Finding $\operatorname{E}(X)$ given that $\operatorname{E}(X) = \operatorname{var}(X)$, $\operatorname{E}(Y) = \operatorname{var}(Y)$, and $Y=3X+3$

Question

$\newcommand{\E}{\operatorname{E}}\newcommand{\v}{\operatorname{var}}$Suppose a random variable $X$ is such that its expected value is equal to its variance. If $Y= 3X+ 3$ is also a random variable having its expected value equal to its variance, what must the value of $\E(X)$ be?

Attempted Solution:

I'm making use of the following formulas:

$\E(aX+b) = a\E(X) + b$

$\v(aX+b) = a^2\v(X)$

We're given $\E(X) = \v(X)$ and $\E(Y) = \v(Y)$.

$\Rightarrow$ $\E(Y) = \v(Y)$

$\Rightarrow$ $\E(3X+3) = \v(3X+3)$

$\Rightarrow$ $3\E(X)+3 = 3^2\v(X) = 3^2\E(X)$

$\Rightarrow$ $3 = 6\E(X)$

$\Rightarrow$ $\E(X) = {1 \over 2}$

I think I did this correctly but I just wanted to make sure.

Answer 1

Yes, this is correct. $\qquad\qquad$

Answer 2

Yep correct now you can evaluate also the variance and expected value of Y...