Expectation of reciprocal of $(1-r^{2})$ 
If $r$ is the coefficient of correlation for a sample of $N$ independent observations from a bivariate normal population with population coefficietn of correlation zero, then $E(1-r^2)^{-1}$ is
  (a) $\quad\frac{(N-3)}{(N-4)}$

I tried finding expectation from the density function but then realised that I was solving with the density function of $r$ and not it's square. I don't know the density function of $r^{2}$. I am stuck. Kindly help.
 A: From the problem statement, you are given that a sample of $N$ observations are made from a bivariate normal population with correlation coefficient equal to zero.  Under these assumptions, the probability density function (PDF) for $r$ simplifies greatly to: 
\begin{eqnarray*}
f_{R}(r) & = & \frac{\left(1-r^{2}\right)^{\frac{N-4}{2}}}{\text{Beta}\left(\frac{1}{2},\,\frac{N-2}{2}\right)}
\end{eqnarray*}
over the support from $-1 < r < 1$ and zero otherwise (see here).
Now, since we know the PDF of $r$, we can readily find the expected value of the transformation $g(r)=\left(1-r^{2}\right)^{-1}$ by applying the Law of the Unconscious Statistician.  This yields:
\begin{align}
E[(1-r^2)^{-1}] & =  \intop_{-\infty}^{\infty}g(r)f_{R}(r)dr \\
 & =  \intop_{-1}^{1}\frac{\left(1-r^{2}\right)^{-1}\left(1-r^{2}\right)^{\frac{N-4}{2}}}{\text{Beta}\left(\frac{1}{2},\,\frac{N-2}{2}\right)}dr\\
 & =  \frac{1}{\text{Beta}\left(\frac{1}{2},\,\frac{N-2}{2}\right)}\intop_{-1}^{1}\left(1-r^{2}\right)^{\frac{N-4}{2}-\frac{2}{2}}dr\\
 & =  \frac{1}{\text{Beta}\left(\frac{1}{2},\,\frac{N-2}{2}\right)}\intop_{-1}^{1}\left(1-r^{2}\right)^{\frac{N-6}{2}}dr \\
 & =  \frac{1}{\text{Beta}\left(\frac{1}{2},\,\frac{M}{2}\right)}\intop_{-1}^{1}\left(1-r^{2}\right)^{\frac{M-4}{2}}dr&&\mbox{(Letting $M=N-2$)}
\end{align}

You should note that the last integrand looks similar to the PDF of $r$, $f_R(r)$, and is simply missing a normalizing constant $1/\text{Beta}\left(\frac{1}{2},\,\frac{M-2}{2}\right)$.  Multiplying the top and bottom by $\text{Beta}\left(\frac{1}{2},\,\frac{M-2}{2}\right)$ (which is simply equal to 1 and does not change the last line) and rearranging terms gives:
\begin{eqnarray*}
\frac{\text{Beta}\left(\frac{1}{2},\,\frac{M-2}{2}\right)}{\text{Beta}\left(\frac{1}{2},\,\frac{M}{2}\right)}\intop_{-1}^{1}\underbrace{\frac{\left(1-r^{2}\right)^{\frac{M-4}{2}}}{\text{Beta}\left(\frac{1}{2},\,\frac{M-2}{2}\right)}}_{\text{PDF of $f_R(r)$ integrates to 1}}dr & = & \frac{\text{Beta}\left(\frac{1}{2},\,\frac{M-2}{2}\right)}{\text{Beta}\left(\frac{1}{2},\,\frac{M}{2}\right)}\\
\end{eqnarray*}.
Now, we can simply expand out this last term by definition of the Beta function to yield:
\begin{eqnarray*}
\frac{{{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{M-2}{2}\right)}}}{{{\Gamma\left(\frac{1}{2}+\frac{M-2}{2}\right)}}}\frac{{{\Gamma\left(\frac{1}{2}+\frac{M}{2}\right)}}}{{{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{M}{2}\right)}}} & = & \frac{{{\Gamma\left(\frac{M-2}{2}\right)\Gamma\left(\frac{1}{2}+\frac{M}{2}\right)}}}{{{\Gamma\left(\frac{1}{2}+\frac{M-2}{2}\right)}}\Gamma\left(\frac{M}{2}\right)}
\end{eqnarray*}

To simplify further, we must make use of an identity of the Gamma function.  The recursion identity of the Gamma function states that for $\alpha \gt 0$, $\Gamma(a+1)=a\Gamma(a)$. Since $M > 0$, we can apply this recursion identity to the $\Gamma\left(\frac{1}{2}+\frac{M}{2}\right)$ term in the numerator and the $\Gamma\left(\frac{M}{2}\right)$ term in the denominator to get:

\begin{eqnarray*}
\frac{{{\Gamma\left(\frac{M-2}{2}\right)\left(\frac{1}{2}+\frac{M}{2}-1\right)\Gamma\left(\frac{1}{2}+\frac{M}{2}-1\right)}}}{{{\Gamma\left(\frac{M-1}{2}\right)}}\left(\frac{M}{2}-1\right)\Gamma\left(\frac{M}{2}-1\right)} & = & \frac{{{\Gamma\left(\frac{M-2}{2}\right)\left(\frac{M-1}{2}\right)\Gamma\left(\frac{M-1}{2}\right)}}}{{{\Gamma\left(\frac{M-1}{2}\right)}}\left(\frac{M-2}{2}\right)\Gamma\left(\frac{M-2}{2}\right)}\\
 & = & \frac{{{\frac{M-1}{2}}}}{\frac{M-2}{2}}\\
 & = & {{\left(\frac{M-1}{2}\right)}}\left(\frac{2}{M-2}\right)\\
 & = & \frac{M-1}{M-2}
\end{eqnarray*}

Finally, if we replace $M$ with $N-2$ from our original definition of $M$, we obtain:
\begin{eqnarray*}
\frac{(N-2)-1}{(N-2)-2} & = & \frac{N-2-1}{N-2-2}\\
 & = & \frac{N-3}{N-4}
\end{eqnarray*}
and this completes the proof.
