# How to calculate Expectation of Multivariate Gaussian with x drawn from another Gaussian distribution?

Say, there is a $$n$$-dimension multivariate Gaussian, $$g(x) = N(x:\mu, \Sigma)$$
where $$\mu$$ is $$n$$-dim mean vector, and $$\Sigma$$ is $$n \times n$$-dim covariance matrix.

I would like to calculate "expectation $$\mathbb{E}$$ "of this gaussian $$g(x)$$ with the weight of another multivariate gaussian distribution defined by $$n$$-dim mean vector $$\mu_e$$ and $$n \times n$$-dim covariance matrix $$\Sigma_e$$ as below.

$$\displaystyle \mathop{\mathbb{E}}_{x\sim N(\mu_e,\Sigma_e)}N(x:\mu,\Sigma)= ?$$

Does anyone know how to calculate this expectation?
Or at least approximated computation? like summation of infinitesimal divisions?

The reason I need to calculate this is because originally I wanted to calculate likelihood of particular value $$x_0$$ for gaussian distribution $$g(x)$$ like a likelihood of weight $$x_0$$ given a weight distribution of whole students. But then I needed to do that for the case where there is an "observation error" in $$x_0$$ with old weighting machine with variance $$\Sigma_s$$, so $$x_0$$ is no longer constant but random variable from other gaussian distribution $$x\sim N(\mu_e,\Sigma_e)$$.

Thank you

---------------EDITED ---------

After the comments, I found out $$\displaystyle \mathop{\mathbb{E}}_{x\sim N(\mu_e,\Sigma_e)}N(x:\mu,\Sigma)= \int_{-\infty }^{\infty} g(x)\cdot f(x) dx$$ where $$g(x)= N(\mu,\Sigma)$$ and $$f(x)= N(\mu_e,\Sigma_e)$$.
This is a product of two multivariate Gaussians and I can imagine how to solve this in case of 1 dimensional Gaussian $$n=1$$, but I am not sure multivariate Gaussian when ,say, $$n=100$$ to compute this integral even with approximate form.. If anyone knows, your advice is appreciated. Thanks

--------------- EDITED ------------

I think I found answer below.
http://www.tina-vision.net/docs/memos/2003-003.pdf
At "3 The Product of n Multivariate Gaussian PDFs",They use different notation. dimension $$n$$ -> $$d$$, and $$n=2$$ because product of 2 Gaussians.

• The way you’ve written this doesn’t make sense. Do you mean the mean $\mu$ of $x$ has a second multivariate normal distribution? I don’t understand it the way you’ve phrased this. Once you’ve taken expectation of $x$ it’s no longer a random variable so how can it be normal? – Xiaomi Oct 17 '18 at 10:56
• Yes, once you take expectation of $x$ of a certain distribution , it is constant. But what I am asking above is the way you take expectation. Usually expectation means "mean", which is you sample $x$ from "uniform distribution". Here what I am saying is you sample not uniformly but by Gaussian distribution to get expectation. – JimSD Oct 17 '18 at 13:19
• Expectation does not generally mean you sample $x$ from a "uniform distribution". It's more like a weighted mean, where the weights come from the density function of the random variable (if it exists). For example, $x \sim N(\mu, 1)$, then $E (x) = \int x ~ f(x) dx$, where $f(x)$ is the probability density function of a Gaussian. – Kori K Oct 17 '18 at 13:59
• I think I understand what you are trying to get at. My initial explanation was not correct or confusing, so I edited. Thank you. – JimSD Oct 18 '18 at 0:12