We are given that
\[ \lambda^2 \mid y \sim \text{Gamma}\left( \frac{n}{2} + a, \; b^* \right) \]
in the shape-rate parameterization.
That is, if \(X \sim \text{Gamma}(\alpha,
\beta)\), then the PDF is
\[ f(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x}, \quad x > 0, \]
with mean \(\alpha/\beta\) and variance \(\alpha/\beta^2\).
We have \(\sigma^2 =
\frac{1}{\lambda^2}\).
We need to show:
\[ E[\sigma^2 \mid y] = \frac{2b^*}{n + 2a - 2}, \] \[ \mathrm{Var}(\sigma^2 \mid y) = \frac{2}{(n+2a-2)^2 (n+2a-4)} \quad \text{(when \( b^* = 1/2 \))}. \]
For \(X \sim \text{Gamma}(\alpha, \beta)\) with \(\alpha > 1\):
\[ E[X^{-1}] = \frac{\beta}{\alpha - 1}. \]
Derivation:
\[ E[X^{-1}] = \int_0^\infty x^{-1} \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x} dx = \frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty x^{\alpha-2} e^{-\beta x} dx. \]
Let \(t = \beta x\), \(dx = dt/\beta\), \(x = t/\beta\):
\[ E[X^{-1}] = \frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty \left( \frac{t}{\beta} \right)^{\alpha-2} e^{-t} \frac{dt}{\beta} = \frac{\beta^\alpha}{\Gamma(\alpha)} \cdot \frac{1}{\beta^{\alpha-1}} \Gamma(\alpha - 1). \]
Since \(\Gamma(\alpha) = (\alpha-1)\Gamma(\alpha-1)\), we get:
\[ E[X^{-1}] = \frac{\beta}{\alpha - 1}. \]
Here \(\alpha = \frac{n}{2} + a\), \(\beta = b^*\), so
\[ E[(\lambda^2)^{-1} \mid y] = \frac{b^*}{\frac{n}{2} + a - 1}. \]
Since \(\sigma^2 = 1/\lambda^2\):
\[ E[\sigma^2 \mid y] = \frac{b^*}{\frac{n}{2} + a - 1} = \frac{2b^*}{n + 2a - 2}. \]
For \(X \sim \text{Gamma}(\alpha, \beta)\) with \(\alpha > 2\):
\[ E[X^{-2}] = \frac{\beta^2}{(\alpha - 1)(\alpha - 2)}. \]
Derivation:
\[ E[X^{-2}] = \frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty x^{\alpha-3} e^{-\beta x} dx. \]
Let \(t = \beta x\), same method:
\[ E[X^{-2}] = \frac{\beta^\alpha}{\Gamma(\alpha)} \cdot \frac{1}{\beta^{\alpha-2}} \Gamma(\alpha - 2) = \frac{\beta^2 \, \Gamma(\alpha - 2)}{\Gamma(\alpha)}. \]
Using \(\Gamma(\alpha) = (\alpha-1)(\alpha-2)\Gamma(\alpha-2)\):
\[ E[X^{-2}] = \frac{\beta^2}{(\alpha - 1)(\alpha - 2)}. \]
\[ E[\sigma^4 \mid y] = E[(\lambda^2)^{-2} \mid y] = \frac{(b^*)^2}{(\alpha - 1)(\alpha - 2)}, \] with \(\alpha = \frac{n}{2} + a\).
Compute:
\[ \alpha - 1 = \frac{n}{2} + a - 1 = \frac{n + 2a - 2}{2}, \] \[ \alpha - 2 = \frac{n}{2} + a - 2 = \frac{n + 2a - 4}{2}. \]
Thus:
\[ E[\sigma^4 \mid y] = \frac{(b^*)^2}{\left( \frac{n+2a-2}{2} \right)\left( \frac{n+2a-4}{2} \right)} = \frac{4(b^*)^2}{(n+2a-2)(n+2a-4)}. \]
\[ \mathrm{Var}(\sigma^2 \mid y) = E[\sigma^4 \mid y] - \left( E[\sigma^2 \mid y] \right)^2. \]
\[ = \frac{4(b^*)^2}{(n+2a-2)(n+2a-4)} - \left( \frac{2b^*}{n+2a-2} \right)^2. \]
\[ = \frac{4(b^*)^2}{(n+2a-2)(n+2a-4)} - \frac{4(b^*)^2}{(n+2a-2)^2}. \]
Factor \(\frac{4(b^*)^2}{n+2a-2}\):
\[ = \frac{4(b^*)^2}{n+2a-2} \left[ \frac{1}{n+2a-4} - \frac{1}{n+2a-2} \right]. \]
Simplify the bracket:
\[ \frac{1}{n+2a-4} - \frac{1}{n+2a-2} = \frac{(n+2a-2) - (n+2a-4)}{(n+2a-4)(n+2a-2)} = \frac{2}{(n+2a-4)(n+2a-2)}. \]
Thus:
\[ \mathrm{Var}(\sigma^2 \mid y) = \frac{4(b^*)^2}{n+2a-2} \cdot \frac{2}{(n+2a-4)(n+2a-2)}. \]
\[ = \frac{8(b^*)^2}{(n+2a-2)^2 (n+2a-4)}. \]
If the problem assumes \(b^* = \frac12\) (a common choice in some conjugate priors), then
\[ 8(b^*)^2 = 8 \times \frac14 = 2. \]
Therefore:
\[ \mathrm{Var}(\sigma^2 \mid y) = \frac{2}{(n+2a-2)^2 (n+2a-4)}. \]
\[ \boxed{E[\sigma^2 \mid y] = \frac{2b^*}{n+2a-2}} \] \[ \boxed{\mathrm{Var}(\sigma^2 \mid y) = \frac{8(b^*)^2}{(n+2a-2)^2 (n+2a-4)}} \]
If \(b^* = 1/2\), then the variance simplifies to \(\frac{2}{(n+2a-2)^2 (n+2a-4)}\).