# Exponential distribution with exponential parameter

Let X be an exponential random variable with parameter Λ, which is itself a random variable with distribution Exponential(β). Find a formula for the conditional density of Λ given X=x, and identify its type (i.e. its name and parameter(s)).

I made some good progress but then got stuck. I started with Bayes' Rule, and wrote

$$f(Λ|x)=\frac {f(x|Λ)f(Λ)}{f(x)}$$ Now the numerator expressions seemed simple enough. $$f(x|Λ) = Λe^{-Λx}$$, and $$f(Λ) = βe^{-βΛ}$$. I'm having a hard time finding an expression for $$f(x)$$. I thought about integrating over the possible values of Λ: $$f(x) = \int_o^\inftyΛe^{-Λx}dΛ$$ but I know this is incorrect. How to approach?

Method 1) $$f(\lambda |x)=\frac {f(x|\lambda)f(\lambda)}{f(x)} \propto f(x|\lambda)f(\lambda)$$

$$=\lambda e^{-\lambda x} \beta e^{-\beta \lambda} \propto \lambda e^{-\lambda (x+\beta)}$$

so $$\lambda \sim \Gamma(2,x+\beta)$$ Gamma_distribution

$$f(\lambda|x)=\frac{(x+\beta)^2}{\Gamma(2)}\lambda^{2-1} e^{-\lambda (x+\beta)}$$

Method 2)

$$f(\lambda |x)=\frac {\lambda e^{-\lambda x} \beta e^{-\beta \lambda}}{f(x)}$$

$$=\frac {\beta \lambda e^{-\lambda (x+\beta)} }{\beta \frac{\Gamma(2)}{(x+\beta)^2}}$$

$$=\frac{(x+\beta)^2}{\Gamma(2)}\lambda^{2-1} e^{-\lambda (x+\beta)}$$

Sine $$f(x)=\int f(x,\lambda) d\lambda=\int f(x|\lambda)f(\lambda) d\lambda$$

$$=\int \beta \lambda e^{-\lambda (x+\beta)} d\lambda$$ $$=\beta\int \lambda^{2-1} e^{-\lambda (x+\beta)} d\lambda$$

$$=\beta \frac{\Gamma(2)}{(x+\beta)^2}$$

• Can you explain what you did, I don't follow Apr 2, 2020 at 18:53
• You do not need calculate $f(x)$ since it is look like a constant in density $f(\lambda|x)$. so I calculate the core of density(every part that depend on $\lambda$) and then change it to valid density. Apr 2, 2020 at 20:29