# Conditional density function with gamma and Poisson distribution

It is Question 17 in Ross's book (Introduction to Probability Models-11th edition)

Let $Y$ be a gamma random variable with parameters $(s, \alpha)$. That is, its density is

$$f_Y(y) = Ce^{-\alpha y}y^{s-1} \quad \forall\,y > 0,$$

where $C$ is a constant that does not depend on $y$. Suppose also that the conditional distribution of $X$ is given that $Y = y$ is Poisson with mean $y$. That is,

$$P(X=i\mid Y=y) = \frac{e^{-y} y ^ i}{i!} \quad \forall\, i \ne 0$$

Show that the conditional distribution of $Y$ given that $X = i$ is the gamma distribution with parameters $(s+i,\alpha+1)$.

Here is my steps:

1. Find the joint density function

$$f_{X,Y}(i,y) = f_{X|Y}(i,y) f_Y(y) = e^{-y} y^i/i! \cdot Ce^{-\alpha y} y^{s-1} = Ce^{-(\alpha + 1)y} \cdot y^{s+i-1}/i!.$$

1. Find $f_X(i)$ by $$\int_0^\infty f_{X,Y}(i,y) \,dy = \int_0^\infty Ce^{-(\alpha + 1)y} \cdot y^{s+i-1}/i! \,dy ?$$

2. Find $f_{Y|X}(y,i)$ by $f_{X,Y}(i,y) / f_X(i)$.

Then I am stuck at step 2, do not know how to proceed with that integral.

Just continue step (3).

\begin{align} f_{Y \mid X}(y,i) &= \frac{f_{X,Y}(i,y)}{f_X(i)} \\ &= \large\frac{\frac{Ce^{-(\alpha + 1)y} \cdot y^{s+i-1}}{i!}} {\int_0^\infty \frac{Ce^{-(\alpha + 1)y} \cdot y^{s+i-1}}{i!} \,dy} \\ &= \underbrace{\frac{1}{\int_0^\infty e^{-(\alpha + 1)y} \cdot y^{s+i-1} \,dy}}_{\text{constant independent of }y} \cdot e^{-(\alpha + 1)y} \cdot y^{s+i-1} \end{align}

This is a constant multiplied by $e^{-(\alpha + 1)y} \cdot y^{s+i-1}$.

Recall that gamma distribution with parameters $(s+i,\alpha+1)$ has density $C_{(s+i,\alpha+1)} e^{-(\alpha + 1)y} \cdot y^{s+i-1}$ for all $y>0$. By the very definition of density function,

$$\int_0^\infty C_{(s+i,\alpha+1)} e^{-(\alpha + 1)y} \cdot y^{s+i-1} \,dy = 1.$$

Observe that

$$\int_0^\infty \frac{1}{\int_0^\infty e^{-(\alpha + 1)y} \cdot y^{s+i-1} \,dy} \cdot e^{-(\alpha + 1)x} \cdot x^{s+i-1} \,dx = \frac{\int_0^\infty e^{-(\alpha + 1)x} \cdot x^{s+i-1} \,dx}{\int_0^\infty e^{-(\alpha + 1)y} \cdot y^{s+i-1} \,dy}=1.$$

So $C_{(s+i,\alpha+1)} = \dfrac{1}{\int_0^\infty e^{-(\alpha + 1)y} \cdot y^{s+i-1} \,dy}$ and $f_{Y \mid X}$ is the density function of $\Gamma(s+i,\alpha+1)$.

• Thanks. My questions seem stupid. 1. when deriving fY|X(y,i)fY|X(y,i), treat fX(i)fX(i) as a constant, because only i is defined and given? Is this always true? Please notice there is an i in exponent in that integral. 2.when Is the constant 'C' in your answer the same as the constant 'C' in $f_Y(y)$? Jan 19 '18 at 3:35
• @evergreenhomeland (1) $f_{Y \mid X}(y,i) = f_Y(y \mid X = i)$. This means you condition on $X = i$. (Recall: $P(A \mid B)$ is read as probability of $A$ given $B$ in words.) (2) I'm sorry for the ambiguity. The constant $C$ in my answer is for $X \mid Y \sim \Gamma(s+i,\alpha + 1)$. I'm going to revise my answer so that it reads differently from $Y \sim \Gamma(s,\alpha)$ in the question. Jan 19 '18 at 4:11