Show that if $X_k \sim N(a, \sigma^2)$ then $\sum_{k=1}^n X_k \sim N(na, n \sigma^2)$

Question

Suppose that, for $1 \leq k \leq n$, $X_k$ are identical independently distributed random variables with $X_k \sim N(a, \sigma^2)$.

I am try to show that $$ S_n := \sum_{k=1}^n X_k \sim N(na, n \sigma^2) $$

Showing that $$ \mathbb{E} [S_n] = na \hspace{10mm} \text{ and } \hspace{10mm} \operatorname{Var} [S_n] = n \sigma^2 $$ is trivial, so it just remains to show that $S_n$ is normally distributed. To do this, I am trying to compute the characteristic function of $S_n$, for which I have attempted to do using the following rule: $$ \mathbb{E} [ \exp (1 - i \lambda X) ] = 1 + i \lambda \mathbb{E}[X] - \frac{\lambda^2 \mathbb{E}[X^2]}{2} + o(\lambda^2) \tag{$*$} $$

This has been my attempt so far: $$ \mathbb{E} [ \exp (1 - i \lambda S_n) ] = \mathbb{E} [ \exp (1 - i \lambda \sum_{k=1}^n X_k) ] \\ = \prod_{k=1}^n \mathbb{E} [ \exp (1 - i \lambda X_k) ] \\ = \left( \mathbb{E} [ \exp (1 - i \lambda X_k) ] \right)^n \\ = \left( 1 + i \lambda \mathbb{E} [X_k] - \frac{\lambda^2}{2} \mathbb{E} [X^2] \right)^n \\ = \left( 1 + i \lambda a - \frac{\lambda^2}{2} (\sigma^2 + a^2) \right)^n \tag{$**$} $$

However, I am not sure of how to proceed from here. The solutions to this question are given, but extremely vague. They do, however, tell me that I need to show that $(**)$ tends to $$ \exp \left( i \lambda a n - \frac{\lambda^2 \sigma^2 n}{2} \right) $$ as $n \rightarrow \infty$.

Can anyone help me to complete this proof?

Answer 1

The moment-generating function of the sum of $n$ IID random variables is equal to the MGF of a single random variable raised to the $n^{\rm th}$ power; i.e., if $$S_n = \sum_{i=1}^n X_i,$$ then $$M_{S_n}(t) = (M_X(t))^n$$ where each $X_i \sim X$ are IID. Then we find $$M_{S_n}(t) = \left(\exp\left(\mu t + \frac{\sigma^2 t^2}{2}\right)\right)^n = \exp\left(n \mu t + \frac{n \sigma^2 t^2}{2}\right) = \exp\left(\mu^* t + \frac{(\sigma^*)^2 t}{2}\right),$$ where $$\mu^* = n \mu, \quad \sigma^* = \sigma \sqrt{n},$$ thus $$S_n \sim \operatorname{Normal}(n \mu, n \sigma^2)$$ as claimed.

Answer 2

$\newcommand{\e}{\operatorname{E}}$The characteristic function is \begin{align} t\mapsto \e\left(e^{itS_n}\right) & = \e\left( \exp\left( it\sum_{k=1}^n X_n \right) \right) = \overbrace{ \e\left( \prod_{k=1}^n \exp(itX_n) \right) = \prod_{k=1}^n \e(\exp(itX_n)) }^\text{because of independence} \\[10pt] & = \left( \e(\exp(itX_1) \right)^n = \left( \exp\left( ita + \frac 1 2 (t\sigma)^2 \right) \right)^n = \exp\left( nita + n\frac 1 2 (t\sigma)^2 \right) \end{align}

If two distributions have the same characteristic function, are they the same distribution? A theorem that you should now cite says the answer is "yes".

Alternatively, you can do this by computing a convolution.