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.
At "3 The Product of n Multivariate Gaussian PDFs",They use different notation. dimension $n$ -> $d$, and $n=2$ because product of 2 Gaussians.

  • $\begingroup$ 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? $\endgroup$ – Xiaomi Oct 17 '18 at 10:56
  • $\begingroup$ 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. $\endgroup$ – JimSD Oct 17 '18 at 13:19
  • $\begingroup$ 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. $\endgroup$ – Kori K Oct 17 '18 at 13:59
  • $\begingroup$ I think I understand what you are trying to get at. My initial explanation was not correct or confusing, so I edited. Thank you. $\endgroup$ – JimSD Oct 18 '18 at 0:12

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