问题及解答

Using the definition of variance, prove that
\[
V(aX+b)=a^2\cdot\sigma_X^2
\]

[Hint] With $h(X)=aX+b$, $E[h(X)]=a\mu+b$ where $\mu=E(X)$.

First, it is easy to prove that $E[aX+b]=a\mu+b$.

\[
\begin{split}
E[h(X)]&=E[aX+b]=\sum_{x\in D}(ax+b)\cdot p(x)\\
&=a\sum_{x\in D}xp(x)+b\sum_{x\in D}p(x)\\
&=a\mu+b
\end{split}
\]

If let $Y=aX+b$, then $\mu_Y=a\mu+b$. Let $D_2$ be the set of variable $Y$. That is

\[D_2=\{aX+b\mid X\in D\}\]

Then, we have

\[
\begin{split}
V(aX+b)&=V(Y)\\
&=\sum_{y\in D_2}(Y-\mu_Y)^2\cdot p(y)\\
&=\sum_{x\in D}\bigl(ax+b-(a\mu+b)\bigr)^2\cdot p(x)\\
&=\sum_{x\in D}a^2(x-\mu)^2\cdot p(x)\\
&=a^2\sum_{x\in D}(x-\mu)^2\cdot p(x)\\
&=a^2\sigma_X^2
\end{split}
\]

Here, we note that $p(y)=p(ax+b)=p(x)$.