# independant random variables ?

• Mar 7th 2011, 01:18 PM
abrahamtim
independant random variables ?
Suppose that X and Y are independent random variables with the same geometric distribution, $P(X = k) = P(Y = k) = pq^(k-1)$where $q = 1 - p$

Find
$P( X = k | X + Y = n + 1) n = 2,3,....$
• Mar 7th 2011, 04:43 PM
mr fantastic
Quote:

Originally Posted by abrahamtim
Suppose that X and Y are independent random variables with the same geometric distribution, $P(X = k) = P(Y = k) = pq^(k-1)$where $q = 1 - p$

Find
$P( X = k | X + Y = n + 1) n = 2,3,....$

I suggest you first work with concrete cases. eg. n = 2 and k = 0, 1, 2, 3. etc. Then see if you can generalise.
• Mar 8th 2011, 02:29 PM
Moo
Hello,

Let's call f the function : $f(x)=pq^{x-1}$

We can notice that $\displaystyle p\frac{f(x)}{f(y)}=f(x-y)$

From Bayes Formula, it appears that
\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 $\{X+Y=n+1\}$ is the disjoint union $\displaystyle \bigcup_{j=1}^n \{X=j,Y=n-j+1\}$

Hence
\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 \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 ? (Itwasntme)