characteristic function when adding Gaussian RV.s

Question

$X_i$ and $Y$ are random variables:

$$X_i \sim \text{Gaussian}(\mu_i,~ \sigma^2_i)$$

$$Y = a_1 X_1 + a_2 X_2 + \cdots + a_n X_n$$

They say to find the Characteristic equation of Y, so:

$$\Psi(\omega) = E[e^{j\omega Y}]$$

$$\Psi(\omega) = E[e^{j\omega (a_1 X_1 + a_2 X_2 + \cdots + a_n X_n)}]\tag{1}$$

$$\Psi(\omega) = \Psi_{X_1,~ \cdots, X_n}(\omega a_1,~ \cdots, \omega a_n)\tag{2}$$

How did they get from step (1) to step (2)?

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I imagine they probably do something like this:

$$\Psi(\omega) = E[~e^{j\omega a_1 X_1}~e^{j\omega a_2 X_2}~ \cdots~ e^{j\omega a_n X_n}~]$$

But its still doesn't match the result (2)...

If I assumed independence of $X_i$ with each other... then I would get:

$$\Psi(\omega) = E[~e^{j\omega a_1 X_1}]~E[e^{j\omega a_2 X_2}]~ \cdots~ E[e^{j\omega a_n X_n}]$$

I don't know if they are independent... but it still doesn't match the result (2).

Answer 1

Let $X=(X_1,\ldots, X_n)$, $a=(a_1,\ldots, a_n)$ and $\omega\in\mathbb{R}$.

The characteristic function of $X$ at $\omega a=(\omega a_1, \ldots, \omega a_n)$ equals the characteristic function of $Y$ at $\omega$:

$$\Psi_X(\omega a)=E[e^{i\langle \omega a,X\rangle}] = E[e^{i\omega\sum_{j=1}^na_jX_j}]=\Psi_Y(\omega)$$

Note that the characteristic function is defined for different dimensions.

Answer 2

I'll be just a bit more explicit than is the answer already appearing here.

First we have $ \Psi_Y(b) = \operatorname E(e^{jb Y}). $

But when $Y$ is a vector-valued, rather than scalar-valued, random variable, then the argument $b$ is also a vector and the product $bY$ is taken to be a dot-product. In line $(1)$ in your question you see a dot-product. In line $(2)$ you see the characteristic function of the random vector $(X_1,\ldots,X_n)$ evaluated at the vector $(\omega a_1,\ldots,\omega a_n).$