# Conditional probability of joint geometric random variables

• Dec 5th 2011, 05:51 PM
I-Think
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?
• Dec 5th 2011, 06:25 PM
harish21
Re: Conditional probability of joint geometric random variables
Quote:

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?

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
• Dec 5th 2011, 06:45 PM
I-Think
Re: Conditional probability of joint geometric random variables
So the denominator becomes
$\displaystyle \sum_{x=1}^{n-1} p(1-p)^{x-1} p ({1-p})^{n-x-1}= \sum_{x=1}^{n-1} p^2({1-p})^{n}$
$\displaystyle =p^2(1-p)^n$

• Dec 5th 2011, 07:20 PM
harish21
Re: Conditional probability of joint geometric random variables
Quote:

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

No. try the calculations again.
• Dec 5th 2011, 08:15 PM
I-Think
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 $\displaystyle \frac{1}{n-1}?$
• Dec 5th 2011, 08:28 PM
harish21
Re: Conditional probability of joint geometric random variables
Quote:

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

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

(Yes)