Calculating Expectation and Variance

Question

I think I have an idea of how to solve these problems, but I keep getting stuck. Any help would be greatly appreciated! Also, sorry I know I messed up the formatting with the exponents in the problems.

1. If $E[X] = 2$ and $var(X) = 3$, what is $E[(X+1)^2]$? Do I literally plug this $2$ into $x+1$ to get $3$ then square it to get $9$?

2. If $E[X] = 2$ and $var(X) = 3$, what is $var(3X+2)$? Same thing, do I plug in 3 to get 11?

3. Let $X$ be a Bernoulli random variable with $p=.25$. Find $E[4X^5-1]$

Do I plug in $1/4$ to the formula? I'm unsure of all these as you can tell.

Answer 1

For the first one: start with $(X+1)^2$. Expand it as always, as $(X+1)^2=X^2+2X+1$ (note that this not an equality of numbers, but an equality of functions, since $X$ is a random variable and therefore a function). Take the means and use the linearity of the mean value: $E((X+1)^2)=E(X^2+2X+1)=E(X^2)+2E(X)+1$ (note that $E(c)=c$, if $c$ is constant). Now you need $E(X^2)$, which you get from $V(X)$: I assume you are aware of the classic formula $V(X)=E(X^2)-E(X)^2$, which gives you the desired expression $E(X^2)=V(X)+E(X)^2$. Plug in all the above to your equality, and you get $E((X+1)^2)=12.$ The rest are left as an exercise.

Answer 2

Expand the expression inside the expectation.

Use linearity of expectation.

$E[aX + b] = aE[X] + b$

and

$Var[X] = E[X^2] - E[X]^2$

You know the values of $E[X]$ and $Var[X]$

So its basically solved once you have done this.