Mean and Variance of a Random Variable

Transcript

0:00Introduction

Hello everyone! Let me start with a small game. You pay ten rupees to throw two dice. If the total is seven or eleven, you get thirty rupees. Thirty rupees for ten, that is three times your money! So, would you play this game again and again? Hmm, the prize looks big, but how often do you actually win? Today we learn the mean and the variance of a random variable, and they settle this. This idea comes up often in JEE problems. So, let's begin.

0:31The mean: a probability-weighted average

First, recall that a random variable, capital X, gives a number to each outcome. It takes values x i with probabilities p i; the small i at the bottom is a subscript that numbers the values. And the big Greek letter, capital sigma, means add them all up; together, the p i add up to one. Okay? Now let me ask you this. For the average value of X, should every value count equally, or should a more likely value count more? The more likely value should count more, yes, exactly! So multiply each value by its probability, and add. This is the mean. We write it with the Greek letter mu, or as E of X, the expectation of X, and it is the sum of x i times p i. Got it? In the same way, E of X squared is the sum of x i squared, times p i. The small two on top is a power, so square each value first, then multiply by its probability. Clear? Now, a quick question. To get E of X squared, can we simply find the mean, and then square it? No, we cannot, and this is a common trap. We square the values first, and only then take the mean. The crossed equals sign says that E of X squared is not mu squared in general. The gap between the two is what we study next. With me?

1:56Variance: how spread out X is

So here is the variance. Sigma squared, the small Greek letter sigma with a two on top, is the variance, and the three letters VAR are short for variance. The variance of X equals E of X squared, minus mu squared. Okay? In words, it is the mean of the squares, minus the square of the mean. Remember the order. Right? Turn it around, and sigma squared plus mu squared equals E of X squared. So when a question gives you the mean and the variance, you get E of X squared at once.

2:31Shift and scale: aX+b

Got it? Now, what if we multiply X by a constant, the letter a, and then add another constant, the letter b? The mean simply follows along. E of, a times X plus b, equals a times E of X, plus b. Okay? But what about the variance? If every value moves up by the same amount b, does the spread change? It does not change at all, that is correct! Everything moves together, so b drops out. But multiplying by a stretches the spread, so the variance of, a times X plus b, equals a squared times the variance of X, right? Let us try it at once. Suppose E of X is five, and the variance of X is four. Find E of, two X plus three, and the variance of, two X plus three. For the mean, two times five, plus three, gives thirteen. For the variance, the three drops out and the two gets squared, so two squared times four gives sixteen. The dot between the numbers means times. Clear?

3:38Independent variables and fair games

Now take two random variables, X and Y, that are independent. You know the variance of each one. So what is the variance of X minus Y? If you said the variance of X minus the variance of Y, that is the trap. For independent X and Y, the variance of X plus or minus Y equals the variance of X, plus the variance of Y, okay? The plus-minus sign covers both cases. Why plus, even for a minus? Minus Y is negative one times Y, and our scaling rule squares that negative one, which gives plus one. So the variances add; just remember, we use this rule for independent variables. Right? And now one last idea. We call a game fair if the expected gain is zero. Our dice game is coming soon. Got it?

4:30Example 1: mean and variance of the score on a fair die

Let's do the first example together. Find the mean and the variance of the score on a fair die. Pause the video and try it first. Done? Okay, let's check. The score X takes the values one to six; the three dots mean, and so on. The die is fair, so each value has probability one sixth. With me? For the mean, multiply each value by one sixth and add. One plus two, all the way to six, is twenty one, so mu is twenty one over six, which is seven over two. The die can never show seven over two, and that is fine; the mean need not be one of the values. Right? Now E of X squared. Square each value: one, four, nine, sixteen, twenty five, thirty six. They add up to ninety one, so E of X squared is ninety one over six. Okay? Now the variance. From ninety one over six, what exactly do we subtract? We subtract mu squared, yes, exactly! Seven over two, squared, is forty nine over four, not seven over two itself. Take twelve as the common denominator. Ninety one over six becomes one hundred and eighty two over twelve, and forty nine over four becomes one hundred and forty seven over twelve. So the variance is thirty five over twelve. Clear?

5:59Example 2: is the Rs 10 two-dice game fair?

Now, back to our game from the start. The player pays ten rupees, written R s on screen, to throw two dice, and receives thirty rupees if the total is seven or eleven. Is the game fair? Pause the video and try it first. Done? Okay, let's check. The total seven comes in six ways, from one and six up to six and one; each pair in round brackets is the first die, then the second. Now, in how many ways can the total be eleven? It comes in two ways, five and six, and six and five. If you said only one way, be careful; the two dice are different, so the order matters. Okay? Six plus two is eight favourable pairs, out of thirty six equally likely ordered pairs, so classical probability works. P of seven or eleven, the probability of seven or eleven, is eight over thirty six, which is two over nine. The two totals cannot happen together, so we add the ways. Right? Now, what is the expected gain of the player? Remember, the ten rupees is paid in every game. Thirty times two over nine gives twenty over three, the expected amount received. Subtract the ten rupees paid, and the expected gain is twenty over three minus ten, which is minus ten over three. Clear? The expected gain is not zero, so the game is not fair: an expected loss of ten over three rupees. Three times your money sounded great, but on average you win only two games out of nine!

7:35Quick revision: E(X), Var(X), aX + b, X - Y and fair games

So, quick revision. First, the mean is the sum of x i times p i, and the variance is E of X squared, minus mu squared. Second, E of, a times X plus b, is a times E of X, plus b, while the variance picks up only the factor a squared. Third, for independent X and Y, variances add, even for X minus Y; and a fair game has expected gain zero. Now go and try the practice problem on this page. All the best!