I'm trying to express $\mathrm{Var}(\mu_x - \mu_y)$ in terms of $\rho$, $\sigma_x$ and $\sigma_y$, where $\mu$ denotes the mean of the random variable.

Firstly:\begin{align*} \mathrm{Var}\left(\sum_{i=1}^n x_i - \sum_{i=1}^n y_i\right) &= \mathrm{Var}\left(\sum_{i=1}^n(x_i-y_i)\right) \\ &=\sum_{i=1}^n \mathrm{Var}(x_i) + \sum_{i=1}^n\mathrm{Var}(y_i) - \sum_{j=1}^n\sum_{i=1}^n 2\mathrm{Cov}(x_i, y_j) \\ &= n\sigma^2_x + n\sigma^2_y - n^2(2\rho\sigma_x\sigma_y) \end{align*} So \begin{align*} \mathrm{Var}\left(\frac 1n\sum_{i=1}^n x_i - \frac 1n\sum_{i=1}^n y_i\right) &= \mathrm{Var}\left(\frac 1n\left(\sum_{i=1}^n(x_i - y_i)\right)\right) \\ &= \frac 1{n^2}\left(n\sigma^2_x + n\sigma^2_y - n^2(2\rho\sigma_x\sigma_y)\right) \\ &= \frac 1n(\sigma^2_x + \sigma^2_y - n2\rho\sigma_x\sigma_y) \end{align*} However, in my applied econometrics textbook, the Wald test for $\mu_x = \mu_y$ uses $\mathrm{Var}(\mu_x - \mu_y) = \frac1n(\sigma^2_x + \sigma^2_y - 2\rho\sigma_x\sigma_y)$, so where have I made a mistake?

  • $\begingroup$ The population means are constants, so the variance of their difference is 0 $\endgroup$
    – Glen_b
    Apr 5, 2019 at 9:21
  • $\begingroup$ Yes, should I edit my question? I was concerned that the answer would not align if I did that. $\endgroup$
    – ahorn
    Apr 5, 2019 at 10:16

1 Answer 1


Your textbook probably assumes the data to be independent across individuals, thus $\mathrm{Cov}(x_i,y_j)=0$ for $i\neq j$. Also, you mean the sample mean and not the mean. The mean of a random variable is just a number and its variance is zero.

  • $\begingroup$ (+1) I was about to point the discrepancy (or the need for a Bayesian perspective). $\endgroup$
    – Xi'an
    Apr 4, 2019 at 18:22
  • $\begingroup$ Should I put hats in my question, or leave it? $\endgroup$
    – ahorn
    Apr 4, 2019 at 19:21

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