Conditional probability of joint geometric random variables

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December 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
$P[X=i|X+Y=n]$

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

Correct?
December 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
$P[X=i|X+Y=n]$

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

Correct?

careful with your denominator...

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

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

$\;=\; \sum_{x=1}^{n-1}p(1-p)^{x-1}p(1-p)^{n-x-1}\;=\;.......$ ,which is your denominator
December 5th 2011, 06:45 PM
I-Think

Re: Conditional probability of joint geometric random variables

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?
December 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

So the answer is 1?

No. try the calculations again.
December 5th 2011, 08:15 PM
I-Think

Re: Conditional probability of joint geometric random variables

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

And the final answer is $\frac{1}{n-1}?$
December 5th 2011, 08:28 PM
harish21

Re: Conditional probability of joint geometric random variables

Quote:

Originally Posted by I-Think

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

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

(Yes)