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The conditional probability mass function.

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

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

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

This implies

$f\{X鈭�N=n\}=\frac{{位}^n{x}^{n-1}{e}^{-位x}}{螕n};x>0$

It is required to find$f\{N鈭�X=x\}$

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

Now find $f(N鈭�X)$

$\begin{array}{r}f(N鈭�X)=f(X鈭�N=n)脳f\left(N\right)\\ =\frac{{位}^n{x}^{n鈭�1}{e}^{鈭�位x}}{\mathrm{螕}n}脳q^{n}p\end{array}$

$\begin{array}{rl}f(X=x)& ={鈭�}_0^{\mathrm{鈭�}}f(X鈭�N=n)脳f\left(N\right)\\ f(X=x)& ={鈭�}_0^{\mathrm{鈭�}}\frac{{位}^n{x}^{n鈭�1}{e}^{鈭�位x}}{\mathrm{螕}n}脳q^{n}p\\ f(X=x)& ={\mathrm{Lim}}_{n鈫�\mathrm{鈭�}}鈦�\left(\frac{{位}^n{x}^{n鈭�1}q{q}^n{e}^{位x}{e}^{鈭�位x}x鈭�1}{\ln 鈦�\left(q\right)+\ln 鈦�\left(x\right)+\ln 鈦�\left(位\right)}\right)\frac{e^{鈭�位x}p}{x\mathrm{螕}n}\text{(using Maple software)}\end{array}$

On simplification

$f(X=x)={\mathrm{Lim}}_{n鈫�\mathrm{鈭�}}鈦�\left(\frac{{位}^n{x}^{n鈭�1}{q}^{n}x鈭�1}{\ln 鈦�\left(q\right)+\ln 鈦�\left(x\right)+\ln 鈦�\left(位\right)}\right)\frac{e^{鈭�位x}p}{x\mathrm{螕}n}$

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

$\begin{array}{r}f(N鈭�X=x)=\frac{f(N鈭�X)}{f\left(X\right)}\\ =\frac{2^{鈥测赌�}{q}^{鈥测赌�\mathrm{鈥�}}}{鈭�h}脳g^{\mathrm{鈥�}}\mathrm{辫谈}脳\frac{[\ln 鈦�(q)+\ln 鈦�(x)+\ln 鈦�(位\left)\right]x\mathrm{丑谈}}{\left[2^n{x}^{鈥测赌�}{g}^{\mathrm{鈥�}}x鈭�1\right]g^{鈥测赌�}\mathrm{辫谈}}\\ =\underset{n鈫�\mathrm{鈭�}}{lim}鈥�\left[\frac{[\ln 鈦�(q)+\ln 鈦�(x)+\ln 鈦�(位\left)\right]x}{x鈭�1}\right];x>1\end{array}$