This post asks "why a familiar and widely used estimator of sample covariance has expected value zero, in a situation where the variables involved are characterized by non-zero and equal pair-wise covariance"?

Specifically, the set up is as follows: we have a sequence of identically distributed random variables $\{X_1,...X_n\}$, and another sequence $\{Y_1,...,Y_n\}$ that have also identical distributions, but different than the $X$'s. Moreover, the following holds:

$${\rm Cov}(X_i,Y_j) = {\rm Cov}(X_j,Y_i) \neq 0, \;\forall \{i \neq j\}\cup \{i=j\} \in \{1,...,n\} \tag{1}$$

Note that the above math imply also that

$$ {\rm Cov}(X_i,Y_j) = {\rm Cov}(X_i,Y_i) \tag{2}$$

This is critical for the results to follow.

(Note: Initially I have described the associations above as "equi-cross-correlation" but if you look at the comments of the thread it appears that the term describes something weaker. So I erased all references to it).

Since the elements of each sequence are identically distributed, we have that $E(X_i) = E(X_j) = E(X)$ and $E(Y_i) = E(Y_j) = E(Y)$. Then, in order to have equal pair-wise correlation coefficients, for $i\neq j$ but also for $i=j$, we must have

$$E(X_iY_j) = E(X_jY_i) = E(X_iY_i) = E(XY) \neq 0, \;\forall i,j \in \{1,...,n\}$$

We are told to consider what we know as an unbiased Covariance estimator

$${\rm \hat Cov}(X, Y) = \frac 1{n-1}\sum_{i=1}^n(X_i-\bar X)(Y_i-\bar Y)$$

with $\bar X = \frac 1{n}\sum_{i=1}^nX_i$ and likewise for the $Y$'s.

Expanding the product, we get

$${\rm \hat Cov}(X, Y) = \frac 1{n-1}\sum_{i=1}^nX_iY_i - \frac n{n-1}\left(\frac 1n \sum_{i=1}^nX_i\right) \left(\frac 1n \sum_{i=1}^nY_i\right)$$

$$= \frac 1{n-1}\sum_{i=1}^nX_iY_i - \frac n{n-1}\frac 1{n^2}\left(\sum_{i=1}^n\sum_{j=1}^nX_iY_j\right)$$

Taking the expected value of the estimator

$$E\left[{\rm \hat Cov}(X, Y)\right] = \frac 1{n-1}\sum_{i=1}^nE(X_iY_i) - \frac n{n-1}\frac 1{n^2}\left(\sum_{i=1}^n\sum_{j=1}^nE(X_iY_j)\right)$$

From previously, we have assumed that $E(X_iY_i) = E(X_iY_j) = E(X_jY_i) = E(XY)$. More over the double sum has $n^2$ elements, so we get

$$E\left[{\rm \hat Cov}(X, Y)\right] = \frac 1{n-1}nE(XY) - \frac n{n-1}\frac 1{n^2}n^2E(XY) =0$$

Great. We have "seriously entangled" (and "linearly" so) random variables, and the unbiased sample covariance, an almost "automatic" metric to calculate when getting to know the data, has expected value zero...

Some twisted, "Theater of the Absurd" intuition can be gleaned from the phrase "if we cannot distinguish between the pair $\{X_i, Y_i\}$ and the pair $\{X_i, Y_j\}$, as regards covariance, we "conclude" that said covariance is zero", but for the time being this sounds more absurd than intuitive.

I understand that the situation described by assumptions $(1)$ and $(2)$ may be of rather limited applied interest, even for moderately large $n$, because if we try to translate it into real-world relations, it pictures too many and at the same time too similar associations, to be probable/believable.

But I feel this is not just a "theoretical curiosity" but it may be telling us something useful about the limitations of our tools... something that may be already well-known -but since it is not well-known to me, I decided to post it as a question.

Any ideas or explanations to better understand the above situation?

"Layman" approaches as well as advanced mathematical ones are equally welcome.

  • $\begingroup$ I have been unable to ascertain what the question is. Could you please be more specific about what characteristic of "the above situation" you wish to ask about? Is there any way you could ask your question in one or two lines rather than requiring readers to wade through over a page of mathematics without understanding where it's headed? $\endgroup$
    – whuber
    Commented Mar 20, 2015 at 16:40
  • $\begingroup$ @whuber I can try to re-phrase, but isn't the title of the question clear enough? $\endgroup$ Commented Mar 20, 2015 at 16:41
  • $\begingroup$ @whuber Anyway, I added a paragrph in the beginning. $\endgroup$ Commented Mar 20, 2015 at 16:46
  • $\begingroup$ Thank you, that helps. But what do you mean by "... marginally identically distributed $n$"? Your assertion that "all involved expected values are the same" implies all $2n$ random variables have correlation coefficients of $1$, whence they are a.s. equal! What use is that? $\endgroup$
    – whuber
    Commented Mar 20, 2015 at 17:14
  • $\begingroup$ @whuber The first point was a typo, the second an unnecessarily broad statement. I corrected both. Does the situation still implies pairwise correlation coefficients equal to one? I cannot see it though. Help would be appreciated. $\endgroup$ Commented Mar 20, 2015 at 17:49

