Truncated expectation of the bivariate Gaussian

Question

I am searching for the solution to

$$\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{2\pi\sigma_{X}\sigma_{Y}\sqrt{1-\rho^{2}}}\exp\left(-\frac{1}{2(1-\rho^{2})}\left[\frac{(\bar{x}-\mu_{X})^{2}}{\sigma_{X}^{2}}+\frac{(\bar{y}-\mu_{Y})^{2}}{\sigma_{Y}^{2}}-\frac{2\rho(\bar{x}-\mu_{X})(\bar{y}-\mu_{Y})}{\sigma_{X}\sigma_{Y}}\right]\right)\bar{x}~\mathrm{d}\bar{x}~\mathrm{d}\bar{y}$$

Does anyone know how to derive it? I hit only unsolvable integrals!

Answer 1

the integral you showed is not a truncated gaussian but simply the integral of a joint gaussian over the domain $\mathbb{R}^+ \times \mathbb{R}^+$. To solve it you can factorize $f(x,y)=f(x)f(y|x)$ and observing that both $X$ and $Y|X$ are still gaussian

Answer 2

Let's have a joint Gaussian density, say

$$f_{XY}(x,y)=\frac{1}{2\pi \sigma_X \sigma_Y \sqrt{1-\rho^2}}e^{ -\frac{1}{2(1-\rho^2)} [\frac{(x-\mu_X)^2}{\sigma_X^2}-2\rho\frac{(x-\mu_x)(y-\mu_Y)}{\sigma_X\sigma_Y}+\frac{(y-\mu_Y)^2}{\sigma_Y^2}]}=$$

$$=\frac{1}{2\pi \sigma_X \sigma_Y \sqrt{1-\rho^2}}e^{ -\frac{1}{2(1-\rho^2)} \frac{(x-\mu_X)^2}{\sigma_X^2}}\times e^{ -\frac{1}{2(1-\rho^2)} [-2\rho\frac{(x-\mu_x)(y-\mu_Y)}{\sigma_X\sigma_Y}+\frac{(y-\mu_Y)^2}{\sigma_Y^2}]}$$

Now let's sum and subtract the quantity $\frac{\rho^2(x-\mu_X)^2}{\sigma_X^2}$ in the argument of the second integral to get a perfect square. After some simply alebraic passages you get

$$f_{XY}(x,y)=\frac{1}{\sigma_X\sqrt{2\pi}}e^{-\frac{1}{2\sigma_X^2}(x-\mu_X)^2}\times \frac{1}{\sigma_Y\sqrt{2\pi}\sqrt{1-\rho^2}}e^{-\frac{1}{2\sigma_Y^2(1-\rho^2)}[y-\mu_Y-\rho\frac{\sigma_Y}{\sigma_X}(x-\mu_X)]^2}$$

Now your joint density is factorized in the product of two gaussian...and I think you can go on by yourself

Hint:

The problem is concerned only with the part of the integral with $xf(x)$

Let's see the following example (wich is very similar to yours):

Calculate

$$\int_0^{+\infty}xf(x)dx$$

where $X\sim N(1;1)$

The integral is the following

$$\int_0^{+\infty}x\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-1)^2}dx$$

Set $x-1=y$ (say, standardize the variable) and you get

$$\frac{1}{\sqrt{2\pi}}\int_{-1}^{\infty}ye^{-\frac{1}{2}y^2}dy+\int_{-1}^{\infty}\phi(y)dy$$

The first integral is easy to solve analitically. For the second use the gaussian table as it is $1-\Phi(-1)$

Answer 3

I made two attempts. Both were futile:

$$\int_{0}^{\infty}\int_{0}^{\infty}f_{XY}(x,y)~\mathrm{d}y~x~\mathrm{d}x\\=\int_{0}^{\infty}\frac{1}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2\sigma_{X}^{2}}(x-\mu_{X})^{2}}\int_{0}^{\infty}\frac{1}{\sigma_{Y}\sqrt{2\pi}\sqrt{1-\rho^{2}}}e^{-\frac{1}{2\sigma_{Y}^{2}(1-\rho^{2})}[y-\mu_{Y}-\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})]^{2}}~\mathrm{d}y~x~\mathrm{d}x\\=\int_{0}^{\infty}\frac{1}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2\sigma_{X}^{2}}(x-\mu_{X})^{2}}\int_{\mu_{Y}+\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^{2}}~\mathrm{d}z~x~\mathrm{d}x\\=\int_{0}^{\infty}\frac{1}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2\sigma_{X}^{2}}(x-\mu_{X})^{2}}\left(1-\Phi\left(\mu_{Y}+\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})\right)\right)~x~\mathrm{d}x\\=??? $$

