2

May I please get help with this question?

What is the answer and how do I get to it?

[Within the context of discrete random variables]. Consider a probability distribution where $E(X^2) = 2E(X)$. In this case, the standard deviation is:

A. $\sqrt{3} \times E(X)$

B. $E(X)$

C. $\sqrt{E(X)}$

D. $\sqrt{3E(X)}$

E. None of the above

  • Can you arrive at $E[(X-EX)^2]=...=E[X^2]-(E[X])^2?$ Where are you stuck, exactly? – Avitus Aug 17 '14 at 10:45
  • Yes, I could arrive at that point. But from there how do you get to one of the answers? –  Aug 17 '14 at 10:48
  • As much as I can work it through:

    $Var(X) = E[X^2]−(E[X])^2$

    SUB $E(X^2) = 2E(X)$

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

    FACTORISE

    $Var(x) = E(X)[2 - E(X)]$

    From here, then what? If I take the square route of the variance in order to get the standard deviation, it looks nothing like the answers. I have a strong feeling that the answer is not E. and that there is a way to get one of the other answers.

     –  Aug 17 '14 at 11:04
  • D is wrong answer . isn't it ? – hanugm Aug 17 '14 at 11:12
  • Unfortunately I cannot quantify a strong feeling; I wrote a quick answer, however. – Avitus Aug 17 '14 at 11:14
  • Maybe there are some more conditions on the distribution of $X$. Possibility: it only takes positive integers as value. – drhab Aug 17 '14 at 11:39
  • What I have posted as the question is all the information I can gather. There are no other restrictions other than $E(X^2)=2E(X)$. Would it be possible to work backwards from the answers to get to the question? –  Aug 17 '14 at 11:42
  • What stumps me is the fact that there is addition and subtraction in the variance but none of the answers have addition and subtraction. I'm also stumped why they would put a '3' in there. –  Aug 17 '14 at 11:43

2 Answers2

1

If $X(\omega)=0$ for each $\omega\in\Omega$, i.e. if you are dealing with a constant random variable then $\mathbb E(X)=\mathbb E(X^2)=0$. So indeed you are dealing here with a probability distribution satisfying $\mathbb E(X^2)=2\mathbb E(X)$ and the answers A,B,C,D are all correct, since also the standard deviation is $0$ here.

However, probably there will be other probability distribution that also satisfy $\mathbb E(X^2)=2\mathbb E(X)$ and where (some) of the answers are not correct.

You are dealing here with a 'strange question'. I would never ask it.

drhab
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  • I gathered some extra information about the question. All of the questions are within the context of discrete random variables not constant random variables. It was not on the question itself, but our current topic is discrete random variables. Assuming it is a discrete random variable, what would the answer be then? –  Aug 17 '14 at 11:29
  • A constant random variable is a special discrete random variable. In my answer you could say that $X$ takes the values $0$, $1$, $2$ et cetera, and this with $P[X=0]=1$, $P[X=1]=0=P[X=2]=\cdots$. – drhab Aug 17 '14 at 11:33
  • The question requires that only one answer be selected, however. –  Aug 17 '14 at 11:38
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With the information provided in the OP, one arrives at $Var(X)=E[(X−EX)^2]=E[X^2]−(E[X])^2=2E[X]−(E[X])^2=E[X](2-E[X]).$ Then $\sqrt{Var(X)}=\sqrt{E[X](2-E[X])}$, which is well defined if $0\leq E[X]\leq 2$. The answer is then E.

Edit(after @drhab's remark)

If $X = x$ is a constant r.v. with $E[X^2]=2E[X]$, then $x=E[X]=0$ or $x=E[X]=2$ and $\sqrt{Var(X)}=0$. In the first case, A. B. C. and D. are correct. In the second case, only E. is correct.

Edit 2: examples

I would like to show 2 examples which give different answers.

  1. Let $X$ be the discrete r.v. with values $\{2,0\}$ s.t. $P(X=0)=P(X=2)=\frac{1}{2}$. Then $E[X]=0\cdot\frac{1}{2}+2\cdot\frac{1}{2}=1$ and $E[X^2]=0^2\cdot\frac{1}{2}+2^2\cdot\frac{1}{2}=2=2E[X]$. It follows that $Var(X)=1=E[X]$ and both answers B. and C. are correct.
  2. Let $X$ be the discrete r.v. with values $\{2,0\}$ s.t. $P(X=0)=\frac{1}{3}$, $P(X=2)=\frac{2}{3}$. Then $E[X]=\frac{4}{3}$ and $E[X^2]=\frac{8}{3}=2E[X]$. It follows that $Var(X)=\frac{\sqrt{8}}{3}=\frac{\sqrt{2}}{2}E[X]$. In this case E. is correct.
Avitus
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  • And what if $X=0$ being a constant rv? then all answers (except E) are correct. – drhab Aug 17 '14 at 11:16
  • I did not consider the case of constant r.v.'s, you are right. I edit my answer. Thanks – Avitus Aug 17 '14 at 11:21
  • What exactly are constant r.v.'s? –  Aug 17 '14 at 11:24
  • 1
    A constant random variable $X$ is such that $X(\omega)=x$, for all $\omega\in\Omega$. Its expectation value $E[X]$ is equal to the (only) value $x$, i.e. $E[X]=x$. For more details, please refer to http://en.wikipedia.org/wiki/Degenerate_distribution – Avitus Aug 17 '14 at 11:28
  • I have added 2 examples of discrete random variables which give 2 different answers. In summary, I would consider this question in your text book as "ill-posed / unclear". – Avitus Aug 17 '14 at 13:52