I am interested in a paper by John Walsh (1947), which demonstrates the effect of intraclass correlation on significance testing. The full text is available here.

In outlining a proof, Walsh states "It is easily seen from elementary considerations that

$$ \frac{(\bar{x} - \mu)\sqrt{n}} {\sigma\sqrt{1 + (n - 1)\rho}}$$

has the distribution N(0 , 1)" Where $\rho$ is the intraclass correlation.

Unfortunately, this is not so elementary for me. I would be grateful if someone can explain why this random variable comes from the standard normal distribution.

  • $\begingroup$ It is not possible to answer this until you define your terms! $\endgroup$ Oct 28, 2015 at 10:03
  • $\begingroup$ @kjetilbhalvorsen It's not necessary to define the terms. It's obviously a standardised version of the average. JohnK has a simple but nice proof. $\endgroup$
    – SmallChess
    Oct 28, 2015 at 10:26

1 Answer 1


$$ \begin{align} Var(\bar{X}) = \frac{1}{n^2} Var( \sum_{i=1}^n X_i ) & = \frac{1}{n^2}\left( \sum_{i=1}^n var(X_i) + \sum _{i \neq j} cov\left(X_i , X_j \right) \right) \\ &= \frac{1}{n^2} \left( n \sigma^2 + n (n-1) \rho \sigma^2 \right) \\ & =\frac{\sigma^2 }{n} \left( 1 + \left(n-1 \right) \rho \right) \end{align}$$

where we have used the fact that $cov\left(X_i, X_j \right) = \rho \sigma_i \sigma_j$ and the equal variances of your observations. So standardize by that in order to get the standard normal distribution!

On a side note, Walsh is a genius and his papers are always worth reading no matter how frustrating it can be at times.

  • $\begingroup$ or rather, standardize by dividing by the square root of that. $\endgroup$
    – Glen_b
    Oct 28, 2015 at 10:31
  • $\begingroup$ @Glen_b Yes, I thought that this was implied :) $\endgroup$
    – JohnK
    Oct 28, 2015 at 10:32

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.