0
$\begingroup$

Suppose $X_1$, $X_2$, ... $X_n$, are independent and identically distributed random variables. Each $X_i$ takes only two values +1 or -1 with equal probabilities. Let $S_n$ = $X_1 + X_2 + \cdots + X_n$.

Find the distribution of $S_n$.

I'm confused on how to approach this. Is this a normal distribution?

$\endgroup$
  • 1
    $\begingroup$ Do you mean $S_n = X_1 + X_2 + \ldots + X_n$? Hint: it is related to a binomial distribution. $\endgroup$ – Robert Israel Mar 13 '14 at 6:56
1
$\begingroup$

Write $Y_i = (X_i+1)/2$. Then the $Y_i$ are still i.i.d. and furthermore, $Y_i$ is $0$ with probability $1/2$ and $1$ with probability $1/2$. We can $1$ consider to be ''success'' and $0$ consider to be ''failure'', then we see that $Y_1 + \cdots + Y_n$ (the number of successes) follows a binomial distribution $N(n,\frac12)$. Then $S_n = 2(Y_1 + \cdots + Y_n) - n$ is just a transformation of this distribution. For large $n$ this will approach a normal distribution.

$\endgroup$
  • $\begingroup$ Corrected. Thanks! $\endgroup$ – user133281 Mar 13 '14 at 7:22
  • 1
    $\begingroup$ $S_n$ wouldn't approach the normal distribution, because $\operatorname{Var}S_n=n\to\infty$. $\frac1{\sqrt n}S_n$ would approach the normal distribution. $\endgroup$ – Cm7F7Bb Mar 13 '14 at 7:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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