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

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    $\begingroup$ 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. $\endgroup$ Commented Oct 2, 2017 at 1:23
  • $\begingroup$ Oh I have been using it to show that it's my next step. What would the better notation be? $\endgroup$
    – Remy
    Commented Oct 2, 2017 at 1:24
  • $\begingroup$ Just remove the arrows and use newlines. $\endgroup$
    – Batman
    Commented Oct 2, 2017 at 1:25
  • $\begingroup$ @JohnH Just write the expression. If you must, use bullet points. $\endgroup$ Commented Oct 2, 2017 at 1:25
  • $\begingroup$ Alright, thanks. I'll do that from now on. $\endgroup$
    – Remy
    Commented Oct 2, 2017 at 1:26

2 Answers 2

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

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  • $\begingroup$ Thanks, Batman. $\endgroup$
    – Remy
    Commented Oct 2, 2017 at 1:22
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Yep correct now you can evaluate also the variance and expected value of Y...

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    $\begingroup$ I'll try that for practice $\endgroup$
    – Remy
    Commented Oct 2, 2017 at 1:30

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