My Favorite One-Liners: Part 114

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

I’ll use today’s one-liner whena step that’s usually necessary in a calculation isn’t needed for a particular example. For example, consider the following problem from probability:

Let X be uniformly distributed on \{-1,0,1\}. Find \hbox{Cov}(X,X^2).

The first step is to write \hbox{Cov}(X,X^2) = E(X \cdot X^2) - E(X) E(X^2) = E(X^3) - E(X) E(X^2). Then we start computing the expectations. To begin,

E(X) = (-1) \cdot \displaystyle \frac{1}{3} + 0 \cdot \displaystyle \frac{1}{3} + 1 \cdot \displaystyle \frac{1}{3} = 0.

Ordinarily, the next step would be computing E(X^2). However, this computation is unnecessary since E(X^2) will be multiplied by E(X), which we just showed was equal to 0. While I might calculate E(X^2) if I thought my class needed the extra practice with computing expectations, the answer will not ultimately affect the final answer. Hence my one-liner:

To paraphrase the great philosopher The Rock, it doesn’t matter what E(X^2) is.

P.S. This example illustrates that the covariance of two dependent random variables (X and X^2) can be zero. If two random variables are independent, then the covariance must be zero. But the reverse implication is false.

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