 Dear students, I will be presenting to you an example to show you that for any distribution, the harmonic mean is less than or equal to the geometric mean and the geometric mean is less than or equal to the arithmetic mean. So let a1, a2, so on, so on up to an be a set of positive numbers. So let us create a distribution for a random variable x by placing a weight of 1 over n on each of these numbers a1, a2, so on up to an. We are doing this when we apply 1 over n weight on each of these numbers. This means that we are creating a uniform distribution. Alright, once we have got that, then the mean of the random variable x, the arithmetic mean that will be equal to summation ai over n. This is an elementary formula that we do not do in terms of probability distribution, but we do it in terms of data. In fact, here I will be writing expected value of x and I will once again make you recall that these numbers a1, a2, so on up to n with probabilities 1 over n for each one of them is actually a discrete, you know, the probability distribution of a discrete random variable. Then what is the formula for the expected value of x for any discrete distribution? Do you not know that it is summation x into p of x? So here you apply summation x into p of x. What will it become if we expand it and open it up? x1 into p of x1 plus x2 into p of x2 plus so on so on xn into p of xn. So just apply it. Instead of x1, x2, we are saying a1, a2 and p of x1, p of x2, instead of that we are saying 1 over n. It is none other than the discrete uniform distribution. Alright, so when we do it like this, then obviously when 1 over n is attached to each other, then it will come out common. When it comes common, then you will write it like this. Expected value of x is equal to 1 over n into summation ai where i goes from 1 to n. Okay, so this is the expression for the arithmetic mean of this particular discrete uniform distribution. Now, let us apply the Jensen's inequality and just keep in mind what is the Jensen's inequality. It is as follows, if phi is a convex function on an open interval i and if x is a random variable whose support is contained in i and also if this random variable has a finite expectation, then phi of e of x is less than or equal to the expected value of phi x. If phi should be a convex function, I would like you to note that minus log of x is a convex function and therefore we can apply the Jensen's inequality. So the left hand side of Jensen's inequality is phi of e of x. So if we start using this function instead of phi, minus log, then what will it be? Minus log of e of x which is on the left side. And e of x, we just mentioned earlier, summation ai over n, so you put that there. Minus log of 1 over n into summation ai, that's the left hand side of the Jensen's inequality. So this entity or this quantity will be less than or equal to. What do we have on the right hand side of the Jensen's inequality? The expected value of phi of x. So what do we write here? Less than or equal to. Expected value of minus log of x. Because we are taking the minus log of phi. Now this right hand side on this. Expected value of minus log of x. What is that equal to? That is equal to minus 1 over n summation log of ai. By the same rationale by which I gave you the reason why the expected value of x was equal to summation ai over n. You can see yourself through the same rationale that expected value of minus log of x. First of all, the minus sign will come out of expectations. So minus the expected value of log x. So instead of x, if it is log x, then apply all that rationale. And we get minus 1 over n summation log of ai. Where i goes from 1 to n. Okay, now we are in this form. Now let's move forward a little bit further. Do we not know that summation log ai would mean log of a1 plus log of a2 plus so on and so on plus log of an? So a basic rule here that log of a plus log of b is the same thing as log of a b. So what do we get? Minus 1 over n log of a1 into a2 into a3 so on so on into an. And then apply the second rule of logarithm. This minus 1 over n is attached. Take 1 over n to the power above which is done for logarithm. So what do we get? We have minus log of a1 into a2 into so on so on an. And this entire product raised to 1 over n. So Jensen's inequality. So once you do this much, how can we read it? If we remove the minus sign on both sides, then the less than equal sign will go back and it will become greater than or equal to. So then what do we get? We can read it as follows. The logarithm of 1 over n into summation a i is greater than or equal to the logarithm of a 1, a 2, so on up to n, just called pi of a i, pi being the product sign and i going from 1 to n, raised to 1 over n. That's what we get. Okay, now what we have, that has log on both sides, making logarithm is an increasing function. Therefore, if I remove it from both sides, the greater than sign that will remain as it is. It will flip and it won't be less than. And what we have then? We have summation a i over n greater than or equal to a 1 into a 2, so on, so on into a n raised to 1 over n. So, what we have now, is that I can't say that arithmetic mean is greater than or equal to the geometric mean. Of course, I can say that because as you can see, the expressions that you have, they are the expressions of the arithmetic and the geometric mean respectively. Okay. So, now we have to see a little bit more and show that our inequality means that the arithmetic mean is greater than or equal to the geometric mean and that is greater than or equal to the harmonic mean. So, that harmonic mean part, of course, how do we do that one? We will replace, let us do this, replace a i by 1 over a i. A i is a positive number, so 1 over a i is also a positive number. Then what do we get? We will have 1 over a 1 into 1 over a 2 into so on so on into 1 over a n and this entire product raised to 1 over n, this will be less than or equal to 1 over n summation 1 over a i. So, students, that is fine. Now, if I take this whole thing as a reciprocal, then what do I get? Now I get 1 over the product 1 over a 1, 1 over a 2, so on so on, 1 over n, raised to 1 over n. Abhi thori der pehle less than or equal tha abhi ho jayega, greater than or equal to, kyu ke aap reciprocal le rahe, ab jante hain ke boh sign ulot jayega, greater than or equal to 1 over 1 over n summation 1 over a i. To ab jo isme pehle chis hain usko aap solve karne, and very quickly you will find that first expression is equal to a 1 into a 2 so on so on into a n raised to 1 over n. In other words, it is the geometric mean again, and according to this inequality, this is greater than 1 over summation 1 over a i over n. So, Yejuti said, do you not see that this is the harmonic mean, because after all, what is the definition of the harmonic mean? Haan kaite hain hain ke it is the reciprocal of the arithmetic mean of the reciprocals of the values. To yeh hi hain aur expectation ke saath jab baat karenge, to kyaise define hoti hain? The harmonic mean is 1 over the expected value of 1 over x, so as you can see that expression is exactly that. So ab jab hain dono parts cho kar diye ke they are correct, then obviously we can combine them and we can say that am is greater than or equal to gm and gm is greater than or equal to hm. Kulta kar ke likhinge to sahir hain ke greater than ki jaga pe you will say less than. This is a demonstration, not a very rigorous proof, but a demonstration of this inequality and I have done for you with reference to the discrete uniform distribution.