Show that $p(0 ; \lambda) + ... + p(n; \lambda) = \frac1{n!} \int_{\lambda}^\infty e^{-x}x^n dx,$ where $p(k; \lambda)$ is a Poisson distribution.

Question

Prove $$p(0 ; \lambda) + ... + p(n; \lambda) = \frac1{n!} \int_{\lambda}^\infty e^{-x}x^n dx.$$

$p(k; \lambda)$ is a Poisson distribution with parameter $\lambda$. The left side is equal to $\frac{1}{n!}\sum_{k=0}^n \frac{\lambda^k}{k!}e^{-\lambda} n!$. Thus, $\sum_{k=0}^n \frac{\lambda^k}{k!}e^{-\lambda} n! = \int_{\lambda}^\infty e^{-x}x^n dx$. I found that the right hand side look close to the gamma function: $\Gamma(n+1) = \int_0^\infty e^{-x}x^n dx$, and $n! = \Gamma(n+1)$. I was trying to find a way to use these to solve this question, but I couldn't. I appreciate if you give some help.

Answer 1

$p(k;\lambda)=e^{-\lambda}\frac{\lambda^k}{k!}$ for $k=0,1,2,\dots$.

$\frac{1}{0!}\int_{\lambda}^{\infty}e^{-x}x^0dx =\int_{\lambda}^{\infty}e^{-x}dx = e^{-\lambda} = p(0;\lambda)$

$\frac{1}{1!}\int_{\lambda}^{\infty}e^{-x}x^1dx =\int_{\lambda}^{\infty}e^{-x}xdx = e^{-\lambda}+\lambda e^{-\lambda} = p(0;\lambda)+p(1;\lambda)$

So, the equality holds up to $n=2$. Assume it holds for $n$. Apply integration by parts for $n+1$ letting $u=\frac{x^{n+1}}{(n+1)!}$ and $dv=e^{-x}$. We then get $\int udv = uv - \int vdu$:

\begin{eqnarray*} \int_{\lambda}^{\infty}e^{-x}\frac{x^{n+1}}{(n+1)!}dx &=& \frac{e^{-x}}{-1}\frac{x^{n+1}}{(n+1)!}\Big|_{\lambda}^{\infty} + \int_{\lambda}^{\infty}e^{-x}\frac{x^n}{n!}dx\\ &=& e^{-\lambda}\frac{\lambda^{n+1}}{(n+1)!} + p(n;\lambda)+\cdots+p(0;\lambda)\\ &=& p(n+1;\lambda) + p(n;\lambda)+\cdots+p(0;\lambda)\\ \end{eqnarray*} so the result holds by induction.