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This is exercise 3.2.5 from Probability and Random Processes by Grimmett and Stirzaker:

Let $X_r$, $1 \leq r \leq n$, be independent random variables which are discrete and symmetric about $0$, that is, $X_r$ and $-X_r$ have the same distributions. Show that for all $x$, $P(S_n \geq x) = P(S_n \leq -x)$ where $S_n = \sum_{r = 1}^n X_r$.

Is the following proof correct?

$\textbf{Proof:}$

Let $x \in \mathbb{R}$ be arbitrary. Then,

\begin{eqnarray} P(S_n \geq x) & = & P\left(\sum_{r = 1}^n X_r \geq x\right) \\ & = & P\left(\underset{x_1 + \cdots + x_n \geq x}{\bigcup_{x_1, \ldots, x_n}} \{X_1 = x_1\} \cap \cdots \cap \{X_n = x_n\} \right) \\ & = & \underset{x_1 + \cdots + x_n \geq x}{\sum_{x_1, \ldots, x_n}} P\left(\{X_1 = x_1\} \cap \cdots \cap \{X_n = x_n\}\right) \\ & = & \underset{x_1 + \cdots + x_n \geq x}{\sum_{x_1, \ldots, x_n}} P\left(\{X_1 = x_1\}) \cdots P(\{X_n = x_n\}\right) \textrm{(from independence of variables)}\\ & = & \underset{x_1 + \cdots + x_n \leq -x}{\sum_{x_1, \ldots, x_n}} P\left(\{X_1 = -x_1\}) \cdots P(\{X_n = -x_n\}\right) \textrm{(from symmetry of variables)} \\ & = & \underset{x_1 + \cdots + x_n \leq -x}{\sum_{x_1, \ldots, x_n}} P\left(\{X_1 = -x_1\} \cap \cdots \cap \{X_n = -x_n\}\right) \textrm{(from independence of variables)}\\ & = & P\left(\underset{x_1 + \cdots + x_n \leq -x}{\bigcup_{x_1, \ldots, x_n}} \{X_1 = -x_1\} \cap \cdots \cap \{X_n = -x_n\} \right) \\ & = & P\left(\sum_{r = 1}^n X_r \leq -x\right) \\ & = & P(S_n \leq -x) \end{eqnarray}

as required. $\square$

Is there a quicker/better way of proving the result?

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For $x$ in the support of $X_1+\cdots+X_n$, let $S$ be the set of all $n$-tuples $(x_1,\ldots,x_n)$ in $\operatorname{support}(X_1) \times\cdots\times \operatorname{support}(X_n)$ for which $x_1+\cdots+x_n=x$, and let $T$ be the set of all $n$-tuples $(x_1,\ldots,x_n)$ in the former support for which $x_1+\cdots+x_n=-x$. Then

\begin{align} & \Pr(X_1+\cdots + X_n = x) \\[10pt] = {} & \sum_{(x_1,\ldots,x_n)\in S} \Pr(X_1 = x_1\ \&\ \cdots \ \&\ X_n=x_n) \\[10pt] = {} & \sum_{(x_1,\ldots,x_n)\in S} \Pr(X_1 = -x_1\ \&\ \cdots \ \&\ X_n=-x_n) \\[10pt] = & \sum_{(x_1,\ldots,x_n)\in T} \Pr(X_1 = x_1\ \&\ \cdots \ \&\ X_n=x_n) \tag 1 \\[10pt] = & \Pr(X_1+\cdots+X_n= -x). \end{align}

To justify the equality in $(1)$, you have to write an argument relying on the symmetry of each of the separate distributions. And the reason you can think of them separately is independence.

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  • $\begingroup$ What does support mean in this context? $\endgroup$ – Raj Oct 3 '15 at 1:12
  • $\begingroup$ For discrete distributions, the support is simply the set of all possible value. For absolutely continuous distributions, it is the closure of the set of points at which the density is positive. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 3 '15 at 3:23
  • $\begingroup$ +1. The first (huge) simplification is to consider the probabilities $P(X_1+\cdots+X_n=x)$ instead of the CDF. $\endgroup$ – Did Oct 3 '15 at 8:14
  • $\begingroup$ Thanks for the clarification and the shorter proof. $\endgroup$ – Raj Oct 3 '15 at 13:37
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Everything seems correct.

But yes, there is both quicker and better way, based on the fact (which is easy to prove) that a random variable is symmetric iff its characteristic function is real-valued. Given this, the symmetry of $S_n$ immediately follows from $$ \varphi_{S_n} (t) = \prod_{k=1}^n \varphi_{X_k}(t). $$

It is better because it allows to prove the required property for any random variables, not only for discrete.

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  • $\begingroup$ OK, thanks for the ideas on how to do a better proof. I will try to remember to think about this question again when I start to learn about characteristic functions. $\endgroup$ – Raj Oct 3 '15 at 1:21

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