# On probabilities of expectations

Question

Let $X_1 , X_2, \cdots , X_n$ be independent and identically distributed with $P(X_i =1)=P(X_i=-1)=p$ and $P(X_i=0)=1-2p$ for all $i=1,2,\cdots,n$. Define $$a_n=P(\prod_{i=1}^{n}X_i=1), b_n=P(\prod_{i=1}^{n}X_i=-1), c_n=P(\prod_{i=1}^{n}X_i=0)$$

Find $\lim_{n\to\infty} a_n,\lim_{n\to\infty} b_n,\lim_{n\to\infty} c_n$.

My Approach

Since $(\prod_{i=1}^n X_i=1)=1$ hence $a_n=P(\prod_{i=1}^{n}X_i=1)=P(X=1)=p$ and so $\lim_{n\to\infty}a_n=p$.

Similarly for others. Is this approach correct?

• It is more likely you will get an answer if you show you have done some work or please briefly explain what concept (s) you are having difficulty understanding. Commented Jun 13, 2018 at 6:07
• @TonyHellmuth Updated! Sorry! Commented Jun 13, 2018 at 6:07
• Also do you have any information to believe that the limits have a non-alternating value? Commented Jun 13, 2018 at 6:09
• @TonyHellmuth It's a part of a multiple choice question. Should I post the other options as well? Commented Jun 13, 2018 at 6:10
• Great observation - that is the starting point to attack this question. Consider the case where all of them are non-zero, which help you to compute $c_n$. Next you may need to consider some odd/even issue for $a_n, b_n$.
– BGM
Commented Jun 13, 2018 at 7:41

Since $1-2p$ is a probability, we can assume that $p \leq 1/2$. If $p<1/2$, then $c_n=1-\text{Pr[No$X_i$is zero]=}1-(2p)^n$. Thus the third limit is 1. The sum of the three quantities is 1, so the first two limits are zero.