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

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!

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

• Can you elaborate? I'm still not quite sure where to start...
– CruZ
Sep 5, 2019 at 21:55

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.

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