker Planck equation the prescription also differs in a manner to compensate for this and get the same equation no matter what interpretation I will try to explain this with an example next time.