Q.7.1 - A First Course in Probability

Question

Show that $E\left[{(X - a)}^2\right]$ is minimized at $a = E\left[X\right]$.

Step-by-Step Solution

Answer

Differentiate $E\left[{(X - a)}^2\right]$ respective to $a$.

Step 1: Given Information

$E\left[{(X - a)}^2\right]$ is minimizing at $a=E\left[X\right]$.

Step 2: Explanation

We have that

$E\left((X-a{)}^2\right)=E\left(X^2+a^2-2Xa\right)$

$=E\left(X^2\right)+a^2-2aE\left(X\right)$

Using the differentiation respective to $a$ and equalizing it to zero, we have that

$\frac{d}{da}E\left((X-a{)}^2\right)=2a-2E\left(X\right)$

$=0$

Step 3: Explanation

which yields that $a=E\left(X\right)$.

So, $E\left((X-a{)}^2\right)$ is minimized when $a=E\left(X\right)$ which had to be proved.

Step 4: Final Answer

$E\left((X-a{)}^2\right)$ is minimized when$a=E\left(X\right)$ which had to be proved.