1
$\begingroup$

Let $(X,Y)$ have a bivariate normal density centered at the origin with $\mathbb{E}(X^2)=\mathbb{E}(Y^2)=1$ and $\mathbb{E}(XY)=\rho$. In polar coordinates, $(X,Y)$ becomes $(R,\Phi)$ where $R^2=X^2+Y^2$. Then it can be shown that $\Phi$ has a density given by $$f_{\Phi}(\phi)=\frac{\sqrt{1-\rho^2}}{2\pi(1-\rho\sin2\phi)},\quad 0<\phi<2\pi$$

But I am having difficulty proving that this is indeed a density, more specifically the fact that $\displaystyle\int_0^{2\pi}f_{\Phi}(\phi)\,\mathrm{d}\phi=1$. Applying the usual substitution $t=\tan\phi$ in the integral $\displaystyle\int_0^{2\pi}\frac{1}{1-\rho \sin{2 \phi }}\,\mathrm{d}\phi$ seems to be misleading. Should there be any change in the integration limits for proving that this a density. If so, why?

EDIT.

Another related integral is $I=\displaystyle\int_0^{\pi/2}\frac{\mathrm{d}\theta}{1-\rho\sin{2\theta}}=\frac{\pi-\arccos\rho}{\sqrt{1-\rho^2}}$, which arises while calculating $\mathbb{P}(X>0,Y>0)$ for the above distribution once I transform to polar coordinates. The above probability is then given by $\dfrac{\sqrt{1-\rho^2}}{2\pi}I$. Following the suggestions given in the answers below, I was able to evaluate $I$ correctly and hence calculated the probability by direct integration.

$\endgroup$
1
$\begingroup$

Enforcing the substitution $2\phi \to \phi$ and exploiting the $2\pi$-periodicity of the integrand reveals that

$$\begin{align} \int_0^{2\pi}\frac{1}{1-\rho\sin(2\phi)}\,d\phi&=\frac12\int_0^{4\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi\\\\ &=\frac12 \left(\int_0^{2\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi+\int_{2\pi}^{4\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi \right)\\\\ &=\frac12 \left(\int_0^{2\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi+\int_0^{2\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi \right)\\\\ &=\int_0^{2\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi \tag 1 \end{align}$$

Next, we exploit the $2\pi$-periodicity and the oddness of the sine function in $(1)$ to obtain

$$\begin{align} \int_0^{2\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi& =2\int_0^\pi \frac{1}{1-\rho^2\sin^2(\phi)}\,d\phi\\\\ &=4\int_0^\pi \frac{1}{(2-\rho^2)+\rho^2\cos(\phi)}\,d\phi\tag2 \end{align}$$

Applying the tangent half-angle substitution $t=\tan(\phi/2)$ to the integral on the right-hand side of $(2)$ yields

$$\begin{align} 4\int_0^\pi \frac{1}{(2-\rho^2)+\rho^2\cos(\phi)}\,d\phi&=4\int_0^\infty \frac{1}{(2-\rho^2)+\rho^2 \frac{1-t^2}{1+t^2}}\,\frac{2}{1+t^2}\,dt\\\\ &=4\int_0^\infty \frac{1}{1+(1-\rho^2)t^2}\,dt\\\\ &=\frac{2\pi}{\sqrt{1-\rho^2}} \end{align}$$

$\endgroup$
  • $\begingroup$ Could you shed some light on why does using the substitution $\tan \phi=t$ straightaway makes the domain of integration go from $0$ to $0$. Why can't it be applied directly? $\endgroup$ – StubbornAtom Sep 23 '17 at 3:22
  • 1
    $\begingroup$ Sure. Note that the tangent function is $\pi$-periodic. $\endgroup$ – Mark Viola Sep 23 '17 at 3:23
1
$\begingroup$

