1. ## problem about discrete uniform distribution

Suppose that X and Y are independent discrete random variables with the same probability mass function

pX(k) = pY(k) =pq^(k-1)

where 0<p<1. Show that for any integer n greater than or equal to 2, the conditional probability mass function

p(k) = P(X=k|X+Y=n)

is a discrete uniform distribution

So I do so in this way

P(X=k|X+Y=n)
=P(X=k|Y=n-k)
=P(X=k, Y=n-k)/P(Y=n-k)
=$\displaystyle p(k,n-k)$/P(Y=n-k)
=p2qn-1+p2qn-1+..../pqn-k
=np2qn-1/pqn-k
=nqk-1
which should be wrong in my opinion. So what should be the correct ans?

2. ## Re: problem about discrete uniform distribution

Sorry I dun know how to type in the summation sign summing up the probability from k=1 to n in step 3 ==,

3. ## Re: problem about discrete uniform distribution

$P[k] = P[X=k|X+Y=n] = \dfrac{P[X+Y=n|X=k]P[X=k]}{P[X+Y=n]}$

$P[X+Y=n|X=k] = P[Y=n-k] = p q^{n-k}$

$P[X=k] = p q^k$

$P[X+Y=n] = \sum \limits_{j=1}^n~(p q^j)(p q^{n-j}) = p^2 n q^n$

$P[k] = \dfrac{(p q^{n-k})(p q^k)}{p^2 n q^n} = \dfrac{1}{n}$