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Let $X_1,X_2, \dots$ be iid random variables that take value 1 with probability 1/2 and value -1 with probability 1/2. Let $$S_n = X_1 + ... + X_n,$$ and let X be a Binomial(n, 1/2). Show that 2X − n has the same distribution as $S_n$ .

I am having a lot of trouble applying the central limit theorem to do this. The idea of $S_n$ being a distribution itself is confusing to me.

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    $\begingroup$ What makes you think the Central Limit Theorem is required? $\endgroup$ – mark999 Apr 26 '17 at 1:05
  • $\begingroup$ That is the chapter we are in right now in my statistics class $\endgroup$ – Ravmcgav Apr 26 '17 at 1:17
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Given that $X_{i}=\pm 1$, with probabilities $P(X_{i}=-1)=P(X_{i}=1)=1/2$ and a random variable $S_{n}$ is defined by $S_{}=X_{1}+X_{2}+\cdots +X_{n}$. We are required to show that the random variables $2X-n$ and $S_{n}$ have the same distribution.

Define a random variable $Y_{i}=\dfrac{X_{i}+1}{2}$ which takes values either $0$ or $1$. Now, $$P(Y_{i}=0)=P\left( \dfrac{X_{i}+1}{2}=0\right)=P(X_{i}=-1)=1/2$$ $$P(Y_{i}=1)=P\left( \dfrac{X_{i}+1}{2}=1\right)=P(X_{i}=1)=1/2$$ As the random variables $Y_{i}, i=1,2,\cdots,n$ are iid $Bern(1/2)$, $\sum_{i=1}^{n}Y_{i}\sim Bin(n,1/2)$.

\begin{eqnarray*} \sum_{i=1}^{n}Y_{i}&=& \sum_{i=1}^{n}\left(\dfrac{X_{i}+1}{2}\right)\\ 2\sum_{i=1}^{n}Y_{i}&=& \sum_{i=1}^{n}\left(X_{i}+1\right)=\sum_{i=1}^{n}X_{n}+n\\ 2\sum_{i=1}^{n}Y_{i}&=& S_{n}+n\\ 2 X - n &=&S_{n} \end{eqnarray*}

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Hint: $\frac{1}{2}(X_i + 1)$ is a $\operatorname{Bernoulli}(1/2)$ random variable.

Thus $\frac{1}{2}(X_1+1) + \cdots + \frac{1}{2}(X_n+1) = \frac{1}{2}(S_n+n)$ is a $\operatorname{Binomial}(n,1/2)$ random variable.


More detail:

  • Check that $\frac{1}{2}(X_i+1)$ is a $\text{Bernoulli}(1/2)$ random variable.
  • $\frac{1}{2}(X_1+1) + \cdots + \frac{1}{2}(X_n+1)$ is therefore a $\text{Binomial}(n,1/2)$ random variable (recall that a $\text{Binomial}(n,p)$ random variable is the sum of independent $\text{Bernoulli}(p)$ random variables)
  • Let $X$ denote the above sum; we showed $X \sim \text{Binomial}(n,1/2)$. Then using the definition of $S_n$, we can show $X = \frac{1}{2} (S_n + n)$.
  • So, $S_n = 2X - n$ where $X \sim \text{Binomial}(n,1/2)$..

The central limit theorem is not needed here.

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  • $\begingroup$ Not sure if I totally follow. How does this imply that 2X - n has the same distribution as S_n? $\endgroup$ – Ravmcgav Apr 26 '17 at 1:28
  • $\begingroup$ @Ravmcgav See my edit $\endgroup$ – angryavian Apr 26 '17 at 2:07
  • $\begingroup$ I just still don't really see how the central limit theorem comes into play. Really sorry to keep this chain going I just want to make sure I understand! I see that $\frac{1}{2}(S_n + n)$ Could be manipulated into the expression I want if we set it equal to a new random variable, but I'm not sure how that would help us... $\endgroup$ – Ravmcgav Apr 26 '17 at 3:24

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