# 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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