# Finding the distribution of a sum of random variables

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?

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

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