问题及解答

The expected value of $X$ measures where the probability distribution is centered. We will use the variance of $X$ to measure the amount of variability in (the distribution of) $X$.

Let $X$ have pmf $p(x)$ and expected value $\mu$. Then the variance of $X$ ($X$ 的方差), denoted by $V(X)$ or $\sigma_X^2$, or just $\sigma^2$, is defined by
\[
V(X):=\sum_{D}(x-\mu)^2\cdot p(x)=E\bigl[(X-\mu)^2\bigr]
\]
Prove that
\[
V(X)=E(X^2)-(E(X))^2.
\]

i.e.,

\[V(X)=\sigma^2=\biggl[\sum_{D}x^2\cdot p(x)\biggr]-\mu^2\]
 

Proof.

\[
\begin{split}
\sigma^2&=\sum_{D}(x-\mu)^2\cdot p(x)\\
&=\sum_{D}\Bigl[x^2-2x\mu+\mu^2\Bigr]p(x)\\
&=\sum_{D}x^2\cdot p(x)-2\mu\cdot\sum_{D}x\cdot p(x)+\mu^2\sum_{D}p(x)\\
&=E(X^2)-2\mu\cdot\mu+\mu^2\\
&=E(X^2)-\mu^2
\end{split}
\]

See also Question 2137 .