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.

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

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    $\begingroup$ 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. $\endgroup$ – mlofton Mar 11 '19 at 2:29

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