Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X_1,X_2,...$ be a sequence of independent random variables with $P(X_n = 3^n) = P(X_n = -3^n) = \frac{1}{2}$. Let $S_n = X_1 + ... + X_n$.

  • Compute $E(X_n)$ for each $n$.

My guess for this one is that $E(X_n) = \frac{1}{2}\{\infty - \infty\}$ but that's only because $3^n$ would diverge to $\infty$ and $-3^n$ would diverge to $-\infty$. Is this assumption incorrect or could you tell what I may be missing?

share|cite|improve this question
up vote 0 down vote accepted

There is absolutely no problem in computing $E(X_n)$. By the usual method, this is $0$ for all $n$. No infinities involved.

Each $S_n$ also has mean $0$. (You did not ask about the $S_n$.)

share|cite|improve this answer
Can you expand on that a little more as to how you came to $E(X_n) = 0$? – dmcqu314 Nov 27 '12 at 3:42
$X_n=-a$ with probability $1/2$, and $X_n=a$ with probability $1/2$. We have $a=3^n$, but that's unimportant. So $E(X_n)=(1/2)(-a)+(1/2)(a)=0$. The bad behaviour will come later, when you worry about $\frac{S_n}{n}$. – André Nicolas Nov 27 '12 at 3:48
Ah ok that makes a lot more sense. Thanks! I appreciate it. – dmcqu314 Nov 27 '12 at 3:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.