Vtyjkyuk

Let $X$ and $Y$ be two independent random variables. For $n=1,2,3,...$ we have $$mathbbP(X=n)=p(1-p)^n
\mathbbP(Y=n)=q(1-q)^n$$

For some $p$ and $q$. Find $mathbbP(Xleq Y)$

What I would like to know is:
1. Is my reasoning (the probabilistic part of the question) right?
2. Someone to help me find this limit I have got at the end.
So since these are probabilities we have$p$ and $q$ need to be in $(0,1)$, this follows from the formula for a sum of geometric series. Now, notice that we are interested in finding sum of all $mathbbP(X=n wedge Y=m)\$ such that $nleq m$. So in fact we can take the limit of the partial sums $$lim_mto inftysum_n=1^mmathbbP(X=n wedge Y=m)$$ $X$ and $Y$ are independent so this translates to $$lim_mto inftysum_n=1^mmathbbP(X=n)mathbbP(Y=m)=lim_mto inftyq(1-q)^m-1sum_n=1^m(1-p)^n-1p $$ using geometric series formula we get
$$lim_mto inftyq(1-q)^m-1(1-(1-p)^m) $$

and here I am stuck evaluating this limit. I could write down what I tried, but it took me nowhere closer to the result.

$PXleq Y=sum_m=1^infty sum _n=1 ^m PX=n,Y=m$. You have written limit instead of sum over $m$. Now try to compute $sum_m=1^infty q(1-q)^m-1[1-(1-p)^m]$. (You will have to sum two geometric series and take the difference).

It's possible to do this question without calculating any infinite sums. Hint: work out $mathbb P(X=nmidmin(X,Y)=n)$.