Probability Mass Function manipulation

Question

I have a random variable $A$ which has PMF, $$p_A(a)=\begin{cases}\frac{1}{2}&\text{if }x=1\\\frac{1}{2}&\text{if }x=-1\end{cases}.\\$$

I calculate by hand,$$\mathbb{E}[|A_1A_2|]=1.$$ I also calculate,$$\mathbb{E}[|A_1A_2+A_3A_4|]=1.$$
Then for $$\mathbb{E}[|A_1A_2+A_3A_4+A_5A_6|]=\frac{3}{2}.$$All these calculations are done by brute force (i.e.: for the third example, I generate the $2^6$ combinations for the joint PMF for the six random variables), so I am looking for a way to determine these results in a more compact manner, if possible.

Would appreciate any tips or comments.

Answer 1

Polynomial generating functions are a useful computational tool for dealing with questions of this type. The poly.g.f has the form $p(t)=\sum_k p_k t^k$ where $p_k$ is the probability of the occurrence of the value $k$. The expected value is $p'(1)=\sum_k k p_k$.

Assuming independence of the various $A$'s, you can check that the product $B=A_1A_2$ takes the values $\pm 1$ with equal probability. Thus the polynomial generating function associated to this discrete random variable is the symmetrical expression $B(t)= \frac{ 1}{2} ( t^{-1}+ t^1)$.

The sum of say 7 independent samples of $B$ is described by raising the polynomial $B(t)$ to the seventh power. It can be written as the symmetrical expression $B^7(t)= q(t)+ q(1/t)$ where $ q(t)=\frac{t^7}{128}+\frac{7 t^5}{128}+\frac{21 t^3}{128}+\frac{35 t}{128}$ which was found using Mathematica.

Of course in this case they are simply binomial coefficients.

Note that each power of $t$ has coefficient (probability) equal to that of $1/t$. The generating function for the absolute value of this sum is obtained by identifying those terms whose powers of $t$ have the same absolute value. By symmetry, in this problem that simply reduces to $p(t)=2 q(t)$.

The derivative of the generating function $p(t)=2q(t)$ at $t=1$ is the desired expected value.

Answer 2

Since the stated values appear to be coming under the independence assumption, I assume so.

For $X=A_1A_2+\dots+A_{2n-1}A_{2n}$, note that the following holds:

In all $2^{2n}$ possibilities, $X=n-2i\ (i\in\{0,\dots,n\})$ for ${n \choose i}\cdot 2^n$ times.

So, the required expectation is $$\frac{\sum_{i=0}^{n} |n-2i|{n \choose i}}{2^{n}}$$

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Edit:

Observe the following:

• $X=n$ if no term in the summation is $-1$. This is achieved in ${n\choose 0}2^n$ ways
• $X=n-1$ cannot be achieved
• $X=n-2$ if exactly one of the $n$ terms in summation is $-1$, and rest all are $1$. This is achieved in ${n\choose 1}2^n$ ways
• $\dots$
• $X=-n$ if all terms in summation are $-1$. This is achieved in ${n\choose n}2^n$ ways