1
$\begingroup$

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?

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

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 $S_n$.)

$\endgroup$
3
  • $\begingroup$ Can you expand on that a little more as to how you came to $E(X_n) = 0$? $\endgroup$
    – dmcqu314
    Nov 27, 2012 at 3:42
  • $\begingroup$ $X_n=-a$ 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 $\frac{S_n}{n}$. $\endgroup$
    – André Nicolas
    Nov 27, 2012 at 3:48
  • $\begingroup$ Ah ok that makes a lot more sense. Thanks! I appreciate it. $\endgroup$
    – dmcqu314
    Nov 27, 2012 at 3:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .