This problem has probably been solved somewhere but I could not find it. We have $n$ Bernoulli random trials $X_i$ with different occurrence probabilities, $\mathrm{Pr}[X_i=1]=p_i>p_{\min}>0$ for $i=1,2,\dots,n$ and some constant $p_{\min}$ what is the probability of having even number of ones when $n$ is very large? I guess that this probability is equal to $\frac{1}{2}$ but how can I prove this? I know how to prove this for the case when all $p_i$'s are equal but how can we prove it for different $p_i$'s? (In case this helps to solve the problem we can assume that $0<p_i \le 0.5$.)
2
-
$\begingroup$ If the probabilities $p_i$ are very small, then in more than half the cases the total number will be even $0$. You have to at least make some assumptions on $p_i$. $\endgroup$– Lev BorisovCommented Apr 1, 2016 at 1:22
-
$\begingroup$ Thanks for you point. I added a minimum constraint on $p_i$'s. $\endgroup$– mhsnkCommented Apr 1, 2016 at 2:23
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
The probability that an even number of trials succeed is exactly $$ \frac12\biggl( 1 + \prod_{i=1}^n (1-2p_i)\Biggr).$$ This is a standard elementary application of probability generating functions.
answered Apr 1, 2016 at 2:31
-
$\begingroup$ I notice that Will Sawin also posted this solution, but deleted it. Actually this question is too elementary for MO. $\endgroup$ Commented Apr 1, 2016 at 2:33