Normal Distribution – Understanding Expectation of Truncated Normal

Question

Suppose I have a bivariate normal $(x,y)\sim \mathcal N(0,\Sigma)$ .

Is there an easy formula for the expectation $\mathrm E(x\mid y>0)$ ?

Answer 1

Let $\phi(\cdot)$ and $\Phi(\cdot)$ be the PDF and CDF of standard normal distribution, as usual.

Suppose $\Sigma=\left[\begin{matrix}\sigma_1^2&\rho \sigma_1\sigma_2\\\rho\sigma_1\sigma_2&\sigma_2^2\end{matrix}\right]$ is the dispersion matrix of $(X,Y)$.

By definition,

\begin{align} E\left[X\mid a<Y<b\right]&=\frac{E\left[XI(a<Y<b)\right]}{P(a<Y<b)} \\&=\frac1{P(a<Y<b)}\iint_{\mathbb R^2} x \mathbf1_{a<y<b}f_{X,Y}(x,y)\,\mathrm{d}x\,\mathrm{d}y \\&=\frac1{P(a<Y<b)}\int_a^b\int_{-\infty}^{\infty} xf_{X,Y}(x,y)\,\mathrm{d}x\,\mathrm{d}y \\&=\frac{1}{P(a<Y<b)}\int_a^b \left\{\int_{-\infty}^{\infty} xf_{X\mid Y}(x\mid y)\,\mathrm{d}x \right\}f_Y(y)\,\mathrm{d}y \\&=\frac{1}{P(a<Y<b)}\int_a^b E\left[X\mid Y=y\right]f_Y(y)\,\mathrm{d}y \end{align}

This of course is same as saying $$E\left[X\mid a<Y<b\right]=\int_a^b E\left[X\mid Y=y\right]f_{Y\mid a<Y<b}(y)\,\mathrm{d}y$$

Now $X$ conditioned on $Y=y$ has a $N\left(\frac{\rho\,\sigma_1}{\sigma_2}y,(1-\rho^2)\sigma_1^2\right)$ distribution, so that

$$E\left[X\mid Y=y\right]=\frac{\rho\,\sigma_1}{\sigma_2}y$$

In this case, we thus have

\begin{align} E\left[X\mid Y>0\right]&=\frac{\rho\,\sigma_1}{\sigma_2}\int_0^\infty \frac{yf_Y(y)}{P(Y>0)}\,\mathrm{d}y \\&=\frac{\rho\,\sigma_1}{\sigma_2}E\left[Y\mid Y>0\right] \end{align}

Finally,

\begin{align} E\left[Y\mid Y>0\right]&=\frac{1}{P(Y>0)}\int_0^\infty \frac{y}{\sigma_2}\phi\left(\frac{y}{\sigma_2}\right)\mathrm{d}y \\&=2\sigma_2\int_0^\infty t\phi(t)\,\mathrm{d}t \\&=2\sigma_2\int_0^\infty (-\phi'(t))\,\mathrm{d}t \\&=2\sigma_2\,\phi(0) \\&=\sqrt{\frac{2}{\pi}}\sigma_2 \end{align}

Hence for $(X,Y)\sim N(\mathbf 0,\Sigma)$, we have the simple formula

$$\boxed{E\left[X\mid Y>0\right]=\sqrt{\frac{2}{\pi}}\rho\,\sigma_1}$$

A general formula where the means of $X$ and $Y$ are not zero, and $Y$ ranging from some $a$ to $b$ can also be found in a similar manner. That formula would involve $\phi$ and $\Phi$, whereas for the current one, we get the values at $\phi$ and $\Phi$ directly.