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?


1 Answer 1


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

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

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

  • $\begingroup$ Can you explain what you did, I don't follow $\endgroup$
    – Dmlawton
    Apr 2, 2020 at 18:53
  • $\begingroup$ 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. $\endgroup$
    – Masoud
    Apr 2, 2020 at 20:29

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