Probability Distribution, where $E(X^2) = 2E(X)$

Question

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

Answer 1

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.

Answer 2

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.