Q.6.14 - A First Course in Probability

Question

Let N be a geometric random variable with parameter p. Suppose that the conditional distribution of X given that N = n is the gamma distribution with parameters n and λ. Find the conditional probability mass function of N given that X = x. 

Step-by-Step Solution

Answer

Conditional probability mass function is $f(N\mid X=x)=\underset{n\to \infty }{\lim }\left[\frac{\left[\ln \right(q)+\ln (x)+\ln (\lambda \left)\right]x}{x-1}\right];x>1$

Step 1:To find

The conditional probability mass function.

Step 2: Explanation

It is given that $N ~geometric\left(p\right)$

$f(N=n)=q^{n}p$

The conditional distribution function of X is $f\{X\mid N=n\}~Gamma$distribution$(n,\lambda )$

This implies

$f\{X\mid N=n\}=\frac{{\lambda }^n{x}^{n-1}{e}^{-\lambda x}}{\Gamma n};x>0$

It is required to find $f\{N\mid X=x\}$

Therefore, this can be obtained with $f(N\mid X=x)=\frac{f(N\cap X)}{f\left(X\right)}$

Now find $f(N\cap X)$

$\begin{array}{r}f(N\cap X)=f(X\mid N=n)\times f\left(N\right)\\ =\frac{{\lambda }^n{x}^{n-1}{e}^{-\lambda x}}{\mathrm{\Gamma }n}\times q^{n}p\end{array}$

Step 3: To find the value of f ( X = x )

$\begin{array}{rl}f(X=x)& ={\int }_0^{\mathrm{\infty }}f(X\mid N=n)\times f\left(N\right)\\ f(X=x)& ={\int }_0^{\mathrm{\infty }}\frac{{\lambda }^n{x}^{n-1}{e}^{-\lambda x}}{\mathrm{\Gamma }n}\times q^{n}p\\ f(X=x)& ={\mathrm{Lim}}_{n\to \mathrm{\infty }}\left(\frac{{\lambda }^n{x}^{n-1}q{q}^n{e}^{\lambda x}{e}^{-\lambda x}x-1}{\ln \left(q\right)+\ln \left(x\right)+\ln \left(\lambda \right)}\right)\frac{e^{-\lambda x}p}{x\mathrm{\Gamma }n}\text{(using Maple software)}\end{array}$

On simplification

$f(X=x)={\mathrm{Lim}}_{n\to \mathrm{\infty }}\left(\frac{{\lambda }^n{x}^{n-1}{q}^{n}x-1}{\ln \left(q\right)+\ln \left(x\right)+\ln \left(\lambda \right)}\right)\frac{e^{-\lambda x}p}{x\mathrm{\Gamma }n}$

Now the conditional probability mass function$f\{N\mid X=x\}$

$\begin{array}{r}f(N\mid X=x)=\frac{f(N\cap X)}{f\left(X\right)}\\ =\frac{2^{\prime \prime }{q}^{\prime \prime \mathrm{\prime }}}{\mid h}\times g^{\mathrm{\prime }}\mathrm{p̸}\times \frac{[\ln (q)+\ln (x)+\ln (\lambda \left)\right]x\mathrm{h̸}}{\left[2^n{x}^{\prime \prime }{g}^{\mathrm{\prime }}x-1\right]g^{\prime \prime }\mathrm{p̸}}\\ =\underset{n\to \mathrm{\infty }}{lim} \left[\frac{[\ln (q)+\ln (x)+\ln (\lambda \left)\right]x}{x-1}\right];x>1\end{array}$