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Suppose we have a joint normal distribution: $X\sim N(\mu_X,\sigma_X^2)$ and $Y^*\sim N(\mu_{Y^*},\sigma_{Y^*}^2)$, and the covariance $cov(X,Y^*)=\rho\sigma_X\sigma_{Y^*}$. Define $Y\equiv Y^*$ if $Y^*>X$ and 0 if otherwise.

How do I compute $E(Y|Y\neq 0)$?

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1 Answers1

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Let $f(x,y^*)$ denote the joint density of $X$ and $Y^*$. Define $I_{(Y^*>X)}$ to be an indicator random variable such that it equals one when $Y^*>X$ and zero otherwise.

First, we have $$ E(Y\mid Y\neq 0) = E(Y^*\mid Y^*>X)=E(Y^*\mid I_{(Y^*>X)}=1). $$ Second, we have $$ P(I_{(Y^*>X)}=1)=\int_{-\infty}^\infty\int_{y^*>x} f(x,y^*)dy^*dx. $$ and $$ P(Y^*=y^*,I_{(Y^*>X)}=1)=\int_{y^*>x} f(x,y^*)dx. $$ Third, the expectation can be calculated by \begin{equation}\label{eq_1} \frac{\int_{-\infty}^\infty y^*\cdot\int_{y^*>x} f(x,y^*)dx dy^*}{\int_{-\infty}^\infty\int_{y^*>x} f(x,y^*)dxdy^*}. \end{equation}

Next, by the property of conditional normal distribution, $$f(x, y^*) = f(x\mid y^*) \cdot f(y^*),$$ where $f(x\mid y^*)$ is the density of $\mathcal{N}\left(\mu_X+\frac{\sigma_{X}}{\sigma_{Y^*}} \rho\left(y^*-\mu_{Y^*}\right),\left(1-\rho^2\right) \sigma_{X}^2\right)$ and $f(y^*)$ is the density of $Y^*$.

Therefore, $$ \int_{y^*>x} f(x,y^*)dx = \int_{y^*>x} f(x\mid y^*) f(y^*)dx = CDF_{x\mid y^*}(y^*)\cdot f(y^*), $$ where $CDF_{x\mid y^*}$ is the CDF of $\mathcal{N}\left(\mu_X+\frac{\sigma_{X}}{\sigma_{Y^*}} \rho\left(y^*-\mu_{Y^*}\right),\left(1-\rho^2\right) \sigma_{X}^2\right)$.

Finally, $$\frac{\int_{-\infty}^\infty y^*\cdot\int_{y^*>x} f(x,y^*)dx dy^*}{\int_{-\infty}^\infty\int_{y^*>x} f(x,y^*)dxdy^*} = \frac{\int_{-\infty}^\infty y^*\cdot CDF_{x\mid y^*}(y^*)\cdot f(y^*) dy^*}{\int_{-\infty}^\infty CDF_{x\mid y^*}(y^*)\cdot f(y^*)dy^*}, $$ where recall that $f(y^*)$ is the density of $\mathcal{N}\left(\mu_{Y^*}, \sigma_{Y^*}^2\right)$.

augustin4
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  • But how do I compute the integrals? I want to express it using $\mu_X,\mu_{Y^*},\rho,\sigma_X,\sigma_Y$, and the PDF, CDF of standard normal. – Accept Cookies Oct 19 '23 at 13:09
  • @AcceptCookies I have made an attempt and added more details. – augustin4 Oct 19 '23 at 13:42
  • But how to express it in terms of standard normal PDF and CDF? Now we're expressing in terms of the CDF of $X|Y^*$. – Accept Cookies Oct 19 '23 at 20:23
  • @AcceptCookies $CDF_{x \mid y^}\left(y^\right) = \Phi((y^-\mu_{x \mid y^})/\sigma_{x \mid y^})$, where $\Phi$ is the CDF of $\mathcal N(0,1)$, $\mu_{x \mid y^} = \mu_X+\frac{\sigma_X}{\sigma_{Y^}} \rho\left(a-\mu_{Y^}\right)$ and $\sigma_{x \mid y^*} = \sqrt{\left(1-\rho^2\right) }\sigma_X$. – augustin4 Oct 19 '23 at 22:58
  • @AcceptCookies You may also change the variable in the integrals $\int_{-\infty}^{\infty} y^* \cdot C D F_{x \mid y^}\left(y^\right) \cdot f\left(y^\right) d y^$ and $\int_{-\infty}^{\infty} C D F_{x \mid y^}\left(y^\right) \cdot f\left(y^\right) d y^$ by $z=(y^-\mu_{Y^}) / \sigma_{Y^}$ (remember to multiply the Jaccobian $\sigma_{Y^}$), and this will transform $f(y^*)$ to the density of the standard normal distribution. – augustin4 Oct 19 '23 at 23:02
  • what is the $a$ in the CDF of $x|y^*$? – Accept Cookies Oct 25 '23 at 18:39
  • @AcceptCookies Sorry, it is a typo. It should be $y^$ instead (recall that it is the density of $X$ conditional on $Y^=y^*$). – augustin4 Oct 26 '23 at 20:25