Averages for independent variables Suppose that you have two variables that are independent. The case of height and weight we used above, is not a good example of this, because you might expect that a taller chipmunk would tend to weigh more. However we can think of something else, let's take two blades of grass at opposite sides of a lawn. Call the length of the first blade $x$ and the second one $y$. You'd expect their lengths to be quite independent, That means $P(x,y)dx dy = (P(x) dx)(P(y) dy)$. Now we see that $$\langle x y\rangle = \int\int x y P(x,y) dx dy = \int\int x y P(x)P(y) dx dy$$ (1.33) $$= \int x P(x) dx \int y P(y) dy = \langle x \rangle \langle y \rangle$$ This says that for independent variables, the average of the product is the product of the average. This property makes independent variables very important conceptually in the study of probability. Let's see how we can apply it to examples of problems where the variables aren't independent. Consider the example we went through above of the two dice in section 1.5.6. There we calculated $\langle xT \rangle$: the average of the result on the first flip times the total. Note that these two variables are not independent. If the first coin lands tails, you pay $1. That will effect the total amount of money that you make. The trick is to find a set of variables that are independent. In this case that's not too hard. Call the result of the second flip $y$. The results for the first and second flip are independent. So now we express the total in terms of these independent variables: $T=x+y$. So $$\langle xT \rangle = \langle x(x+y) \rangle = \langle x^2+xy \rangle = \langle x^2\rangle +\langle xy \rangle$$ (1.34) Now since $x$ can only take values $\pm 1$, $\langle x^2\rangle = \langle 1 \rangle = 1$. In the second term we can use the independence of $x$ and $y$ to say $\langle x y \rangle = \langle x \rangle \langle y \rangle = 0$. So $$\langle xT \rangle = 1$$ (1.35) as we obtained before.