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At the end of section 2 of this paper, the authors mention the integral:

$$\frac{\alpha}{Z}\int N(x_i | \mu, \sigma I) N(\mu | 0, \rho I) d \mu$$

(Note: $x_i$ and $\mu$ are $d$-dimensional.)

In the first part of section 3.1, they claim that "by a straightforward calculation," that integral evaluates to

$$\frac{\alpha}{Z} (2\pi(\rho + \sigma))^{-d/2}\exp\biggr(-\frac{1}{2(\rho + \sigma)}||x_i||^2\biggr).$$

I have tried (a few times) to obtain this result by actually evaluating the integral and completing the square by hand, but I've obtained

$$\frac{\alpha}{Z} \biggr(2\pi\frac{\sigma \rho}{\rho + \sigma} \biggr)^{d/2}\exp\biggr(-\frac{1}{2(\rho + \sigma)}||x_i||^2\biggr),$$

and I've not been able to find my mistake. I trust the paper over myself, so I must have done something incorrectly. Could someone show how to properly evaluate the integral step-by-step to obtain the paper's solution?

jeg
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  • The paper appears to integrate w.r.t a probability distribution. https://math.stackexchange.com/questions/380785/what-does-it-mean-to-integrate-with-respect-to-the-distribution-function talks about such integrals. There's definitely some mistake in your solution, since the normal distribution has a term of $(2\pi)^{-\frac d2}$ which appears in the paper's solution and not in yours. – Wolfuryo Mar 21 '24 at 01:08

1 Answers1

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Let $y_i=x_i-\mu$ then $y_i \sim N(0,\sigma I)$. The distribution of $x_i$ is the same as the distribution of $y_i+\mu$ where $y_i$ and $u$ are independent. The distribution of the sum of two independent normal random variables is a normal distribution. We also have that $\sigma^2(X)+\sigma^2(Y)=\sigma^2(X+Y)$ for two independent random variables $X$, $Y$. Therefore $x_i = y_i+\mu \sim N(0,(\sigma +\rho) I)$, hence the formula. That's why it's straightforward

ioveri
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  • Thank you! So to make your comment more explicit: You want to take the original integral and rewrite it as $$\int N(x_i - \mu | 0, \sigma I) N(\mu | 0, \rho I) d\mu.$$ This now looks like the convolution of two normals, which is how we'd write the joint distribution of their sum. Because we know that the mean and variance of the sum are the sums of the individual means and variances, in our case, we have a mean of $0 + 0$ and a variance of $\rho + \sigma$. The conclusion in the paper follows.

    Is this correct?

     – jeg Mar 21 '24 at 03:26
  • Yes, it would be clearer that way. – ioveri Mar 21 '24 at 10:01
  • And, for completeness, I realized I didn't include the fact that the convolution is how we'd write the joint distribution of the sum of independent RVs. – jeg Mar 21 '24 at 10:55