# Expectation of truncated normal

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

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

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

This of course is same as saying $$E\left[X\mid a

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.

• You can write the answer in terms of the components of $\Sigma$ alone: it's simple. – whuber Jul 13 '18 at 23:29
• @whuber For the given question, yes we get the answer in terms of the elements of $\Sigma$ only. What I was referring to before the edit was the general formula, which would naturally involve the pdf and cdf of a standard normal variable. – StubbornAtom Jul 14 '18 at 7:05
• thanks for completing the answer, I posted mine below before seeing yours – andy Jul 14 '18 at 10:51

Let's generalize the question so that a key idea can be revealed.

Let $Y$ and $R$ be independent random variables. Define $$X=f(Y)+R$$ for a specified (measurable) function $f$. For any $a\lt b,$ what is the conditional expectation $$E[X\mid a\le Y \le b]?$$

(This states that $f$ is the regression of $X$ on $Y$ with i.i.d. additive errors.)

The independence assumption makes this an easy question to answer, because the random variables $f(Y)$ and $R$ will still be independent, whence (by "taking out what is known")

$$E[X\mid a\le Y \le b] = E[f(Y)+R\mid a\le Y \le b] = E[f(Y)\mid a \le Y \le b] + E[R].\tag{*}$$

This is very general. Let's specialize to the case $E[R]=0$ and $f$ is a linear transformation $$f(y) = \alpha\, y$$ for some constant number $\alpha.$ In this case $(*)$ simplifies to

$$E[X\mid a\le Y \le b] = \alpha\, E[Y\mid a \le Y \le b].\tag{**}$$

The right hand side has a direct expression when the distribution of $Y,$ $F_Y,$ has a density $f_y,$ for then (by the definitions of conditional probability and expectation)

$$E[Y\mid a \le Y \le b] = \frac{1}{F(b)-F(a)}\int_a^b y f_y(y) \mathrm{d}y.$$

The result is now immediate and obvious to anyone who has studied bivariate regression. (A good non-mathematical reference with all the necessary details is Freedman, Pisani, and Purves, Statistics [any edition].) The next section of this post is a review for those who might not have encountered this theory.

In the case of the question, where $(x,y)$ is bivariate Normal with mean $(0,0),$ standard deviations $\sigma_x$ and $\sigma_y,$ and correlation $\rho,$ we know that the distribution of $(x,y)$ is the same as the distribution of $(X,Y)$ constructed as above where $Y$ has a Normal$(0,\sigma_y^2)$ distribution and $R$ independently has a Normal$(0,\tau^2)$ distribution (with $\tau$ to be found). The proof of this lies in the observations that

1. Because $Y$ and $R$ are independent, the variance of $X = \rho\,Y+R$ equals $$\rho^2 \operatorname{Var}(Y) +\operatorname{Var}(R) =\rho^2\sigma^2_y + \tau^2.$$ Consequently, if we set $$\tau^2 = \sigma_x^2 - \rho^2\sigma_y^2,$$ the variance of $X$ will equal the variance of $x.$

2. The covariance $(X,Y)$ is $$\operatorname{Cov}(X,Y) = \operatorname{Cov}(\alpha\,Y+R,Y) = \alpha\operatorname{Var}(Y) = \alpha\,\sigma_y^2.$$ If we set $$\alpha = \rho\frac{\sigma_x}{\sigma_y}$$ then the correlation of $X$ and $Y$ will be $$\rho(X,Y) = \frac{\operatorname{Cov}(X,Y)}{\sigma_x\sigma_y} = \frac{\alpha\,\sigma_y^2}{\sigma_x\sigma_y} = \frac{\rho\frac{\sigma_x}{\sigma_y} \sigma_y^2}{\sigma_x\sigma_y} = \rho.$$

Observations (1) and (2) establish that $(X,Y)$ and $(x,y)$ have the same moments through second order, which means they have identical bivariate Normal distributions.

Applying the main result $(**)$ gives

$$E[X\mid a\le Y \le b] = \alpha\, E[Y\mid a \le Y \le b] = \rho\frac{\sigma_x}{\sigma_y}E[Y\mid a \le Y \le b].$$

We can go a little further towards simplification by recognizing that $Y/\sigma_y$ is a standard Normal variate. Thus,

$$E[X\mid a\le Y \le b] = \rho\,\sigma_x E\left[\frac{Y}{\sigma_y}\mid \frac{a}{\sigma_y} \le \frac{Y}{\sigma_y} \le \frac{b}{\sigma_y} \right].$$

This shows that for fixed $a,b,$ the answer is directly proportional to $\rho\, \sigma_x.$ Evaluating the constant is a matter of manipulating integrals of a standard Normal variable, but doing so is primarily an exercise in Calculus rather than of statistical interest.

I will conclude with the expression that I actually began with when solving this problem. Notice that the answer can also be written $$E\left[\frac{X}{\sigma_x}\mid a\le Y \le b\right] = \rho\, E\left[\frac{Y}{\sigma_y}\mid \frac{a}{\sigma_y} \le \frac{Y}{\sigma_y} \le \frac{b}{\sigma_y} \right].$$

The left hand side is the conditional expectation of a standard Normal variate $X/\sigma_x$ while the right hand side is just $\rho$ times the expectation of a truncated standard Normal variate $Y/\sigma_y.$

If you get used to standardizing variables automatically--which amounts to choosing a particularly convenient unit of measurement for them--then simply by visualizing the regression of $X$ against $Y$ you will say to yourself

oh yes, since after standardization $X$ is just a multiple $\rho$ of $Y$, then the conditional expectation of $X$ must be $\rho$ times the corresponding expectation of $Y.$

That is the heart of the matter.

• This provides good intuition rather than merely calculating the conditional expectation directly as I would have done. – StubbornAtom Jul 14 '18 at 21:33

I am posting the complete answer for reference: $$\begin{eqnarray*} E(X|a where $$\rho_{xy} = E(xy) / (\sigma_x\sigma_y) =\sigma_{xy} / (\sigma_x \sigma_y)$$.

The second-to-last step is derived from the properties of the univariate truncated normal distribution, easily found on its wikipedia page. A reference for the first step is here

With $$a=0,b=\infty$$ we have $$E(X,Y>0) = \rho_{xy}\sigma_x\frac{\phi(0)}{\Phi(0)}= \rho_{xy}\sigma_x\cdot \frac{2}{\sqrt{2\pi}}= \frac{\sigma_{xy}}{\sigma_y}\cdot \frac{2}{\sqrt{2\pi}}$$ Thanks for pointing in the right direction.

• There must be a typographical error because the $\sigma_y/\sigma_y$ term cancels, leaving the answer independent of $\sigma_y$ altogether. – whuber Jul 14 '18 at 14:04
• But $\rho_{xy}$ contains $\sigma_y$ – andy Jul 14 '18 at 15:36
• Don't you have to standardize $a$ and $b$? – Waldir Leoncio Jan 3 '19 at 11:37