 Dear students, let me put forth to you the following question. Where would the mean of a symmetric distribution lie? I think many of you will have answered the question. You must have said that it will be in the middle. Yes, it is correct. If it is a symmetric distribution, the mean is actually in the middle of the distribution. But let us try to prove this algebraically. Let us do a proper formal proof so that it is established. Alright, question kya hai? Let a random variable x of the continuous type have a pdf f of x whose graph is symmetric with respect to x is equal to c. I am not saying that the origin of 0 is symmetric. In fact, I have said that the origin of 0 is symmetric that value is equal to c, that point is equal to c. So now this is the situation and what we have to show is that the expected value of x, the mean of x is equal to c. That is the point of symmetry. The mean of this variable also lies there. So how will we proceed? Let me give you a few hints to answer this question. We just said that we want to show that e of x is equal to c. So we can modify that and write it in a way that we may show that e of x minus c is equal to 0. I am giving you a hint that you should show that expected value of x minus c is equal to 0. That is the same thing as expected value of x is equal to c. And the next hint is that the expected value that you will get, obviously we are talking about continuous variable, so we will integrate the formula of our expected value. If we put that, then the integral will come. So you should break it. Break it into two parts. Write that integral as the sum of two integrals. The first one going from minus infinity to c and the second one going from c to plus infinity. After this, the next hint is that we will apply some transformation which will solve this easily. The first integral is let y be equal to c minus x. And the second integral is let z be equal to x minus x. x minus c. And apart from that, very importantly, use symmetry condition. And what is the symmetry condition? If a curve is symmetric around the value c, then the symmetry condition is given by f of c minus y is equal to f of c plus y. That is y. If the variable is being called y, then the algebraic version is f of c minus y will be equal to f of c plus y if it is symmetric around the value c. This is the first integral in which you have substituted y equal to c minus x. Then you apply symmetry condition. Now that I have given you all the hints, let us try to implement them one by one. First of all, let us write that integral the expected value of x minus c is equal to the integral from minus infinity to infinity of what, of which expression? The expression x minus c multiplied by f of x. Now, we are going to break it into two parts. So it will be equal to the integral from minus infinity to c of x minus c into f of x plus the integral from c to infinity of x minus c into f of x. Of course, it is necessary to write dx in front of it because the integral is not complete unless you write dx with it. But when you say it, all you don't have to say it. It is understood that if you are taking the integral, then you are writing dx at the end. All right. Now, as I gave you the next hint, that now these two integrals are here, so first of all, let us substitute it. Let y be equal to c minus x. So as soon as we substitute or transformation is a better word, as soon as we apply transformation, you know that we have to see the change of limits and we have to see the conversion of dx. Obviously, if y is equal to c minus x, then dy by dx will be equal to 0 minus 1, or not minus 1. And that means that dy is equal to minus dx. This is the first thing. The second thing is the change of limits, as x tends to minus infinity, y will tend to c minus infinity, i.e. c plus infinity, i.e. infinity. Let me say it again. If x is tending to minus infinity, y is tending to plus infinity. And the second limit of the integral was the lower limit, what was the upper limit? The first integral was c. So as x tends to c, y will tend to c minus c, i.e. y will tend to 0. So this situation has come, as far as the first integral is concerned. Now the transformation applied on the second integral, you can do that too. What is the transformation applied on the second integral? Let z or z be equal to x minus c. So then dz by dx will be equal to 1 minus 0, i.e. 1. So that means dz is equal to dx. And the change of limits, as x tends to c, z tends to 0. You can do it, it works out very simply. And as x tends to infinity, z also tends to infinity. Okay, now we are ready. Now we will put this in it. The transformation we have done, now apply the two integrals. So when we do that, then as you can see on the screen, first its converted version is minus infinity to 0. Note that I did not say 0 to infinity. Infinity to 0, and then what is inside it? Minus y into f of c minus y dy. And then the second plus, what is its situation? 0 to infinity z into f of c plus z dz. Now solve it for the first one. You know that if you want to interchange the limits of the integral, you want to make the lower limit upper and upper. So what happens? If the sign is first applied to it, then it will be over. If it was first plus, then it will be minus. And if it was first minus, then it will be plus. So let's apply that. By doing all that, what is the final answer? The expected value of x minus c is equal to minus 0 to infinity. That minus is over, but there was another minus inside it. So we are still having a minus. We reversed the limits. Infinity to 0. Now 0 to infinity is over, which is what we need and which is correct. So minus 0 to infinity y into f of c minus y dy plus 0 to infinity z into f of c plus z dz. Okay. Now last tip that I gave you, let us leave the second integral as it is. Just now let us focus on this first integral. The first integral is, we apply this symmetry condition inside it. Because we know from the start that our f function, that is a symmetric function, symmetric around the point c, then apply that algebraic condition in the first integral. And what was that condition? f of c minus y is equal to f of c plus y. So now put this first integral in it and it is in front of you on the screen and what does it come out to be equal to? Minus 0 to infinity y into f of c plus y. We wrote a plus instead of minus. Now after that, this expression for my dear students, you can also write this way that it is equal to minus 0 to infinity z into f of c plus z dz. That is why I called it z instead of y. Why? Because I am applying the concept of the dummy variable. Whenever your integral is a definite integral, it is not an indefinite integral, then the variable inside it, whether you call it y or z, or x or w, or whatever you call it, my dear students, it does not matter. Because the upper limit has to be applied, the lower limit has to be applied later. That variable has to disappear from the middle. So that is what is the concept of the dummy variable. Apply that here. Put z instead of y where it is written. After that, what is the final expression that we now have? The expected value of x minus c is equal to minus 0 to infinity z into f of c plus z dz plus 0 to infinity z into f of c plus z dz. If we are doing minus and plus one thing, then obviously the answer is equal to 0. So what we wanted to prove is that it has been proved. Expected value of x minus c is equal to 0, implying that expected value of x is equal to c. If our function c is symmetric, then the mean value of that variable, expected value of x is also equal to the same value. Expected value of x is equal to c.