Resulting distribution from inner product between vector which entries are normally distributed and another vector

Question

I need to understand the following part of lectures notes (below). How did we obtain distribution of inner product outcome (in red box)? It is probably some obvious rule that I do not know..

Thank you!

Answer 1

More context would be nice, but in general, if $X_1, X_2, \dots, X_N$ are independent Gaussians, with:

$$ X_i \sim N(\mu_i, \sigma^2_i) $$

Then write $X= [X_1, X_2, \dots, X_n]^T$. Any linear combination of the $X_i$ is also Gaussian, so for any vector of constants $\theta = [\theta_1, \dots, \theta_n]^T$, we have that:

$$ \theta^TX \sim N(\mu, \sigma^2) $$

where: \begin{align*} \mu &= E[\theta^TX] = E[\theta_1X_1 + \theta_2 X_2 +\dots \theta_n X_n ] \\ &= \theta_1 E[X_1] +\dots + \theta_n E[X_n] \\ &= \theta_1 \mu_1 +\dots + \theta_n \mu_n \end{align*} and \begin{align*} \sigma^2 &= V(\theta^TX)\\ &= V(\theta_1X_1 + \theta_2 X_2 +\dots \theta_n X_n)\\ &= V(\theta_1X_1) + V(\theta_2 X_2) +\dots V(\theta_n X_n) && \text{(independence)}\\ &=\theta_1^2V(X_1) + \theta_2^2V( X_2) +\dots \theta_n^2V(X_n)\\ &=\theta_1^2\sigma^2_1 + \theta_2^2 \sigma^2_2 +\dots \theta_n^2 \sigma^2_n\\ &= \sum_{j=1}^n \theta_j^2 \sigma_j^2 \end{align*}

In your case, $\mu_i = 0$ for all $i$, so $\mu = 0$. And $\theta_i = A_{i,j}$ and $\sigma_i^2 = 1/k$ for all $i$.