0
$\begingroup$

The question

Given independent random variables $$X_1,\cdots,X_n \sim N(\theta,\phi)$$

with joint prior density $$\pi(\theta,\phi)\propto \frac1\phi$$

Show that the posterior distribution of $t(\theta)=\frac{\sqrt n}{s}(\theta - \bar x)$ follows a $t$-distribution

What I have done so far

I have found the posterior density to be

$$\pi(\theta,\phi | \vec x ) \propto \phi^{-\frac n2-1} \exp \biggl [-\frac{1}{2\phi}\biggl((n-1)s^2+n(\bar x-\theta)^2\biggr)\biggr] $$

where $$s^2=\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\bar x)^2$$

Integrating over $\phi \in (0,\infty)$ with reference to the Inverse Gamma distribution, we get the marginal distribution of $\theta$

$$\pi(\theta|\vec x) \propto \biggl(\frac{2}{(n-1)s^2+n(\bar x-\theta)^2}\biggr)^{n/2}$$

And then I am stuck...

For reference

The posterior density $\pi(\theta,\phi|\vec x)$ is certainly correct because it was given in the question.

Inverse gamma distribution with parameters $(\alpha,\beta)$:

$$f(y)=\frac{\beta^\alpha}{\Gamma(\alpha)y^{\alpha+1}}e^{-\beta/y} \; \; \; \; \; y>0$$

$t$-distribution with parameter $r$:

$$f(y) \propto \biggl(1+\frac{y^2}{r}\biggr)^{-\frac{r+1}{2}} \; \; \; \; y \in \Bbb R$$

$\endgroup$
4
  • $\begingroup$ Don't you want to use $\propto$ instead of $\alpha$? $\endgroup$ Commented Mar 30, 2018 at 13:49
  • $\begingroup$ How did u make that symbol? $\endgroup$ Commented Mar 30, 2018 at 13:50
  • $\begingroup$ You may want to consider using propto ($\propto$) or varpropto ($\varpropto$). detexify is useful in these matters. $\endgroup$ Commented Mar 30, 2018 at 13:50
  • $\begingroup$ @glowstonetrees Backslash propto. $\endgroup$ Commented Mar 30, 2018 at 13:50

1 Answer 1

Reset to default
1
$\begingroup$

Suppose we know the density function $f_x$ of a random variable $X$, we can work out the density function of $Y = a(X - b)$ for constants $a$ and $b$. If $F_X$ denotes the distribution function of $X$, i.e. $F(x) = P(X \leq x)$ then $$ F_Y(y) = P(a(X - b) \leq y) = P\left(X \leq \frac{1}{a}y + b\right) = F_X\left(\frac{1}{a}y + b\right) $$ Then $f_X(x) = F'_X(x)$ therefore $f_y(y) = \frac{1}{a} f_x\left(\frac{1}{a} y + b \right)$

In your case $X = \theta$ and $Y = t(\theta) = \frac{\sqrt{n}}{s}(X - \bar{x})$. So plugging $\frac{s}{\sqrt{n}}y + \theta$ into $\pi(\theta \mid \vec{x})$ and simplifying gives $$ f_{Y}(y \mid \vec{x}) \propto \left( \frac{(n-1)s^2 + n(\bar{x} - \frac{s}{\sqrt{n}} y - \bar{x})^2 }{2} \right)^{\frac{-n}{2}} \propto \left( 1 + \frac{y^2}{n-1} \right)^{\frac{-n}{2}}$$ So in fact $t(\theta)$ follows a $t$ distribution on $n - 1$ degrees of freedom.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.