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