$X$ is a random variable such that $E(X^2)=E(X)=1$. Find $E(X^{100}).$

My attempt: Assuming $X$ is discrete, we have $\sum x_i\mathbb P(X=x_i) = \sum x_i^2\mathbb P(X=x_i) = 1.$ We have something like $x_1p_1+\cdots+x_np_n=x_1^2p_1+\cdots+x_n^2p_n=1$. How do I find out $\sum x_i^{100}\mathbb P(X=x_i)$? I am completely blank. And, I am feeling uneasy as $X$ could well be continuous. How do I proceed then?

  • 2
    $\begingroup$ Hardy's answer is the most simple and elegant way to go, but you can also solve it in using for example Cauchy-Schwarz (here for the discrete case, but the argument is general). Apply CS on the vectors $\sqrt{p_i}$ and $\sqrt{p_i}x_i$: $\sum x_ip_i \leq \sum x_i^2p_i$ and since we have equality the vectors are co-linear and it follows that $x_i \equiv 1$ whenever $p_i > 0$. $\endgroup$
    – Winther
    Apr 9 '15 at 3:54

$$ 1 = \operatorname{E}(X^2) = (\operatorname{E}(X))^2 + \operatorname{var}(X) = 1 + \operatorname{var}(X). $$ Therefore $$ \operatorname{var}(X)=0. $$ So $\Pr(X=1)=1$.

Can you do the rest?

  • 1
    $\begingroup$ Yes. Then $E(X^{100})=1$. But I didn't understand how $var(X)=0 \implies P(X=1)=1$. $\endgroup$
    – saubhik
    Apr 9 '15 at 3:35
  • $\begingroup$ $\operatorname{var}(X)=E[(X-EX)^2]$, so if $\operatorname{var}(X)=0$ then $[X-E(X)]^2$ must be zero, so $X$ is constant and equal to its expectation. $\endgroup$
    – grand_chat
    Apr 9 '15 at 3:49
  • $\begingroup$ If the variance of a random variable is $0$, then it takes only one value. ${}\qquad{}$ $\endgroup$ Apr 9 '15 at 16:40

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