Please help. I have to find out conditional distribution for X given X-Y=n where X and Y are independent random variables representing the no. of failure preceeding the first success.
Asked
Active
Viewed 726 times
1 Answers
1
We are dealing wit the probability distribution of the number $X$ of failures before the first success, supported on the set $\{0,1, 2, 3, ...\}$. The pmf of $X$ is then $$P(X=n)=(1-p)^np.$$ where $0<p\le1$ is the probability of success of the Bernoulli trials involved.
We have two independent Bernoulli distributed random variable: $X$ and $Y$ and we want to compute $P(X=n\mid X-Y=k).$ By definition $$P(X=n\mid X-Y=k)=\frac{P(X=n\cap X-Y=k)}{P(X-Y=k)}.$$
Now,
$$P(X=n\cap X-Y=k)=\sum_{i=0}^{\infty}P(X=n\,\cap\ X-Y=k\ \mid Y=i)(1-p)^ip=$$ $$=p\sum_{i=0}^{\infty}P(X=n\,\cap\ X=k+i\ \mid Y=i)(1-p)^i=$$ $$=\begin{cases} P(X=n)(1-p)^{n-k}p&\text{ if } k\le n\\ 0&\text{ if } k>n \end{cases}=$$ $$=\begin{cases} (1-p)^np(1-p)^{n-k}p&\text{ if } k\le n\\ 0&\text{ if } k>n \end{cases}=$$ $$=\begin{cases} (1-p)^{2n-k}p^2&\text{ if } k\le n\\ 0&\text{ if } k>n. \end{cases}$$
And, the same way $$P(X-Y=k)=\sum_{i=0}^{\infty}P(X-Y=k\ \mid \ Y=i)(1-p)^ip=$$ $$=\sum_{i=0}^{\infty}P(X=k+i)(1-p)^ip=$$ $$=\sum_{i=0}^{\infty}(1-p)^{k+2i}p^2=p^2(1-p)^k\sum_{i=0}^{\infty}\left[(1-p)^2\right]^i=$$ $$=\frac{p^2(1-p)^k}{1-(1-p)^2}.$$
So,
$$P(X=n\mid X-Y=k)=\begin{cases} (1-(1-p)^2)(1-p)^{2(n-k)}&\text{ if } k\le n\\ 0&\text{ if } k>n. \end{cases}$$
zoli
- 20,452