Solution
= Solution
{c}
Use the <gamma density> and the normalization integral for shape $n-1$:
$$
\mathbb E[Y^{-1}]=\frac{\theta^n}{(n-1)!}\int_0^\infty y^{n-2}e^{-\theta y}\,dy
=\frac{\theta^n}{(n-1)!}\frac{(n-2)!}{\theta^{n-1}}
=\boxed{\frac\theta{n-1}}.
$$
The same argument, valid for $n\geq3$, gives $\mathbb E[Y^{-2}]=\theta^2/[(n-1)(n-2)]$.