How can I prove that the sum of $X_1, X_2, \ldots,X_n$ random variables, all of which have normal distributions $N(\mu_i, \sigma_i)$, is a random variable that is itself normally distributed with mean $$\mu =\sum_{i=1}^n \mu_i$$
and variance
$$\sigma^2 = \sum_{i=1}^n \sigma_i^2$$
Edit: I forgot to add that this was with the assumption that all $X_1, X_2,\ldots,X_n$ are independent.
Best Answer
I figured this out on the transit ride to work this morning. I used the moment-generating function, which may be the same thing as the "Probability-generating function" that is recommended in the answer above? I'm not sure. Anyway...
If we let $Y = X_{1} + X_{2} + \space \cdots \space + X_{n}$ , then the moment-generating function of $Y$ is given by:
$$ M_{Y}(t) = E[e^{t(X_{1} \space + \space X_{2} \space + \space \cdots \space + \space X_{n})}] = \prod_{i=1}^{n}E[e^{tX_{i}}] = \prod_{i=1}^{n}M_{X_{i}}(t) $$
If we then find
$$M_{X_{i}}(t) = \exp[\mu_{i}t + \frac{\sigma_{i}^2t^2}{2}]$$
then we have
$$ M_{Y}(t) = \prod_{i=1}^{n}\exp[\mu_{i}t + \frac{\sigma_{i}^2t^2}{2}] $$
then by properties of exponents we have
$$ M_{Y}(t) = \prod_{i=1}^{n}\exp[\mu_{i}t + \frac{\sigma_{i}^2t^2}{2}] = \prod_{i=1}^{n}\exp[\mu_{i}t]\exp[\frac{\sigma_{i}^2t^2}{2}] = \exp[t(\sum_{i=1}^{n}\mu_{i}) + \frac{t^2}{2}(\sum_{i=1}^{n}\sigma_{i}^2)] $$
Because
$$ M_{Y}(t) = \exp[t(\sum_{i=1}^{n}\mu_{i}) + \frac{t^2}{2}(\sum_{i=1}^{n}\sigma_{i}^2)] $$ this implies that Y is normally distributed with mean $\space\sum_{i=1}^{n}\mu_{i}$ and variance $ \space \sum_{i=1}^{n}\sigma_{i}^2\space$