Discrete Independent random variables X and Y

Question

Let $X$ and $Y$ be independent random variables, taking values in the positive integers and having the same mass function $f(x)=2^{-x}$ for $x=1,2,..... \infty$
Find $P(Y\gt X)$.

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Here's what I did,
$P(Y \gt X) = \sum_{x=1}^{\infty} \{P(Y \ge x,X \le x) - P(Y=x,X=x)\}$
I think this equation is correct because; for any arbitary $x$, we must have $X$ smaller than it, and $Y$ greater than it (that is the first part). Then I subtract the probability that both of them are equal to $x$ (the second part).

$\sum_{x=1}^{\infty}P(Y=x,X=x) = \sum_{x=1}^{\infty} P(Y=x)P(X=x) = \sum_{x=1}^{\infty} 4^{-x} = \frac 13$

$P(Y \ge x,X \le x)=P(Y \ge x)P(X \le x) = P(X \le x) .(1-P(Y\le x) + P(Y=x)) $

After this, I solved for $P(X \le x)$ and $P(Y \le x)$ using the mass distribution function.

Next I just solved simplified the equations until I got the answer.
I know there is something wrong as I get the probability as $1$.
Any help would be appreciated.

Answer 1

As others have suggested, you could use that $P(Y<X) = P(X<Y)$ since they are IID. But your initial step is wrong as there is a whole lot of double counting happening. Instead of that try writing(more along your approach) $$P(Y>X) = \sum_{x=1}^{\infty}P(X = x, Y>x) = \sum_{x=1}^{\infty}P(X=x)(1 - P(Y \leq x))$$ Since the variables are independent.

Answer 2

EDITED

$$P(Y>X)=$$ $$=P(Y=2)P(X=1)+P(Y=3)P(X=1)+\cdots$$

$$+P(Y=3)P(X=2)+P(Y=4)P(X=2)+\cdots$$

$$+P(Y=4)P(X=3)+P(Y=5)P(X=3)+\cdots$$ $$\cdots=$$

$$=2^{-3}+2^{-4}+\cdots$$ $$+2^{-5}+2^{-6}+\cdots$$ $$+2^{-7}+2^{-8}+\cdots$$ $$\cdots=$$

$$=2^{-2}+$$ $$=2^{-4}+$$ $$=2^{-6}+$$ $$\cdots$$ $$=\frac13.$$

As $$P(X=Y)=P(X<Y)=P(Y<X).$$

Answer 3

Hint: $\mathsf P(Y<X)+\mathsf P(Y=X)+\mathsf P(Y>X)=1$ and $\mathsf P(Y>X)=\mathsf P(Y<X)$ so...$$\mathsf P(Y>X)=\tfrac 12-\tfrac 12\sum_{x=1}^\infty 2^{-2x}$$