Question

Consider the simplified Bayesian model for normal data The joint posterior pdf is ful, σ21 x)a(σ2,-/2-1 expl_jy.tx, _aPI The marginal posterior pdfs of μ and σ 2 can be obtained by integrating out the other variable (8.30) y@1 x) α (σ2)-m;,-1/2 expl-- Σ.-tri-x)2 (8.31) d. Let q1 and q2 be they/2 and 1-y/2 quantiles of (8.31). Show that the 1-γ credible interval (gi,q2) is identical to the classic confidence interval (5.19) (with ar replaced by y). Hence, a (1-α) stochastic confidence interval for σ2 is (5.19) e. By using (8.30) and (8.6) show that -(v+D/2 where v 1. Verify that, in view of (2.23), this means that Definition 8.2. (Inverse-Gamma Distribution). A random variable Z is said to have an inverse-gamma distribution with shape parameter α > 0 and scale parameter A >0 if its pdf is given by (8.6) 1/X with X ~ Gamma(α, λ). This is the pdf of the random variable Z = Definition 2.19. (Student's t Distribution). A random variable X is said to have a Student's t distribution with parameter v >0 if its pdf is given by where Γ denotes the gamma function. We write X ~ tv. For integer values the parameter v is referred to as the degrees of freedom of the distribution. f. Let q1 and q2 be the y/2 and 1-y/2 quantiles of fu l x). Show that the 1-γ credible interval 41,42) is identical to the classic confidence interval (5.18) (with α replaced by γ). (5.18)

Answer 1

d)Note the sample variance is  .

n- 1

From the given density, the distribution of sample variance is such that $\fn_jvn \frac{(n-1)S_X^2}{\sigma ^2}\sim \chi ^2_{n-1}$ . Here

$P\left ( \chi ^2_{\alpha /2,n-1}<\chi ^2<\chi ^2_{1-\alpha /2,n-1} \right )=1-\alpha$. So the
(1-0) 100%
confidence interval for $\sigma ^2$ can be found as

(n - 1)S2 σ2 (n- 1)S? X1-a/2,n-1 χο/2,n-1

Or in interval notation,  

(n-1)S(n-1)S 2

e) Here has t-distribution with $df=n-1$ . If you set $x=\frac{\mu -\overline{x}}{s_x/\sqrt{n}}$ , then  
sx/Vn
. Also

$P\left ( t_{\alpha /2,n-1}<T<t_{1-\alpha /2,n-1} \right )=1-\alpha$

So the
(1-0) 100%
confidence interval for $\mu$ can be found as

$t_{\alpha /2,n-1}<\frac{\mu -\overline{x}}{s_x/\sqrt{n}}<t_{1-\alpha /2,n-1}\\ -t_{1-\alpha /2,n-1}<\frac{\mu -\overline{x}}{s_x/\sqrt{n}}<t_{1-\alpha /2,n-1}\\ {\color{Blue} \overline{x}\pm t_{1-\alpha /2,n-1}\frac{s_x}{\sqrt{n}}}$