This is Lemma 11.3 from the book the theory of linear models and multivariate analysis , which says Let $U\sim N_p(\theta,I)$ and K have a Poission distribution with mean $||\theta||^2/2$, then $E\frac{1}{||U||^2}=E\frac{1}{p-2+2K}$.
The proof is like: Let $V|K\sim \chi^2_{p+2K}(0) $, then by Lemma 1.8 of the book, $V\sim \chi_p^2(||\theta||).$ Also, by the defintion of a $\chi^2$ distribution, $||U||\sim \chi_p^2(||\theta||^2).$ Thereore $$E\frac{1}{||U||^2}=E\frac{1}{V}=E(E\frac{1}{V}|K)=E\frac{1}{p-2+2K}$$
However, I couldn't figure out why $E(\frac{1}{V}|K)=\frac{1}{p-2+2K}$, Can someone give me some help?
The question is to understand why, if the distribution of $V_n$ is central chi-squared with $n\gt2$ degrees of freedom, then $E[V_n^{-1}]=(n-2)^{-1}$.
One can go back to the density $f_n$ of $V_n$, which is defined by $f_n(x)=c_n^{-1}x^{(n/2)-1}\mathrm e^{-x/2}$ with $c_n=2^{n/2}\Gamma(n/2)$ and note that $$ E[V_n^{-1}]=\int_0^\infty x^{-1}f_n(x)\mathrm dx=\int_0^\infty c_n^{-1}x^{(n/2)-2}\mathrm e^{-x/2}\mathrm dx. $$ One recognizes in the last integrand a multiple of $f_{n-2}(x)$, hence $$ E[V_n^{-1}]=c_n^{-1}c_{n-2}=\frac{\Gamma((n/2)-1)}{2\Gamma(n/2)}=\frac1{n-2}. $$ Another approach is to rely on the MGF of $V_n$, which is such that $$ E[\mathrm e^{-tV_n}]=(1+2t)^{-n/2}, $$ at least for every $t\gt0$, and to use the identity $$ E[V_n^{-1}]=\int_0^\infty E[\mathrm e^{-tV_n}]\mathrm dt, $$ hence $$ E[V_n^{-1}]=\int_0^\infty \frac{\mathrm dt}{(1+2t)^{n/2}}=\left.\frac{-1}{(n-2)(1+2t)^{n/2-1}}\right|_0^\infty=\frac1{n-2}. $$