Gaussian estimate conditioned to stay positive

Question

Given a centered gaussian variable $X$ with variance $\sigma^2$ it is very easy to compute the conditional expectation $$\mathbb {E}[X|X <0]=-\sqrt {2/\pi}\sigma. $$ Now I want to generalize this result for a 2-dimensional centered gaussian vector $(X,Y)$ with a fixed covariance covariance Matrix such that $\mathbb {E}[X^2]=\mathbb {E}[Y^2]=\sigma>0$ and $\mathbb {E}[XY]=c>0$. Can I compute or give an estimate of $$\mathbb {E}[X|X <0,Y<0]? $$

Below you will find a very nice solution of this problem. Still open the problem to generalize the idea for a $n-$dimensional centered gaussian vector, that is, how to compute or estimate $$\mathbb {E}[X_1|X_1<0,X_2<0,\dots,X_n<0]?$$

There should be a negative sign before $\sqrt{2/\pi}\sigma$. — Dhruv Kohli, Jun 19 '17 at 08:59

Dhruv Kohli · Answer 1 · 2017-06-19T08:45:26.873

1

Suppose $f_{X,Y}(x,y; \sigma,c)$ is the bivariate normal pdf with mean $(0,0)$ and $c = \frac{\rho}{\sigma^2}$ and $f_{X}(x;\sigma)$ and $f_{Y}(y;\sigma)$ are the corresponding marginal normal pdf's with mean $0$.

$$\begin{align}\color{red}{E(X,X<0|Y<0)} &= \int\limits_{-\infty}^{0}xf_{X|Y}(x|y<0)dx \\\\ &= \int\limits_{-\infty}^{0}x \frac{f_{X,Y}(x,y<0)}{f_{Y}(y<0)}dx \\\\ &= \int\limits_{-\infty}^{0} x \frac{\int\limits_{-\infty}^{0}f_{X,Y}(x,y)dy}{\int\limits_{-\infty}^{0}f_{Y}(y)dy}dx \\\\ &= \int\limits_{-\infty}^{0}x\frac{\int\limits_{-\infty}^{0}f_{X,Y}(x,y)dy}{0.5}dx \\\\ &= \int\limits_{-\infty}^{0}2xf_{X}(x)\Phi(\frac{-\rho x}{\sqrt{1-\rho^2}})dx\end{align}$$

where $\Phi$ is the normal cdf with parameters, mean $ = 0$ and variance $=\sigma^2$.

Now, based on this answer, let the RHS be $h(\rho)$. Then using Leibniz rule, we get,

$$\begin{align}\frac{\partial h(\rho)}{\partial \rho} &= \int\limits_{-\infty}^{0}2xf_{X}(x)f_{X}(\frac{-\rho x}{\sqrt{1-\rho^2}})\left(\frac{-x}{(1-\rho^2)^{3/2}}\right)dx \\\\ &= \frac{-2}{2\pi\sigma^2(1-\rho)^{3/2}} \int\limits_{-\infty}^{0} x^2\exp\left(\frac{-x^2}{2\sigma^2(1-\rho^2)}\right)dx \\\\ &= \frac{-\sigma}{\sqrt{2\pi}}\end{align}$$

So, $h(\rho) = \frac{-\sigma\rho}{\sqrt{2\pi}} + K$, where $h(0) = \frac{-\sigma}{\sqrt{2\pi}} \implies K = \frac{-\sigma}{\sqrt{2\pi}}$.

$$E(X,X<0|Y<0) = h(\rho) = \frac{-\sigma(\rho+1)}{\sqrt{2\pi}}$$

$$E(X,X<0|Y<0) = E(X\mathbb{I}(X<0)|Y<0) = \frac{E(X\mathbb{I}(X<0)\mathbb{I}(Y<0))}{P(Y<0)}$$

$$E(X|X<0,Y<0) = \frac{E(X\mathbb{I}(X<0)\mathbb{I}(Y<0))}{P(X<0,Y<0)} = \frac{E(X,X<0|Y<0)P(Y<0)}{P(X<0,Y<0)}$$

$$\implies E(X|X<0,Y<0) = \frac{h(\rho)\frac{1}{2}}{\Phi_{X,Y}(0,0)}$$

where $\Phi_{X,Y}$ is the bivariate normal cdf with mean $(0,0)$, variance $\sigma_X^2 = \sigma_Y^2 = \sigma^2$ and covariance $c = \rho \sigma^2$.

