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Let $X$ be poisson distribution with parameter $\lambda$ and $Y$ geometric with parameter $p$. Find $P(Y>X)$. ($\bf independent$)

Attempt

We know $p_X(x) = \frac{ e^{-\lambda} \lambda^x }{x!} $ and $p_Y(y) = (1-p)^{y-1} p$. Now, we have the region $\{(x,y) : y>x \}$ for integers $x,y$. We want to sum over this region. My thought is to first sum $y$ from $1$ to $\infty$ and $x$ is between $y$ and $\infty$. Thus,

$$ P(Y>X) = \sum_{y=1}^{\infty} \sum_{x=y}^{\infty} \frac{ e^{-\lambda} \lambda^x }{x!} (1-p)^{y-1} p = pe^{-\lambda} \sum_{y=1}^{\infty} (1-p)^{y-1} \sum_{x=y}^{\infty} \frac{ \lambda^x}{x!} = p e^{-\lambda} \frac{1}{1-1+p} \sum_{x=y}^{\infty} \frac{ \lambda^x}{x!} = e^{- \lambda} \sum_{x=y}^{\infty} \frac{\lambda^x}{x!}$$

Here is where I get stuck. My answer key says it should $e^{- \lambda p }$. But I cant get to this answer. Any help?

James
  • 3,997

1 Answers1

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\begin{align} P(Y>X)&=\sum_{n=0}^\infty P(Y>n)P(X=n)\\ &=\sum_{x=0}^\infty (1-p)^n\left(\frac{ e^{-\lambda} \lambda^n }{n!}\right)\\ &=e^{-\lambda}\sum_{n=0}^\infty \frac{(q\lambda)^n }{n!};\quad q=1-p\\ &=e^{-\lambda}e^{q\lambda}\\ &=e^{q\lambda-\lambda}=e^{-(1-q)\lambda}\\ &=e^{-\lambda p} \end{align}

Karn Watcharasupat
  • 4,323