If $X_1,\cdots,X_n$ all have the same variance equal to 1, then $0\leq \mbox{Var}[\bar{X}]\leq 1$ where $\bar{X}=(X_1 + \cdots + X_n)/n$. The upper bound is attained if $\mbox{Cov}[X_k,X_l]=1$ for all $k, l$, for instance if $X_1 = X_2 = \cdots = X_n$.

But what is the lower bound? Clearly, not all covariances can be simultaneously equal to -1. Is the lower bound attained when $X_{k+1}=-X_k$ for $k=1, \cdots, n-1$? This is true if $n$ is even (in that case $\mbox{Var}[\bar{X}]=0$), but what if $n$ is odd?

  • $\begingroup$ Your statement for n being even doesn't not hold in general. For example in the case when ${X_i}'s$ are all positive. $\endgroup$ May 17, 2019 at 6:12
  • $\begingroup$ True, but here I was interested about the absolute minimum. So we can consider Gaussian variables. $\endgroup$ May 17, 2019 at 6:17
  • $\begingroup$ Ok. If you're further willing to assume that ${X_i}'s$ are independent then via Cramer-Rao lower bound you can claim that the Variance of $\bar{X}_n$ is at least $\frac{1}{n}$. $\endgroup$ May 17, 2019 at 6:56
  • $\begingroup$ The $X_i's$ are strongly auto-correlated in my case, but if $n$ is odd, that's also the lowest I have obtained so far: $\frac{1}{n}$. $\endgroup$ May 17, 2019 at 6:59
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    $\begingroup$ In that case you can always get a 0 covariance by letting $Cov(X_i,X_j)=\frac{-1}{n-1}$ for all $i \neq j$ $\endgroup$ May 17, 2019 at 7:08

1 Answer 1


There is many ways in which the lower bound of 0 can be obtained, they all represent multivariate distributions of $X$ where $X_1+X_2+\dotsm+X_n$ is constant (with probability one.) So they will be singular distributions, with a support which is not all of $\mathbb{R}^n$.

Let the covariance matrix of the random ector $X$ be $\Sigma$. Then we can compute the variance of the mean $\bar{X}_n$ as $(\frac1{n})^2 1^T \Sigma 1$ where $1$ represents the $n$-vector of all ones. So the lower bound is obtained if $1$ is an eigenvector of $\Sigma$ with eigenvalue zero, so $\Sigma$ is only positive semidefinite.

But in a comment the OP adds that he has a strongly autocorrelated time-series. In that case maybe more can be said. For an example, say the series is (second-order) stationary autoregressive of order 1. Then the covariance matrix $\Sigma$ will be a symmetric Toeplitz matrix, more spcifically tridiagonal with 1's along the diagonal and the autocorrelation (at lag 1) $\alpha$ along it, below and above. Then we can calculate the variance of the mean as $$ (\frac1{n})^2 1^T \Sigma 1=\frac1{n}\left(1+2\alpha \frac{n-1}{n}\right) $$ and the requirement that the variance be nonnegative means that $\alpha \ge -\frac12 \frac{n}{n-1}$ and equality will result in a variance of zero. But if this really are observations from a stationary time series, then a value for $\alpha$ is possible only if it is possible for all $n$, which in this case leads to $\alpha \ge -\frac12$. In that case the minimum value of the variance of $\bar{X}_n$ (for an AR of order 1) becomes $(\frac1{n})^2$. A similar analysis should be possible for higher-order AR processes.

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    $\begingroup$ This is becoming very interesting, and it actually answers a deeper question that I had in mind: for an AR time series, can the variance of the mean $\mbox{Var}[\bar{X}_n]$ be an order of magnitude lower than $\frac{1}{n}$? You positively answer that question. $\endgroup$ May 18, 2019 at 16:33
  • $\begingroup$ Even deeper, I mentioned (without proof) a general result implicitly implying that the variance in question is asymptotically of the form $A n^{-B}$ with $B\in [0, 1]$. In your answer, you reached the upper bound $B = 1$. The theorem in question (well, a conjecture or empirical observation) is stated in section 1 in my article "Confidence Intervals Without Pain" (see this link). Application to time series is discussed in section 2.3. in the same article. So here we have reached a milestone in trying to prove that theorem. $\endgroup$ May 18, 2019 at 16:44

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