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find expectation

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.

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    $\begingroup$ Please add the self-study tag to homework or self-study questions (even though it's obvious from the picture that it is!) $\endgroup$ – jbowman Dec 29 '18 at 18:12
  • $\begingroup$ You can at least try using this theorem. $\endgroup$ – StubbornAtom Dec 29 '18 at 20:04
  • $\begingroup$ I tried but don't know the PDF of $r^{2}$. $\endgroup$ – Harry Dec 30 '18 at 3:57
  • $\begingroup$ Hint: the law of the unconscious statistician could come in handy here. $\endgroup$ – StatsStudent Dec 30 '18 at 4:09
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    $\begingroup$ @user46697 If you had a look at the wiki page I linked to, you will see that you do not require the distribution of $r^2$ or even $(1-r^2)^{-1}$ to calculate $E(1-r^2)^{-1}$; the pdf of $r$ suffices. $\endgroup$ – StubbornAtom Dec 30 '18 at 6:28
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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.

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