# $E(X)=0, E(X^2)=2,E(X^4)=4$ MSQ

Suppose $X$ is a random variable such that $E(X)=0, E(X^2)=2,E(X^4)=4$. Then which are true?

A. $E(X^3)=0$

B. $P(X\geq 0)=\dfrac{1}{2}$

C. $X\thicksim N(0,2)$

D. $X$ is bounded with probability $1$

I tried to use $X=(X-\mu)+\mu$ and then expand $X^4$, but since $\mu=0$, nothing is making any sense to me. I know for sure that C is not an answer. But don't know how to solve others. Please help!

• What is the actual problem? Is it to select which of those must be true? – Aaron Montgomery Oct 30 '17 at 15:14
• @AaronMontgomery yes. – Stat_prob_001 Oct 30 '17 at 15:15
• Define Y=X^2 and use your trick on Y – madprob Oct 30 '17 at 15:21

Hint: The information about the second and fourth moments is important. To see why, let $Y = X^2$; notice that the variance of $Y$ is $0$, implying that $Y$ is a constant.

See what mileage you can get from that hint; more hints are given in the spoiler boxes below.

Further hint 1:

Since $X^2$ is constant, this means that $|X|$ can only be one possible value. In other words, $X$ is either $0$ with probability 1, or it is $\pm c$ for some constant $c$.

Further hint 2:

In reality, the only possibility for $c$ that fits with the above information is that $c = \pm \sqrt 2$, and each must occur with probability $1/2$ in order for $E[X]$ to be $0$. Now that you know the full distribution of your variable, you can easily check which of the other conditions must be true.

• wow! thanks a lot! – Stat_prob_001 Oct 30 '17 at 15:28

$$0= 4-2^2 = \operatorname{E}(X^4) - (\operatorname{E}(X^2))^2 = \operatorname{var}(X^2)$$ Therefore $X^2$ is constant, so $\Pr(X= \sqrt 2\text{ or } X=-\sqrt 2) = 1.$ Since $\operatorname{E}(X)=0,$ that means $\Pr(X=\sqrt 2) = \Pr(X=-\sqrt 2) = 1/2.$

So $X$ is bounded and $\operatorname{E}(X^3)=0$ and $\Pr(X\ge0) = 1/2.$