I am wondering what is the $E[\textbf{a}\cdot \textbf{b}]$ and $var[\textbf{a}\cdot \textbf{b}]$ where $\textbf{a}, \textbf{b}$ are independent random vectors. That is as a vector whose elements are random variables. There are n elemetns in the vector. Each element in vector is assumed to be random sample from a normal distribution with mean 0 and variance $\sigma^{2}=1/n$. and $\cdot$ denotes dot product.

I read somewhere that \begin{equation} \begin{aligned} E(\tilde{\textbf{a}}\cdot \tilde{\textbf{b}})&=E(\sum_{i=1}^{n} a_{i} b_{i})\\ &= n E(XY)= 0 \end{aligned} \end{equation}

\begin{equation} \begin{aligned} var(\tilde{\textbf{a}}\cdot \tilde{\textbf{b}})&=var(\sum_{i=1}^{n} a_{i} b_{i})\\ &=n~var(XY) \\ &=\dfrac{1}{n} \end{aligned} \end{equation}

How we can say $var(\sum_{i=1}^{n} a_{i} b_{i}) =n~var(XY)$ or $E(\sum_{i=1}^{n} a_{i} b_{i}) =n~E(XY)$. Does anyone have an idea on this?

  • 1
    $\begingroup$ I think that the Matrix Cookbook will help you out here, specifically the section on General Statistics and Probability $\endgroup$
    – call-in-co
    Commented Dec 11, 2017 at 17:13
  • 3
    $\begingroup$ Are you familiar with properties of expectation? The relevant one is linearity. This is not a matrix problem--it requires only the most straightforward application of linearity. $\endgroup$
    – whuber
    Commented Dec 11, 2017 at 17:14
  • $\begingroup$ Thank you for your reply @whuber. I am familiar with properties of expectation more or less, but I was not sure that is correct: $var(\sum_{i=1}^{n} a_{i} b_{i})=E((\sum_{i=1}^{n} a_{i} b_{i})^{2})- (E(\sum_{i=1}^{n} a_{i} b_{i}))^{2} =n(E((XY)^{2})-(E(XY))^{2})$. It shouldn't be $n^{2} (E((XY)^{2})-(E(XY))^{2})$? $\endgroup$
    – Niki
    Commented Dec 11, 2017 at 17:30
  • $\begingroup$ I have provided proof here: github.com/BAI-Yeqi/Statistical-Properties-of-Dot-Product/blob/… :) $\endgroup$
    – Jake2099
    Commented Nov 23, 2021 at 5:25
  • $\begingroup$ a.b will be a scalar value. How does it make sense to get the mean and variance of a scalar? Sorry for dumb question... $\endgroup$
    – gag123
    Commented Aug 21, 2023 at 3:47

1 Answer 1


\begin{equation} \begin{aligned} E(\sum_{i=1}^{n} a_{i} b_{i}) &=\sum_{i=1}^{n}E( a_{i} b_{i}) \text{, due to linearity}\\ &= n E(XY) \text{ , due to i.i.d}\\ \end{aligned} \end{equation} Note that variance of sum of independent variables is equal to the sum of their variance. \begin{equation} \begin{aligned} var(\sum_{i=1}^{n} a_{i} b_{i}) &=\sum_{i=1}^{n}var( a_{i} b_{i}) \\ &=n~var(XY) \text{, due to i.i.d}\\ \end{aligned} \end{equation}

Here $X,Y$ are independent are follows distribution $N(0,\sigma^2)$. You will have to use the property that $X$ and $Y$ are independent to evalute $E(XY)$ and $var(XY)$.

  • $\begingroup$ How should be $var(\sum_{x} (\sum_{i} a_{x-i}b_{i} \cdot \sum_{i'} a_{x-i'}c_{i'}))$? this is not correct $n^{3}var(X^{2}YZ)$, am I right? $\endgroup$
    – Niki
    Commented Dec 11, 2017 at 19:00
  • $\begingroup$ where does $c$ come from? $\endgroup$ Commented Dec 11, 2017 at 19:09
  • $\begingroup$ Oh,I forgot to write, $\textbf{c}, \textbf{a},\textbf{b} $ are independent random vectors. It is another equation that I want to know its variance. $\endgroup$
    – Niki
    Commented Dec 11, 2017 at 19:36
  • $\begingroup$ @SiongThyeGoh hey could you please explain the second step in expectation for which the reason is i.i.d ? Thanks $\endgroup$
    – Siddharth
    Commented Jan 15, 2019 at 3:38
  • $\begingroup$ $E(a_1b_1)=E(a_2b_2) = \ldots=E(a_nb_n)=E(XY)$, then we sum them up. $\endgroup$ Commented Jan 15, 2019 at 3:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.