If $E(W) = \mu$ and $Var(W) = \sigma^2$

Question

This is one of two HW problems that I'm positing that I have no clue how to go about.

If $E(W) = \mu$ and $\operatorname{Var}(W) = \sigma^2$ show that $E\left(\frac{W - \mu}{\sigma}\right) = 0$ and $\operatorname{Var}\left(\frac{W - \mu}{\sigma}\right) = 1$

For the first section, based on something out of the book, I'm thinking I should be able to do

$$E\left(\frac{W}{\sigma} - \frac{\mu}{\sigma}\right) = E\left(\frac{W}{\sigma}\right) - \frac{\mu}{\sigma} = \frac{\mu}{\sigma} - \frac{\mu}{\sigma} = 0$$

I'm not sure if that's correct, and I have no clue on the following part.

Answer 1

The first part is true as you have shown due to the linearity of expectation. For the second part use the definition of variance:

$$\operatorname{var}(W) = E\left[(W-\mu)^2\right] = E[W^2] - E[W]^2$$

Then $$\operatorname{var}\left(\frac{W-\mu}{\sigma}\right) = E\left[\left(\frac{W-\mu}{\sigma}\right)^2\right] - E\left[\frac{W-\mu}{\sigma}\right]^2$$

From the first part you know that $$E\left[\frac{W-\mu}{\sigma}\right] = 0$$

So $$\operatorname{var}\left(\frac{W-\mu}{\sigma}\right) = E\left[\left(\frac{W-\mu}{\sigma}\right)^2\right] = \frac{1}{\sigma^2}E\left[(W-\mu)^2\right]$$

But $$\operatorname{var}(W) = E\left[(W-\mu)^2\right] = \sigma^2$$

by the definition of $W$ so $$\operatorname{var}\left(\frac{W-\mu}{\sigma}\right) = 1$$