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Suppose that $\mathbf{y} \sim N(\mathbf{n},\sigma^2\mathbf{I})$ and $\mathbf{n} \sim N(\boldsymbol{\mu},\boldsymbol{\Sigma})$. I want to integrate the following:

$$\int [\mathbf{y}\mid\mathbf{n},\sigma^2][\mathbf{n} \mid \boldsymbol{\mu},\boldsymbol{\Sigma}] \, d\mathbf{n},$$ where $[\cdot]$ indicates a probability distribution.

I take it that $\mathbf{y}$ and $\mathbf{n}$ have multivariate normal distributions, so I really want to solve the integral:

$$\int (2\pi)^{-\frac{1}{2}(k+k)} \det(\sigma^2\mathbf{I})^{-1/2} \det(\boldsymbol{\Sigma})^{-1/2} \exp\left[-\frac{1}{2}((\mathbf{y}-\mathbf{n})^T (\sigma^2\mathbf{I})^{-1}(\mathbf{y}-\mathbf{n}) + (\mathbf{n}-\boldsymbol{\mu})^T(\boldsymbol{\Sigma})^{-1}(\mathbf{n}-\boldsymbol{\mu})\right] \, d\mathbf{n},$$

where $k$ is the dimension of y and of n.

However, I don't know what to do next. Is there some standard trick that mathematicians use to complete this integration? I take it that the result is a multivariate normal. Thank you.

Ron Snow
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    At this point where you say "a probability distribution" you probably mean a probability density function. – Michael Hardy Jun 18 '20 at 18:47
  • @MichaelHardy That is correct. How would you attempt getting the resulting multivariate normal distribution from this integral? – Ron Snow Jun 18 '20 at 18:51
  • @HansEngler You are correct. My apologies. $\mathbf{n}$ and $\mathbf{y}$ have the same dimensions – Ron Snow Jun 18 '20 at 19:04
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    Write the expression in the [ ] brackets in the form $(\mathbf{n} - \mathbf{c})^TB^{-1}(\mathbf{n} - \mathbf{c}) + \gamma$ for a suitable matrix $B$, suitable $\mathbf{c}$, and suitable $\gamma$. This is essentially an exercise in completing the square. This is now again a multiple of the density of a normal distribution, and you can integrate. – Hans Engler Jun 18 '20 at 19:10
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    https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Cumulative_distribution_function –  Jun 18 '20 at 19:19
  • @HansEngler I understand the idea, but I'm failing to see how to put that into practice when I have a sum of two of those terms. Additionally, I must admit that I particularly don't know how to combine the two inverted matrices in this exponential term. – Ron Snow Jun 18 '20 at 19:20
  • @Renard I actually noticed the similar notation under the "Interval" subsection on that page. However, the fact that there are two exponential terms make the solution unclear to me. – Ron Snow Jun 18 '20 at 19:22
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    $$ \begin{align} & \text{Instead of } \mathbf{y} \sim N(\mathbf{n}, \sigma^2\mathbf{I}), \text{ I'd write }\mathbf{y} \mid \mathbf n \sim N(\mathbf{\mu},\sigma^2\mathbf{I}). \ & \text{From the above it follows that }\mathbf y- \mathbf n \mid N \sim N(\mathbf 0, \sigma^2 \mathbf I). \ {} \ & \text{Since the conditional distribution of } \mathbf y - \mathbf n\text{ given } \mathbf n \text{ is} \ & \text{thus seen not to depend on } \mathbf n, \text{ we can draw two conclusions:} \end{align} $$ – Michael Hardy Jun 18 '20 at 21:46
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    $$ \begin{align} & \bullet\quad \mathbf y - \mathbf n \text{ and } \mathbf n \text{ are independent; and} \ \ & \bullet\quad \text{The marginal (or “unconditional'', if you like) distribution} \ & \phantom{\bullet\quad\text{o}} \text{of } \mathbf y - \mathbf n \text{ is the same as this given conditional distribution.} \end{align} $$ – Michael Hardy Jun 18 '20 at 21:47
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    Therefore $\mathbf y$ is the sum of two independent normally distributed random vectors that have these two specified distributions. Hence the sum, $\mathbf y,$ is normally distributed with mean and variance equal respectively to the sum of the means and the sum of the variances: $$ \mathbf y \sim N(\mathbf \mu, \sigma^2\mathbf I + \mathbf\Sigma). $$ – Michael Hardy Jun 18 '20 at 21:47
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    @HansEngler : The poster's statement at the beginning that $\mathbf{y} \sim N(\mathbf{n},\sigma^2\mathbf{I})$ leaves no room for $\mathbf y$ and $\mathbf n$ to fail to have the same dimensions. $\qquad$ – Michael Hardy Jun 18 '20 at 21:49

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