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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} \,dx - P(Y = \alpha -1 )$$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} \,dx - P(Y = \alpha -2 ) - P(Y = \alpha -1 )$$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} \,dx - P(Y = \alpha -(n-1)) \ldots -P(Y = \alpha -2 ) - P(Y = \alpha -1 )$$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} \,dx - [ 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} \,dx - [ 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} \,dx - P(Y \le \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} \,dx$$\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.

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

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

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

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

which will be

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

$ = \frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dx - [ 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} \,dx - 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 $\frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dx$ is an integral of a valid pdf.

Thanks for the help in advance.

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.

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user1916067
user1916067
  • 137
  • 1
  • 9

Relationship between Poisson and Gamma Distribution

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

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

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

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

which will be

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

$ = \frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dx - [ 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} \,dx - 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 $\frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dx$ is an integral of a valid pdf.

Thanks for the help in advance.