Finding the mean and variance of a discrete variable.

Question

This is the question

Suppose that $X$ is a discrete random variable with the mean $4$ and variance $9$. If $Y= -2X+1$. What is the mean and variance of $Y$?

I was able to easily solve the mean $E[Y]$ like so:

$E[Y] = E[-2X+1] = -2E[X] + 1 = -2(4) + 1 = -8 + 1 = -7$

but I got a little confused about solving the variance. This was my attempt

$VAR[Y] = E[Y^2] - (E[Y])^2 = E[(-2X + 1)^2] - (-7)^2 = ?$

My issue is handling the $E[(-2X + 1)^2]$ portion well.

Answer 1

Note that $E[-2X+1]^2=4E[X^2]-2E[X]+1$, and remember that $\operatorname{Var}[X]=E[X^2]-(E[X])^2$.

Or you can use directly the fact that $\operatorname{Var}[aX+b]=a^2\operatorname{Var}[X]$.

Answer 2

The enclosed link gives a wrong result

$$\mathbb{V}[-2X+1]=(-2)^2\mathbb{V}[X]=4\times 9=36$$

in the enclosed link the calculation of $\mathbb{E}[Y^2]$ is wrong. It's 85 and not 101.

I suggest you not to follow that brainstorming, boring and useless for your problem