Linear Combination of Gaussian Random Variables

Theorem

Let $X_1, X_2, X_3, \ldots, X_n$ be independent random variables.

Let $\sequence {\alpha_i}_{1 \mathop \le i \mathop \le n}$ and $\sequence {\mu_i}_{1 \mathop \le i \mathop \le n}$ be sequences of real numbers.

Let $\sequence {\sigma_i}_{1 \mathop \le i \mathop \le n}$ be a sequence of positive real numbers.

Let $X_i \sim \Gaussian {\mu_i} {\sigma^2_i}$ for $1 \le i \le n$, where $\Gaussian {\mu_i} {\sigma^2_i}$ is the Gaussian distribution with parameters $\mu_i$ and $\sigma^2_i$.

Then:

$\ds \sum_{i \mathop = 1}^n \alpha_i X_i \sim \Gaussian {\sum_{i \mathop = 1}^n \alpha_i \mu_i} {\sum_{i \mathop = 1}^n \alpha^2_i \sigma^2_i}$

Proof

Let:

$\ds Z = \sum_{i \mathop = 1}^n \alpha_i X_i$

Let $M_Z$ be the moment generating function of $Z$.

We aim to show that:

$\ds Z \sim \Gaussian {\sum_{i \mathop = 1}^n \alpha_i \mu_i} {\sum_{i \mathop = 1}^n \alpha^2_i \sigma^2_i}$

By Moment Generating Function of Gaussian Distribution and Moment Generating Function is Unique, it is therefore sufficient to show that:

$\ds \map {M_Z} t = \map \exp {\paren {\sum_{i \mathop = 1}^n \alpha_i \mu_i} t + \frac 1 2 \paren {\sum_{i \mathop = 1}^n \alpha^2_i \sigma^2_i} t^2}$

We also have, by Moment Generating Function of Gaussian Distribution, that the moment generating function of $X_i$, $M_{X_i}$, is given by:

$\map {M_{X_i}} t = \map \exp {\mu_i t + \dfrac 1 2 \sigma^2_i t^2}$

We have:

\(\ds \map {M_Z} t\)

\(=\)

\(\ds \prod_{i \mathop = 1}^n \map {M_{X_i} } {\alpha_i t}\)

Moment Generating Function of Linear Combination of Independent Random Variables

\(\ds \)

\(=\)

\(\ds \prod_{i \mathop = 1}^n \map \exp {\mu_i \alpha_i t + \frac 1 2 \sigma^2_i \alpha^2_i t^2}\)

\(\ds \)

\(=\)

\(\ds \map \exp {\sum_{i \mathop = 1}^n \paren {\mu_i \alpha_i t + \frac 1 2 \sigma^2_i \alpha^2_i t^2} }\)

Exponential of Sum

\(\ds \)

\(=\)

\(\ds \map \exp {\paren {\sum_{i \mathop = 1}^n \alpha_i \mu_i} t + \frac 1 2 \paren {\sum_{i \mathop = 1}^n \alpha^2_i \sigma^2_i} t^2}\)

$\blacksquare$