# In a bivariate normal sample, why is the squared sample correlation Beta distributed?

If $$(X_i,Y_i), i = 1, \dots, n$$ are independently bivariate normal distributed, with mean $$(\mu_x , \mu_y)$$ and variances $$(\sigma_x^2, \sigma^2_y)$$ and correlation coefficient $$\rho = 0$$.

Denote $$T = \frac{\sum_i (X_i - \mu_x)(Y_i - \mu_y)} {\sqrt{\sum_i (X_i - \mu_x)^2 \sum_j (Y_j - \mu_y)^2}}$$ the sample correlation. How can I show $$T^2$$ is distributed as a Beta distribution?

I know $$(X_i - \mu_x)^2$$ is a $$\chi^2_1$$, but I can't figure out what are the relations between distributions and algebra manipulation to show it is beta distributed.

• We know that density of correlation coefficient $R$ when $\rho=0$ is $$f_R(r)\propto (1-r^2)^{(n-4)/2}\mathbf1_{|r|<1}$$ By a change of variables $Y=R^2$, it follows that $Y\sim \text{Beta}(\frac{1}{2},\frac{n}{2}-1)$. – StubbornAtom Feb 10 at 20:52