Relationship between Poisson and Gamma Distribution

Question

There are several solutions to this problem but I am interested in the solution in Casella & Burger Pg. 100. The problem shows that if $X$ follows gamma($\alpha$, $\beta$), a random variable and $Y \sim Poisson(x/\beta)$, then $P(X\le x) = P(Y\ge \alpha)$.

In the text, $P(X \le x) = \frac{1} {(\alpha - 2)! \beta^{\alpha - 1}}\int_{0}^{x} t^{\alpha - 2} e^{-t/\beta} \,dt - P(Y = \alpha -1 )$

After second integration by parts (as suggested by the text), I get

$P(X \le x) = \frac{1} {(\alpha - 3)! \beta^{\alpha - 2}}\int_{0}^{x} t^{\alpha - 2} e^{-t/\beta} \,dt - P(Y = \alpha -2 ) - P(Y = \alpha -1 )$

I surmise that at the $n^{th}$ step I will have $P(X \le x) = \frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dt - P(Y = \alpha -(n-1)) \ldots -P(Y = \alpha -2 ) - P(Y = \alpha -1 )$

which will be

$\frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dt - [ P(Y = \alpha -(n-1)) \ldots +P(Y = \alpha -2 ) + P(Y = \alpha -1 )]$

$ = \frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dt - [ P(Y = \alpha -(n-1)) \ldots +P(Y = \alpha -2 ) + P(Y = \alpha -1 )]$

$ = \frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dt - P(Y \le \alpha- 1) $

I want to argue that as $n \rightarrow \infty $ , $\frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta}= 1 $. In which case, $ = 1 - P(Y \le \alpha- 1) = P(Y \ge \alpha) $.

I am not sure however, if $\frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dt$ is an integral of a valid pdf.

Thanks for the help in advance.

Answer 1

You are repeating integration by parts of the term

$$\frac{\int_{0}^{x} t^{\alpha - 1 - n} e^{-t/\beta} \,dt} {(\alpha - 1 - n)! \beta^{\alpha - n}} = - \left[ \frac{(t/\beta)^{\alpha - 1 - n} e^{-t/\beta} } {(\alpha - 1 - n)! } \right]_0^x+\frac{\int_{0}^{x} t^{\alpha - 2 - n} e^{-t/\beta} \,dt} {(\alpha - 2 - n)! \beta^{\alpha - 1- n}} $$

The repeated integration by parts will terminate at the point when $n = \alpha - 1$. The last term becomes

$$\frac{1} {(0)! \beta^{1}}\int_{0}^{x} t^{0} e^{-t/\beta} \,dt = -\left[e^{-t/\beta}\right]_0^x = 1 - P(Y=0)$$