The joint posterior distribution of Reyleigh distribution

Question

Suppose that $X_1 ,\ldots ,X_n$ have the one parameter Rayleigh distribution. The probability density function (pdf) is given by $$ f(x |\theta )=\frac{x}{\theta}\exp(-\frac{x^2}{2\theta}), \quad x >0,\; 0<\theta <\infty .$$ The maximum likelihood estimate (MLE) of $\theta$ is $\hat{\theta}=\frac{T}{2n}$, where $T=\sum_{i=1}^{n}x_i^2$. Let conjugate prior distribution has the form $$g(\theta |\alpha,\beta)\propto \theta^{-(\alpha +1)}\exp(-\frac{\beta}{\theta}),\quad \theta >0.$$ I don't know how to get the joint posterior distribution as follows $$\pi (\theta |T)=\frac{(\beta +\frac{T}{2})^{n+\alpha}\theta^{-(n+\alpha +1)}e^{-\frac{1}{\theta}(\beta +\frac{T}{2})}}{\Gamma (n+\alpha)}.$$ By the Bayes theorem, we have $$\pi (\theta |T)=\frac{p(\theta)p(T|\theta)}{p(T)}$$, where $p(\theta)$ is the prior distribution and $p(T|\theta)$ is likelihood function. But I don't understand how to get $\pi (\theta |T) $ like above.

Answer 1

Consider that your likelihood is the following (never mind $x$ because it will become a constant, not depending on $\theta$)

$$p(\mathbf{x}|\theta)\propto \theta^{-n}e^{-T/(2\theta)}$$

... and your prior is

$$\pi(\theta)\propto \theta^{-(\alpha+1)}e^{-\beta/\theta}$$

Thus, just multiplying them you get that

$$\pi(\theta|\mathbf{x})\propto \theta^{-(n+\alpha+1)}\cdot e^{-\frac{1}{\theta}\left( \beta+ T/2\right)}$$

Now looking at this posterior we immediate recognize the kernel of an Inverse Gamma distribution with parameters:

• Shape: $(n+\alpha)$

• Scale: $\left(\beta+\frac{T}{2}\right)$

As you can see in the link above, this density has exactly the following expression:

$$\pi(\theta|\mathbf{x})=\frac{\left(\beta+\frac{T}{2}\right)^{n+\alpha}}{\Gamma(n+\alpha)}\cdot \theta^{-(n+\alpha+1)}\cdot e^{-\frac{1}{\theta}\left( \beta+ T/2\right)} $$

Which is exactly what you have to show but your $m$ that is a typo, as you have $n$ observations...

In Bayesian Statistics, when possible, the normalizing constant has to be derived recognizing the kernel of your posterior...otherwise it has to be derived by integration...

Answer 2

how do you understand that the likelihood is ...

the likelihood is the product of your densities: $f(x_1)\cdot f(x_2)\cdot \dots\cdot f(x_n)$ that is

$$p(\mathbf{x}|\theta)=\theta^{-n}\cdot \prod_i x_i\cdot e^{-\sum_i x_i^2/(2\theta)}$$

The quantity $\prod_i x_i$ is a constant if you look at it as a function of $\theta$. Your goal is to derive the posterior distribution of $\theta$ thus you can waste it and write directly

$$p(\mathbf{x}|\theta)\propto \theta^{-n}\cdot e^{-\sum_i x_i^2/(2\theta)}$$