Variance: just how fat is your distribution? The average of a probability distribution tells you a lot about the real world. What's the average height of the lawn? What's the average amount of money do you expect to lose playing video poker in Las Vegas? What is the life-expectancy of a chipmunk? But aside from knowing this, you'd like to know how close to that average you expect to be. That is, when Bob looks at some random blade of grass, he might on average get $6 cm$, but he might find that in one instance it's $5.5cm$, another is $6.9 cm$. But does he expect to find $10 cm$? The complete distribution tells you that, but you want just want one number to tell how close, for example $6 \pm 1 cm$. This is a measure of how sharply peaked, or how wide is the distribution. There are lots of different definitions you could come up with, but we'll use the same one that everyone else does, the variance. $$Var(x) = \langle (x- \langle x \rangle)^2 \rangle$$ (1.39) This means we take the average difference between and outcome and the mean, square it, and then average. In terms of discrete probabilities this is: $$Var(x) = \sum_x (x-\langle x\rangle)^2) P(x)$$ (1.40) and for a continuous distribution: $$Var(x) = \int (x-\langle x\rangle)^2) P(x) dx$$ (1.41) This has the units of $x^2$ and is often referred to as $\sigma^2$. The standard deviation is then just $\sigma = \sqrt{Var(x)}$. Note for those that remember their classical mechanics, that the variance corresponds to the moment of inertia of an object of mass 1, about the center of mass. Let's do some examples. What's the variance for the distribution in fig. 1.5.2? That's the case where you flip a coin and you get 1 for heads and -1 for tails. In that case the mean was zero, so applying eqn. 1.40 you have: $$\sigma^2 = (-1-0)^2 {1\over 2}+ (1-0)^2 {1\over 2}= 1$$ (1.42) So the variance in this case is $1$. For this trial you expect to make $0 \pm 1$. Lastly, there's an interesting identity for the variance that sometimes simplifies calculations: $$Var(x) = \langle (x- \langle x \rangle)^2 \rangle =\langle x^2- 2 \langle x \rangle x + \langle x \rangle^2 \rangle$$ (1.43) $$=\langle x^2\rangle- 2 \langle x \rangle \langle x \rangle + \lan... ...le x \rangle^2 + \langle x \rangle^2 = \langle x^2\rangle - \langle x \rangle^2$$ (1.44) Now we'll move on to a more complex example.