and

$$\int_{0}^{\infty}\int_{0}^{\infty}f_{XY}(x,y)~y~\mathrm{d}y~\mathrm{d}x\\ =\int_{0}^{\infty}\frac{1}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2\sigma_{X}^{2}}(x-\mu_{X})^{2}}\int_{0}^{\infty}\frac{1}{\sigma_{Y}\sqrt{2\pi}\sqrt{1-\rho^{2}}}e^{-\frac{1}{2\sigma_{Y}^{2}(1-\rho^{2})}[y-\mu_{Y}-\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})]^{2}}~y~\mathrm{d}y~\mathrm{d}x\\ =\int_{0}^{\infty}\frac{1}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2\sigma_{X}^{2}}(x-\mu_{X})^{2}}\int_{\frac{-\mu_{Y}-\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})}{\sqrt{\sigma_{Y}^{2}(1-\rho^{2})}}}^{\infty}\frac{1}{\sigma_{Y}\sqrt{2\pi}\sqrt{1-\rho^{2}}}e^{-\frac{1}{2}z^{2}}~\left(\sqrt{\sigma_{Y}^{2}(1-\rho^{2})}z+\mu_{Y}+\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})\right)~\sqrt{\sigma_{Y}^{2}(1-\rho^{2})}~\mathrm{d}z~\mathrm{d}x\\ =\int_{0}^{\infty}\frac{1}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2\sigma_{X}^{2}}(x-\mu_{X})^{2}}\int_{\frac{-\mu_{Y}-\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})}{\sqrt{\sigma_{Y}^{2}(1-\rho^{2})}}}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^{2}}~\left(\sqrt{\sigma_{Y}^{2}(1-\rho^{2})}z+\mu_{Y}+\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})\right)~\mathrm{d}z~\mathrm{d}x\\ =\int_{0}^{\infty}\frac{1}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2\sigma_{X}^{2}}(x-\mu_{X})^{2}}\left(\left(\mu_{Y}+\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})\right)\int_{\frac{-\mu_{Y}-\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})}{\sqrt{\sigma_{Y}^{2}(1-\rho^{2})}}}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^{2}}~\mathrm{d}z+\sqrt{\sigma_{Y}^{2}(1-\rho^{2})}\int_{\frac{-\mu_{Y}-\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})}{\sqrt{\sigma_{Y}^{2}(1-\rho^{2})}}}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^{2}}z~\mathrm{d}z\right)~\mathrm{d}x\\ =\int_{0}^{\infty}\frac{1}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2\sigma_{X}^{2}}(x-\mu_{X})^{2}}\left(\left(\mu_{Y}+\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})\right)\left(1-\Phi\left(\frac{-\mu_{Y}-\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})}{\sqrt{\sigma_{Y}^{2}(1-\rho^{2})}}\right)\right)+\sqrt{\sigma_{Y}^{2}(1-\rho^{2})}\left(\frac{-\mu_{Y}-\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})}{\sqrt{\sigma_{Y}^{2}(1-\rho^{2})}}\right)\right)~\mathrm{d}x\\ =-\int_{0}^{\infty}\frac{1}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2\sigma_{X}^{2}}(x-\mu_{X})^{2}}\left(\mu_{Y}+\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})\right)\Phi\left(\frac{-\mu_{Y}-\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})}{\sqrt{\sigma_{Y}^{2}(1-\rho^{2})}}\right)~\mathrm{d}x\\ =\left(\rho\frac{\sigma_{Y}}{\sigma_{X}}\mu_{X}-\mu_{Y}\right)\int_{0}^{\infty}\frac{1}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2\sigma_{X}^{2}}(x-\mu_{X})^{2}}\Phi\left(\frac{-\mu_{Y}-\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})}{\sqrt{\sigma_{Y}^{2}(1-\rho^{2})}}\right)~\mathrm{d}x-\rho\frac{\sigma_{Y}}{\sigma_{X}}\int_{0}^{\infty}\frac{1}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2\sigma_{X}^{2}}(x-\mu_{X})^{2}}x\Phi\left(\frac{-\mu_{Y}-\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})}{\sqrt{\sigma_{Y}^{2}(1-\rho^{2})}}\right)~\mathrm{d}x $$

The second part is solvable but

$$\int_{0}^{\infty}\frac{1}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2\sigma_{X}^{2}}(x-\mu_{X})^{2}}\Phi\left(\frac{-\mu_{Y}-\rho\frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})}{\sqrt{\sigma_{Y}^{2}(1-\rho^{2})}}\right)~\mathrm{d}x$$

is not a solvable intgral.