From this post and following a tip by @Ian in the comments.

If $X_1,\dots,X_n \sim \text{ i.i.d. } N(\mu,\sigma^2)$ with $\displaystyle \bar X= \frac{\sum_{i=1}^n X_i}{n},$ the covariance of the random variables $Y_i=\frac{X_i-\bar X}{\sigma}$ (which I supposed can be read as centered and scaled entries of a random vector) is $-1/n.$

I see that

$$\begin{align} \small \operatorname{cov}\left(Y_i,Y_j\right) = {} & \small \operatorname{cov} \left(\frac{(X_i-\bar X)}{\sigma},\frac{(X_j-\bar X)}{\sigma}\right)\\[2ex] = {} &\small\frac{1}{\sigma^2}E\big[(X_i-\bar X)(X_j-\bar X)\big] - E\big[X_i-\bar X\big]E\big[X_j-\bar X\big]\\[2ex] = {} & \small \frac{1}{\sigma^2}E\big[(X_i-\bar X)(X_j-\bar X)\big]\\[2ex] = {} & \small \frac{1}{\sigma^2}E[(X_i-\mu+\mu-\overline{X})(X_j-\mu+\mu-\overline{X})]\\[2ex] = {} &\small \frac{1}{\sigma^2}\left( E[(X_i-\mu)(X_j-\mu)]+E[(X_i-\mu)(\mu-\overline{X})]+E[(X_j-\mu)(\mu-\overline{X})]+E[(\mu-\overline{X})^2]\right)\\[2ex] = {} &\small\frac{1}{\sigma^2}\left( E[(X_i-\mu)(\mu-\overline{X})]+E[(X_j-\mu)(\mu-\overline{X})]+E[(\mu-\overline{X})^2]\right) \end{align}$$

The last term $E[(\mu-\overline{X})^2=\operatorname{var}(\bar X)=\frac{\sigma^2}{n}$ as derived here.

Does, then

$$E[(X_i-\mu)(\mu-\overline{X})]+E[(X_j-\mu)(\mu-\overline{X})] \overset{?}= \frac{-2\sigma^2}{n} \text{ ?}$$

  • $\begingroup$ Why are you adjusting your sample by the sample mean, but with the population variance? Should you not be adjusting by both the sample mean and sample variance ... or by population mean and population variance? $\endgroup$ – wolfies Nov 15 '17 at 5:25

$\newcommand{\c}{\operatorname{cov}}$You seem to be doing it the hard way.

\begin{align} & \c(X_i - \overline X, X_j - \overline X) \\[10pt] = {} & \c(X_i,X_j) - \c(X_i,\overline X) - \c(\overline X, X_j) + \c(\overline X,\overline X) \\[10pt] = {} & 0 - \c\left( X_i, \frac {X_1+\cdots+X_n} n \right) - \c\left( \frac {X_1+\cdots+X_n} n, X_j \right) + \operatorname{var}(\overline X) \\[10pt] = {} & - \frac{\sigma^2} n - \frac{\sigma^2} n + \frac{\sigma^2} n = \frac{-\sigma^2} n. \end{align}

  • $\begingroup$ Note to self: $cov \left( X_i, \frac {X_1+\cdots+X_n} n \right)=\frac{1}{n} cov \left( X_i, X_1+\cdots+X_n\right)=\frac{1}{n} [cov (X_i, X_1)+\cdots+cov(X_i,X_i)+\cdots+cov(X_i,X_n)]=\frac{1}{n}[0+\cdots+\sigma^2+\cdots+0]$ $\endgroup$ – Antoni Parellada Nov 15 '17 at 8:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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