Suppose that X and Y are independent random variables with the same geometric distribution,$\displaystyle P(X = k) = P(Y = k) = pq^(k-1) $where $\displaystyle q = 1 - p$
Find
$\displaystyle P( X = k | X + Y = n + 1) n = 2,3,....$
Suppose that X and Y are independent random variables with the same geometric distribution,$\displaystyle P(X = k) = P(Y = k) = pq^(k-1) $where $\displaystyle q = 1 - p$
Find
$\displaystyle P( X = k | X + Y = n + 1) n = 2,3,....$
Hello,
Let's call f the function : $\displaystyle f(x)=pq^{x-1}$
We can notice that $\displaystyle \displaystyle p\frac{f(x)}{f(y)}=f(x-y)$
From Bayes Formula, it appears that
$\displaystyle \displaystyle \begin{aligned} P(X=k|X+Y=n+1)&=\frac{P(X=k,X+Y=n+1)}{P(X+Y=n+1)} \\
&=\frac{P(X=k,Y=n-k+1)}{P(X+Y=n+1)} \\
&=\frac{f(k)f(n-k+1)}{P(X+Y=n+1)} \end{aligned}$
(by independence for the last line)
But $\displaystyle \{X+Y=n+1\}$ is the disjoint union $\displaystyle \displaystyle \bigcup_{j=1}^n \{X=j,Y=n-j+1\}$
Hence
$\displaystyle \displaystyle \begin{aligned}P(X+Y=n+1)&=\sum_{j=1}^n P(X=j,Y=n-j+1) \\
&=\sum_{j=1}^n f(j)f(n-j+1) \end{aligned}$
Back to the formula...
$\displaystyle \displaystyle \begin{aligned}P(X=k|X+Y=n+1)&=\frac{f(k)f(n-k+1)}{\sum_{j=1}^n f(j)f(n-j+1)} \\
&=\frac{1}{\sum_{j=1}^n \frac{f(j)}{f(n-k+1)}\cdot\frac{f(n-j+1)}{f(k)}} \\
&=\frac{1}{\sum_{j=1}^n \frac{pf(j-n+k-1)}{pf(k-n+j-1)}} \\
&=\frac 1n \end{aligned}$
Ain't that nice ?