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

$$\newcommand{\E}{\operatorname{E}}\newcommand{\v}{\operatorname{var}}$$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.

• What do you mean by all those arrows? The notation $A\Rightarrow B$ should mean $B$ is a logical consequence of $A.$ Strangely, for decades students have been persisting in using arrows in strange ways like this although that is never taught. Commented Oct 2, 2017 at 1:23
• Oh I have been using it to show that it's my next step. What would the better notation be?
– Remy
Commented Oct 2, 2017 at 1:24
• Just remove the arrows and use newlines. Commented Oct 2, 2017 at 1:25
• @JohnH Just write the expression. If you must, use bullet points. Commented Oct 2, 2017 at 1:25
• Alright, thanks. I'll do that from now on.
– Remy
Commented Oct 2, 2017 at 1:26

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