Assuming $X\sim\mathcal{P}(\lambda)$ and $Y\sim\Gamma(w,1)$ prove that $P(X\ge w)=P(Y\le \lambda)$. How this fact is lead from the connection between the poisson and exponential distributions?
I don't know from where to start. poisson is defined only for discrete situations but the exponential is only for continious situation. How can I prove the fact ?
EDIT: for gamma distribution I wrote that $f_Y(y)=\frac{y^{w-1}e^{-y}}{\Gamma(w)}$, but I have problem with integrating it. About Poisson: its function is $\displaystyle \sum _{w_i=0}^w P(X=w_i)$ which I don't know how to sum into a final expression. How can I continue?
Best Answer
Let $w$ be an integer. Then $$ P(X\geqslant w)=1-P(X\leqslant w-1)=1-e^{-\lambda}\sum_{k=0}^{w-1}\frac{\lambda^k}{k!}. $$ Now, we use that if $\Gamma(a,b)$ denotes the incomplete Gamma function, i.e. $$ \Gamma(a,b)=\int_b^\infty t^{a-1}e^{-t}\,\mathrm dt, $$ then $$ \Gamma(n,b)=(n-1)!e^{-b}\sum_{k=0}^{n-1}\frac{b^k}{k!} $$ provided that $n$ is an integer. Thus (recall that $\Gamma(n)=(n-1)!$ for $n$ integer) $$ P(X\geqslant w)=1-\frac{\Gamma(w,\lambda)}{\Gamma(w)}. $$ Rewriting this expression we arrive at the desired expression $$ \begin{align} P(X\geqslant w)&=\frac{\Gamma(w)-\Gamma(w,\lambda)}{\Gamma(w)}=\frac{1}{\Gamma(w)}\left(\int_0^\infty t^{w-1}e^{-t}\,\mathrm dt-\int_\lambda^\infty t^{w-1}e^{-t}\,\mathrm dt\right)\\ &=\frac{1}{\Gamma(w)}\int_0^\lambda t^{w-1}e^{-t}\,\mathrm dt=P(Y\leqslant\lambda). \end{align} $$