I'm taking an upper level Economics class and one of my assignments asks the question in the title.

I approached it by using one property of expectation: expectation of the sum is equal to expectation of it's parts.

So, I did:

$$E[(2X + 3)^2] = E[4X^2 + 12X + 9] = 4E[X^2] + 12 E[X] + E[9]$$

I didn't get the right answer. I'm not even sure if what I did was the approach.

There's also second part to this question, which also asks for the variance VAR(2X + 5), given VAR(X) = 4. I can only think of one relevant property of the variance and it doesn't help with that question.

Anyone have an idea of how I should approach this?

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    $\begingroup$ You will need information about $E[X^2]$. Is there anything in the original problem statement that can help you figure out that value? $\endgroup$ – whuber Apr 14 '15 at 16:01
  • $\begingroup$ Ha. @whuber corrects the maths while latexing it. $\endgroup$ – conjugateprior Apr 14 '15 at 16:02
  • $\begingroup$ @conjugate I'm not trying to sneak in an answer via an edit: I merely corrected some typographical errors in an intermediate expression. The distinction is important, because a true mathematical error in a question usually gives important clues concerning why the question arose and how to answer it. Therefore, in editing a question, we should usually take care not to fix actual errors. $\endgroup$ – whuber Apr 14 '15 at 16:06
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    $\begingroup$ Is $X$ the same variable in part 1 and 2? $\endgroup$ – ekvall Apr 14 '15 at 18:02
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    $\begingroup$ "I originally thought that E[X^2} for this problem would also be 1" No, except in degenerate cases (when $X$ is a constant), $E[X^2] \gt (E[X])^2$. The difference $E[X^2] - (E[X])^2$ is the variance of $X$. But to repeat whuber's question with slightly different wording "Are you sure you are not told anything else about $X$"? For example, you might not be sharing with us that the information that $X$ is an exponential random variable on the grounds that it is irrelevant, but it really is a very relevant fact. For an exponential random variable with $E[X]=1$, $E[X^2]$ equals $2$. $\endgroup$ – Dilip Sarwate Apr 14 '15 at 18:55

Based on the comments, I assume the that the $X$ in part 1 and 2 is the same. Then we know $\mathbb E X =1,\mathrm{Var}(X)=4$.

You may now find the answer by using the relationship $\mathrm{Var}(X)=\mathbb EX^2- (\mathbb E X)^2$. (Hint: The correct answer is 41.)

I leave the below as an example of why the information in the first part is not sufficient.

Let $Y=1$, $X\sim N(1,1)$. Then $\mathbb E (2Y+3)^2=25$ while $(2X+3)=:Z\sim N(5, 4)$ so that $\mathbb E Z^2 = \mathrm{Var}(Z)+(\mathbb E Z)^2=29$.

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  • $\begingroup$ X looks to be the same variable in part 1/2, and before I read your comment I ended up getting 16 as an answer after realizing that Var (aX+B) = a^2 Var(X). I think that's the same as Var(X)= E[X]^2- (EX)^2? Would I also be wrong in getting 25 for the answer in the first part? The professor might have made a typo in the answer choices he gave us, which has happened a lot before. $\endgroup$ – user1943854 Apr 14 '15 at 18:53
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    $\begingroup$ Re: "Var (aX+B) = a^2 Var(X). I think that's the same as Var(X)= E[X]^2- (EX)^2?" No, they're not the same formula. 25 is not the correct answer for the first question. Have you tried using the formula I gave for the variance together with the given fact $Var(X)=4$ on the expression you have derived in your question? $\endgroup$ – ekvall Apr 14 '15 at 19:04
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    $\begingroup$ You're right, I misread the notes and the formulas aren't the same. Yes, I thought it over some more and from what you gave me, I realized that E[X]^2 = 5. Would that be wrong? If I plug that into part 1 and I get 41, which is D from the answer choices. $\endgroup$ – user1943854 Apr 14 '15 at 19:13
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    $\begingroup$ You are correct! $\endgroup$ – ekvall Apr 14 '15 at 19:15
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    $\begingroup$ Your formula is correct for part 2 $\endgroup$ – ekvall Apr 14 '15 at 19:23

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