# Thread: Conditional probability of joint geometric random variables

1. ## Conditional probability of joint geometric random variables

Suppose that X and Y are independent random variables with the same parameter p.
What is the value of
$\displaystyle P[X=i|X+Y=n]$

Want to make sure my solution is correct
$\displaystyle P[X=i|X+Y=n]=\frac{P[X=i]P[Y=n-i]}{P[X+Y=n]}$
$\displaystyle =\frac{(1-p)^{i-1}p(1-p)^{n-i-1}p}{(1-p)^{n-1}p}$
$\displaystyle =p(1-p)$

Correct?

2. ## Re: Conditional probability of joint geometric random variables

Originally Posted by I-Think
Suppose that X and Y are independent random variables with the same parameter p.
What is the value of
$\displaystyle P[X=i|X+Y=n]$

Want to make sure my solution is correct
$\displaystyle P[X=i|X+Y=n]=\frac{P[X=i]P[Y=n-i]}{P[X+Y=n]}$
$\displaystyle =\frac{(1-p)^{i-1}p(1-p)^{n-i-1}p}{(1-p)^{n-1}p}$
$\displaystyle =p(1-p)$

Correct?

careful with your denominator...

the sum of geometric rvs (for n>=2) is given by:

$\displaystyle P(X+Y=n)=\sum_{x=1}^{n-1} P(X=x, Y=n-x)=\sum_{x=1}^{n-1} P(X=x)\cdot P(Y=n-x)$

$\displaystyle \;=\; \sum_{x=1}^{n-1}p(1-p)^{x-1}p(1-p)^{n-x-1}\;=\;.......$ ,which is your denominator

3. ## Re: Conditional probability of joint geometric random variables

So the denominator becomes
$\displaystyle \sum_{x=1}^{n-1} p(1-p)^{x-1} p ({1-p})^{n-x-1}= \sum_{x=1}^{n-1} p^2({1-p})^{n}$
$\displaystyle =p^2(1-p)^n$

So the answer is 1?

4. ## Re: Conditional probability of joint geometric random variables

Originally Posted by I-Think
So the denominator becomes
\sum_{x=1}^{n-1} p(1-p)^{x-1} p {1-p}^{n-x-1}= \sum_{x=1}^{n-1} p^2{1-p}^{n}
=p^2(1-p)^n

So the answer is 1?
No. try the calculations again.

5. ## Re: Conditional probability of joint geometric random variables

Round 2
Denominator = $\displaystyle p^2(n-1)(1-p)^{n-2}$

And the final answer is $\displaystyle \frac{1}{n-1}?$

6. ## Re: Conditional probability of joint geometric random variables

Originally Posted by I-Think
Round 2
Denominator = $\displaystyle p^2(n-1)(1-p)^{n-2}$

And the final answer is $\displaystyle \frac{1}{n-1}?$