# Expectation of reciprocal of $(1-r^{2})$

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.

• Please add the self-study tag to homework or self-study questions (even though it's obvious from the picture that it is!) – jbowman Dec 29 '18 at 18:12
• You can at least try using this theorem. – StubbornAtom Dec 29 '18 at 20:04
• I tried but don't know the PDF of $r^{2}$. – user46697 Dec 30 '18 at 3:57
• Hint: the law of the unconscious statistician could come in handy here. – StatsStudent Dec 30 '18 at 4:09
• @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. – StubbornAtom Dec 30 '18 at 6:28

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.