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Let $X \sim N_p(\mu_1,\Sigma_1)$ and $Y\sim N_p(\mu_2, \Sigma_2)$, where $\Sigma$ denotes the covariance matrix, and assume $X$ and $Y$ are independent.

Show that $X+Y\sim N_p(\mu_1 + \mu_2, \Sigma_1 + \Sigma_2)$


My thoughts: I am not quite sure how to show this... Am I supposed to multiply a vector from the definition of the multivariate normal distribution or should I use the characteristic function? Any help is appreciated!

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2 Answers 2

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Usually such arguments are done by defining $Z=X+Y$ and showing that the characteristic function of $Z$ has a particular shape.

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  • $\begingroup$ Can you elaborate? I'm still not quite sure where to start... $\endgroup$
    – CruZ
    Sep 5, 2019 at 21:55
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Let $Z = X + Y$ then the distribution of $f_Z$ is the convolution of $f_X$ and $f_Y$:

$$f_Z(z) = \int_{-\infty}^{\infty}f_Y(z - x)f_X(x)dx$$

where

$$f_X(x) = \frac{1}{\sqrt{2\pi}\sigma_X}e^{-(x-\mu_X)^2/(2\sigma^2_{X})}$$

and

$$f_Y(y) = \frac{1}{\sqrt{2\pi}\sigma_Y}e^{-(y-\mu_Y)^2/(2\sigma^2_{Y})}$$

Then substituting into the convolution we have

$$f_{Z}(z) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}\sigma_Y}e^{-(z-x-\mu_Y)^2/(2\sigma^2_{Y})} \frac{1}{\sqrt{2\pi}\sigma_X}e^{-(x-\mu_X)^2/(2\sigma^2_{X})} = \ldots $$

$$=\frac{1}{\sqrt{2\pi(\sigma_X^2 + \sigma_Y^2)}}\exp{\left[ \frac{-(z-(\mu_X + \mu_Y)^2)}{2(\sigma_X^2 + \sigma_Y^2)} \right]}$$

Let me know if you can fill in the details or not, but this should help.

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  • $\begingroup$ The question involves the multivariate normal distribution, so this answers only the $p=1$ case. $\endgroup$ May 17, 2021 at 15:54

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