Find the probability $P(X<Y)$.

Question

Suppose $X$ and $Y$ be two independent Poisson $(\lambda)$ random variables. Find $P(X<Y)$.

My attempt $:$

\begin{align*}P(X<Y) &= \sum\limits_{x=0}^{\infty} \sum\limits_{y=x+1}^{\infty} P(X=x,Y=y) \\ &= \sum\limits_{x=0}^{\infty} \sum\limits_{y=x+1}^{\infty} P(X=x) P(Y=y) \\ &= e^{-2\lambda} \sum\limits_{x=0}^{\infty} \frac {{\lambda}^x} {x!} \sum\limits_{y=x+1}^{\infty} \frac {{\lambda}^y} {y!} \\ &= e^{-2\lambda} \sum\limits_{x=0}^{\infty}\left( \frac{e^{-\lambda}\lambda^x}{x!} - \frac{\lambda^x}{x!} \sum\limits_{y=0}^x \frac {\lambda^y}{y!} \right) \\ &= e^{-2\lambda} - e^{-2\lambda} \sum\limits_{x=0}^{\infty} \left( {{\frac {{\lambda}^x} {x!}} \sum\limits_{y=0}^{x} {\frac {{\lambda}^y} {y!}}} \right) \end{align*}

Now how do I calculate $$\sum\limits_{x=0}^{\infty} \left( \frac {{\lambda}^x} {x!} \sum\limits_{y=0}^{x} \frac {{\lambda}^y} {y!} \right)$$ Please help me in this regard.

Thank you very much.

Answer 1

For the special case of $X$ and $Y$ being identically distributed, you have

$$P(X < Y) + P(Y < X) + P(Y = X) = 1$$ $$2 P(X < Y) + P(X = Y) = 1$$ $$ P( X < Y) = 1/2 (1 - P(X = Y))$$

So it reduces to computing $P(X =Y)$ whose computation appears here Probability that two independent Poisson random variables with same paramter are equal