# Expectation and Variance of dot product of a random vector and random linear combinations of vectors from the same distribution?

Let's say we have a multivariate distribution $$D$$ which generates random $$n$$-dimensional vectors $$x$$ for us ($$x \in R^n$$). We know that the dimensions of vector $$x$$ are correlated, and that each dimension of $$x$$ has a mean of 0 and a standard deviation of 1. Now, let's say we have another random vector $$y$$ (of shape $$(n,1)$$) defined as: $$y = \sum_{i=1}^{M} \alpha^{(i)} x^{(i)}\qquad\\ x^{(i)} \sim D\\ \alpha^{(i)} \sim \mathcal{N}(0,1)$$ where $$x^{(i)}$$ is the $$i$$'th sampled vector from the distribution $$D$$ and has the shape $$(n,1)$$. We sample $$M$$ of these vectors ($$i=1,2,3,...,M$$) where $$M>>n$$. Also, $$\alpha^{(i)}$$ is a sampled scalar from the distribution $$\mathcal{N}(0,1)$$.

What would be the expectation and variance of the dot product between x and y? $$E[x \cdot y]=?\\ Var(x \cdot y)=?$$

Update1: In case this is too difficult to solve for any distribution $$D$$, I would still appreciate it if someone can solve this for when $$D$$ is a multivariate gaussian distribution with a full rank covariance matrix, and assume that $$y = \dfrac{\sum_{i=1}^{M} \alpha_i x^{(i)}}{\sum_{i=1}^{M} \alpha_i}$$.

Correct me if I'm wrong but I think these two assumptions would make things easier because we could treat $$y$$ as $$y \sim D$$.

Edit: Changed notations of question as suggested by whuber and mlofton. Thank you.

• Could you please explain how you can let the index $i$ range up to $100n$ when $x$ only has $n$ components?? – whuber Mar 10 '19 at 20:41
• $x^{(i)}$ is a vector of shape $(n,1)$ sampled from the distribution $D$, and we sample $100n$ of these vectors ($i=1,2,3,...,100n$). – Soroush Mar 10 '19 at 21:29
• Please edit your post to explain and clarify this. As it stands, the notation just makes no sense. – whuber Mar 10 '19 at 22:54

Wouldn't the expectation be just 0, because $$\alpha$$s come from a 0 centered independent Gaussian distribution, which makes $$y$$ identically zero in expectation? Assuming $$x$$ and $$y$$ are independent draws of course, hm?
• Hi Tomas: I think you're right but maybe the OP meant the $\alpha_{i}$ to be fixed constants rather than random variables ? As whuber mentioned, there's an $n$ dimensional rv $X$ but also an $n$ in the sum so it's not so clear what's going on. – mlofton Mar 11 '19 at 2:29