Let $X_{i}\sim N(\mu_{i} ,\sigma^{2}_{i}), 1 \leq i \leq n$, denote $n$ normally
distributed independent random variables. I want to show that $$\sum_{i=1}^n X_i \sim N\left(\sum_{i=1}^n\mu_{i},\sum_{i=1}^n\sigma_{i}^{2}\right),$$
without using convolution integrals or characteristic functions.
[Math] The sum of normally distributed random variables.
probability
Best Answer
The means part is straightforward to show using the linearity of expectation; no convolutions or characteristic functions are needed. So, for ease in calculation, we can take the $X_i$ to be zero-mean normal random variables and just show that $\sum_i X_i \sim N\left(0,\sum_i \sigma_i^2\right)$.
Suppose $X$ and $Y$ are independent standard normal random variables. Then, as described in this answer of mine, $\alpha X + \beta Y$ is a zero-mean normal random variable with variance $\alpha^2+\beta^2$. This proof is based purely on a rotation of axes; no convolutions or characteristic functions are involved. But, $\alpha X$ and $\beta Y$ are zero-mean normal random variables with variances $\alpha^2$ and $\beta^2$ respectively. So, we have that $X_1+X_2 \sim N(0,\sigma_1^2+\sigma_2^2)$, and as Pierre said, the general result $\sum_i X_i \sim N\left(0,\sum_i \sigma_i^2\right)$ follows by induction. Look, Ma! No convolutions and no characteristic functions.