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Let $X$ and $Y$ be two normal random variables with mean zero, variance 1 and correlation $\rho$. Let $Z=sign(XY)$ with $sign(x)=1$ if $x>0$ and $sign(x)=0$ otherwise.

Now, I calculate $\mathbb{E}[Z]$ as follows \begin{eqnarray}\label{exp} \mathbb{E}\left[Z_{t}\right] &=&\mathbb{P}\left( X\geq 0, Y\geq 0\right)+\mathbb{P}\left( X\leq 0, Y\leq 0\right)\notag\\ &=& 2\mathbb{P}\left( X\geq 0, Y\geq 0\right)\notag\\ &=&2\sqrt{1-\rho^{2}}\frac{1}{2\pi}\int_{0}^{+\infty}\int_{0}^{+\infty} \frac{uv}{|u||v|}e^{-\frac{1}{2}(u^{2}+v^{2}-2uv\rho)}dudv\notag\\ &=&2\sqrt{1-\rho^{2}}\bigg[\frac{1}{\sqrt{2\pi}}\int_{0}^{+\infty}\frac{1}{\sqrt{2\pi}} \int_{0}^{+\infty} e^{-\frac{1}{2}(u-\rho v)^{2}}du e^{-\frac{1}{2}(v^2-\rho^2 v^2)} dv\bigg] \notag\\ &=&2\sqrt{1-\rho^{2}}\bigg[\frac{1}{\sqrt{2\pi}}\int_{0}^{+\infty} \frac{1}{\sqrt{2\pi}}\int_{-v\rho}^{+\infty} e^{-\frac{x^{2}}{2}}dx e^{-\frac{1}{2}(v^2-\rho^2 v^2)} dv\bigg]\notag\\ &=&2\sqrt{1-\rho^{2}}\bigg[\frac{1}{\sqrt{2\pi}}\int_{0}^{+\infty} \Phi(v\rho) e^{-\frac{1}{2}(v^2-\rho^2 v^2)} dv\bigg]\notag\\ &=&2\sqrt{1-\rho^{2}}\bigg[\frac{1}{\sqrt{2\pi}\rho}\int_{0}^{+\infty} \Phi(y) e^{-\frac{y^{2}}{2}(\frac{1-\rho^2}{\rho^2})} dy\bigg]\notag\\ &=&2\sqrt{1-\rho^{2}}\left[\frac{1}{\sqrt{2\pi}\rho} \left[\frac{\rho}{\sqrt{2\pi(1-\rho^{2})}}\arctan\left(\frac{\rho} {\sqrt{1-\rho^{2}}}\right) +\frac{\sqrt{2\pi}\rho}{4\sqrt{1-\rho^{2}}}\right] \right]\notag\\ &=&\frac{1}{\pi}\arctan\left(\frac{\rho} {\sqrt{1-\rho^{2}}}\right)+\frac{1}{2}\,. \end{eqnarray}

However, the solution is $\frac{1}{\pi} \arccos (-\rho)$.

Is any step wrong? Thanks for your kindely helps.

steven
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1 Answers1

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Your answer is correct. You only need some inverse trigonometric formulas to get there, see https://en.wikipedia.org/wiki/Inverse_trigonometric_functions.

In particular, since $\tan(\arcsin(x)) = \frac{x}{\sqrt{1-x^2}}$, $$ \frac{1}{\pi} \arctan \left(\frac{\rho}{\sqrt{1-\rho^2}}\right) + \frac{1}{2} = \frac{1}{\pi}\arcsin(\rho)+\frac{1}{2} =\frac{1}{\pi}\left(\arcsin(\rho)+\frac{\pi}{2}\right), $$ and since $\arcsin(x)=-\arcsin(-x)$ and $\arccos(x)=\pi/2-\arcsin(x)$, we get $$ \frac{1}{\pi} \arctan \left(\frac{\rho}{\sqrt{1-\rho^2}}\right) + \frac{1}{2} = \frac{1}{\pi}\left(\arcsin(\rho)+\frac{\pi}{2}\right) = \frac{1}{\pi}\left(\frac{\pi}{2}-\arcsin(-\rho)\right) = \frac{1}{\pi} \arccos(-\rho). $$

Lanc3
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