Expectation of $\mathbf{v}\mathbf{v}^\top$ when $\mathbf{v}$ is an independent Bernoulli vector.

Question

It has been a while since I took probability class, so I dare to ask.

Suppose I have $\mathbf v$, a column vector of length $n$. Each entry is independently chosen from $\{+1, -1\}$ so that for each entry $v_i$, $\Pr[v_i = +1] = \Pr [v_i = -1] = 1/2$.

Then I generate an $n\times n$ matrix $\mathbf{V} = \mathbf{v}\mathbf{v}^\top$. And I want to show that the expectation of $\mathbf V$ is an $n\times n$ identity matrix. Is it true?

My guess is that in the matrix $\mathbf V$,

• the diagonal term $V_{i,i}$ is always 1 because it's the product of the same random variable $v_i$ and $v_i$.
• the off-diagonal term $V_{i,j}$ is the product of two independent zero-mean random variable, so the expectation of the off-diagonal is always zero.

Does it make sense?

Answer 1

Yes.

The diagonal elements have the form $E[v_i^2]=1$ and the off diagonal elements for $i\neq j$, $E[v_i v_j]= E v_i E v_j=0$.