Have a look at this answer to get the value of $\Phi_{X,Y}(0,0)$.

$$\Phi_{X,Y}(0,0) = \frac{1}{2\pi}\arcsin\rho + \frac{1}{4}$$

Finally,

$$E(X|X<0,Y<0) = \frac{\frac{-\sigma(\rho+1)}{\sqrt{2\pi}}}{\frac{1}{\pi}\arcsin\rho + \frac{1}{2}}$$

edited Jun 19 '17 at 08:45

answered Jun 18 '17 at 13:42

Dhruv Kohli

5,216

Is it possible to generalize your argument? That is,for example compute $\mathbb{E}[X|X<0,Y<0,Z<0]$ for a centered gaussian vector $(X,Y,Z)$ with $\mathbb{E}[X^2]=\mathbb{E}[Y^2]=\mathbb{E}[Z^2]=\sigma>0?$ – foubw Jun 18 '17 at 15:18
I don't understand. Can you please elaborate. – Dhruv Kohli Jun 18 '17 at 15:19
From your computations, it seems that the result does not depend on the covariances between $X,Y;Z$. – foubw Jun 18 '17 at 15:23
The result of the original problem does depend on the correlation of $X$ and $Y$ which is $\rho$. – Dhruv Kohli Jun 18 '17 at 15:26
Sure..sorry:):) – foubw Jun 18 '17 at 15:29
From your computations we have the following result: $E(X|X<0,Y<0)>E(X|X<0)$ or similarly $E(X|X>0,Y>0) <E(X|X>0)$. Is that a natural fact? – foubw Jun 18 '17 at 15:38
No. It should be $E(X|X<0,Y<0) \leq E(X|X<0)$ because $\Phi(\frac{\rho x}{\sqrt{1-\rho^2}}) \leq 0.5, , , \forall x \leq 0$ – Dhruv Kohli Jun 18 '17 at 15:43
ok but we have the opposite inequality for $2xf_{X}(x)\Phi(\frac{\rho x}{\sqrt{1-\rho^2}})\geq2xf_{X}(x)\cdot 0.5$ if $x\leq0.$ – foubw Jun 18 '17 at 15:51
Oh yes you are right. I forgot to change the inequality when $x\leq 0 $. So the inequalities you suggested are correct. – Dhruv Kohli Jun 18 '17 at 16:03
It's amazing this result – foubw Jun 18 '17 at 16:03
Please read about condition expectation. – Dhruv Kohli Jun 18 '17 at 16:06
In the first line you are computing: $\int\limits_{-\infty}^{0}xf_{X|Y}(x|y<0)dx=E(X1_{{x<0}}|Y<0)$ and not $E(X|X<0,Y<0).$ Am I worong? – foubw Jun 18 '17 at 17:31
$E(X|X<0,Y<0) = E(X\mathbb{I}(X<0)|Y<0) = \int\limits_{-\infty}^{0}xf_{X|Y}(x|y<0)dx$ – Dhruv Kohli Jun 19 '17 at 03:55
How do you obtain the first equality? You can't split the intersection on the conditionig event, which property are your applaying? – foubw Jun 19 '17 at 07:02
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$E(X,X<0|Y<0)=\frac{E(X\mathbb{I}(X<0)\mathbb{I}(Y<0))}{P(Y<0)} \neq \frac{E(X\mathbb{I}(X<0)\mathbb{I}(Y<0))}{P(X<0,Y<0)} =E(X|X<0,Y<0)$ – foubw Jun 19 '17 at 07:13
Please check the edit. – Dhruv Kohli Jun 19 '17 at 08:45

Gaussian estimate conditioned to stay positive

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