Let $z=e^{i\phi}$ and then $d\phi=\frac{1}{iz}dz$. So $\sin\phi=\frac{1}{2i}(z-\frac1z)$ \begin{eqnarray} &&\int_0^{2\pi}\frac{1}{1-\rho \sin{2 \phi }}\,\mathrm{d}\phi\\ &=&\int_0^{2\pi}\frac{1}{1-\rho \sin{\phi }}\,\mathrm{d}\phi\\ &=&\int_{|z|=1}\frac{1}{1-\rho \frac{1}{2i}(z-\frac1z)}\,\frac{1}{iz}\mathrm{d}z\\ &=&\int_{|z|=1}\frac{2}{2iz-\rho (z^2-1)}\,\mathrm{d}z\\ &=&-\frac{1}{\rho}\int_{|z|=1}\frac{2}{z^2-\frac{2i}\rho z-1}\,\mathrm{d}z\\ &=&-\frac{2}{\rho}\int_{|z|=1}\frac{1}{(z-z_1)(z-z_2)}\,\mathrm{d}z\\ \end{eqnarray} where $$ z_{1,2}=\frac{i}{\rho}\pm \frac{\sqrt{1-\rho^2}i}{\rho} $$ are the two root of $z^2-\frac{2i}\rho z-1=0$ and only $z_2$ is inside $|z|=1$. So \begin{eqnarray} &&\int_0^{2\pi}\frac{1}{1-\rho \sin{2 \phi }}\,\mathrm{d}\phi\\ &=&-\frac{2}{\rho}\int_{|z|=1}\frac{1}{(z-z_1)(z-z_2)}\,\mathrm{d}z\\ &=&-\frac{2}{\rho}2\pi i\frac{1}{z_2-z_1}\\ &=&\frac{2\pi}{\sqrt{1-\rho^2}}. \end{eqnarray}

$\endgroup$
1
$\begingroup$

\begin{align} t & = \tan\varphi \\[10pt] dt & = \sec^2\varphi\,d\varphi = (1+\tan^2\varphi)\,d\varphi = (1+t^2)\,d\varphi, \\[10pt] \text{so } \frac{dt}{1+t^2} & = d\varphi \\[10pt] \text{and } \sin(2\varphi) & = 2\sin\varphi\cos\varphi = 2\sin(\arctan t)\cos(\arctan t) \\[10pt] & \phantom{{}= 2\sin\varphi\cos\varphi} = 2\frac{t}{\sqrt{1+t^2}} \cdot\frac{1}{\sqrt{1+t^2}} = \frac{2t}{1+t^2} \end{align} Then we have $$ \int_0^{2\pi} \frac{d\varphi}{1 - \rho\sin(2\varphi)} = \left(\int_{-\infty}^\infty + \int_{-\infty}^\infty \right) \frac{\left( \dfrac{dt}{1+t^2} \right)}{1 - \rho\left(\dfrac{2t}{1+t^2}\right)} $$ What do I mean by that?? Simply that as $\varphi$ goes from $0$ to $2\pi,$ then $(\cos(2\varphi),\sin(2\varphi))$ goes around the circle twice, and each revolution causes $t$ to run through the whole real line once. We have a function of period $\pi$ integrated over the interval from $0$ to $2\pi.$ Thus we have $$ 2 \int_{-\infty}^\infty \frac{\left( \dfrac{dt}{1+t^2} \right)}{1 - \rho\left(\dfrac{2t}{1+t^2}\right)} = 2\int_{-\infty}^\infty \frac{dt}{1+t^2 -2\rho t}. $$ Now complete the square: $$ t^2 - 2\rho t + 1 = (t^2 - 2\rho t + \rho^2) + 1 - \rho^2 = (t-\rho)^2 + 1-\rho^2. $$ We would like a quadratic polynomial whose constant term is $1,$ so this becomes \begin{align} & (t-\rho)^2 + 1-\rho^2 = (1-\rho^2)\left( \frac{(t-\rho)^2}{1-\rho^2} + 1 \right) \\[10pt] = {} & (1-\rho^2) \left( \left( \frac{t-\rho}{\sqrt{1-\rho^2}} \right)^2 + 1 \right) \\[10pt] = {} & (1-\rho^2) (u^2 + 1) \\[10pt] \text{so } \sqrt{1-\rho^2}\,\,du & = dt. \end{align} Our integral becomes \begin{align} 2\sqrt{1-\rho^2}\int_{-\infty}^\infty \frac{du}{(1-\rho^2) (u^2+1)} = \frac{2\pi}{\sqrt{1-\rho^2}}. \end{align}

$\endgroup$
  • $\begingroup$ @StubbornAtom : Could be. But now that's what you've got. $\endgroup$ – Michael Hardy Sep 22 '17 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.