3 Answers 3


The conditions on the covariances will force the $X_i$ to be strongly correlated to one another, and the $Y_j$ to be strongly correlated to each other, when the mutual correlations between the $X_i$ and $Y_j$ are nonzero. As a model to develop intuition, then, let's let both $(X_i)$ and $(Y_j)$ have an exponential autocorrelation function

$$\rho(X_i, X_j) = \rho(Y_i, Y_j) = \rho^{|i-j|}$$

for some $\rho$ near $1$. Also take every $X_i$ and $Y_j$ to have zero expectation and unit variance. Let $\text{Cov}(X_i,Y_j)=\alpha$. (For any given $n$ and $\alpha$, the possible values of $\rho$ will be limited to an interval containing $1$ due to the necessity of creating a positive-definite correlation matrix.)

In this model the covariance (equally well, the correlation) matrix in terms of $(X_1, \ldots, X_n, Y_1, \ldots, Y_n)$ will look like

$$\begin{pmatrix} 1 & \rho & \cdots & \rho^{n-1} & \alpha & \alpha & \cdots & \alpha \\ \rho & 1 & \cdots & \rho^{n-2} & \alpha & \alpha & \cdots & \alpha \\ \vdots & \vdots & \cdots & \vdots & \vdots & \vdots & \cdots & \vdots \\ \rho^{n-1} & \cdots & \rho & 1 & \alpha & \alpha & \cdots & \alpha \\ \alpha & \alpha & \cdots & \alpha & 1 & \rho & \cdots & \rho^{n-1} \\ \alpha & \alpha & \cdots & \alpha &\rho & 1 & \cdots & \rho^{n-2} \\ \vdots & \vdots & \cdots & \vdots & \vdots & \vdots & \cdots & \vdots \\ \alpha & \alpha & \cdots & \alpha & \rho^{n-1} & \cdots & \rho & 1 \end{pmatrix}$$

A simulation (using $2n$-variate Normal random variables) explains much. This figure is a scatterplot of all $(X_i,Y_i)$ from $1000$ independent draws with $\rho=0.99$, $\alpha=-0.6$, and $n=8$.


The gray dots show all $8000$ pairs $(X_i,Y_i)$. The first $70$ of these $1000$ realizations have been separately colored and surrounded by $80\%$ confidence ellipses (to form visual outlines of each group).

The orientations of these ellipses have a uniform distribution: on average, there is no correlation among individual collections $((X_1,Y_1), \ldots, (X_n,Y_n))$.

Figure 2: histogram of orientations.

However, due to the induced positive correlation among the $X_i$ (equally well, among the $Y_j$), all the $X_i$ for any given realization tend to be tightly clustered. From one realization to another they tend to line up along a downward slanting line, with some scatter around it, thereby realizing a cloud of correlation $\alpha=-0.6$.

We might summarize the situation by saying by recentering the data, the sample correlation coefficient does not account for the variation among the means of the $X_i$ and means of the $Y_j$. Since, in this model, the correlation between those two means is exactly the same as the correlation between any $X_i$ and any $Y_j$ (namely $\alpha$), the expected correlation nets out to zero.

Here is working R code to play with the simulation.

n.sim <- 1000
alpha <- -0.6
rho <- 0.99
n <- 8
mu <- rep(0, 2*n)
sigma.11 <- outer(1:n, 1:n, function(i,j) rho^(abs(i-j)))
sigma.12 <- matrix(alpha, n, n)
sigma <- rbind(cbind(sigma.11, sigma.12), cbind(sigma.12, sigma.11))
min(eigen(sigma)$values) # Must be positive for sigma to be valid.
x <- mvrnorm(n.sim, mu, sigma)
#pairs(x[, 1:n], pch=".")
ell <- function(x, color, plot=TRUE) {
  if (plot) {
    points(x[1:n], x[1:n+n], pch=1, col=color)
    dataEllipse(x[1:n], x[1:n+n], levels=0.8, add=TRUE, col=color,
                center.cex=1, fill=TRUE, fill.alpha=0.1, robust=TRUE)
  v <- eigen(cov(cbind(x[1:n], x[1:n+n])))$vectors[, 1]
  atan2(v[2], v[1]) %% pi
n.plot <- min(70, n.sim)
plot(as.vector(x[, 1:n]), as.vector(x[, 1:n + n]), type="p", pch=".", col=gray(.4),
invisible(sapply(1:n.plot, function(i) ell(x[i,], colors[i])))
ev <- sapply(1:n.sim, function(i) ell(x[i,], color=colors[i], plot=FALSE))
hist(ev, breaks=seq(0, pi, by=pi/10))
  • 2
    $\begingroup$ Thanks, this is an approach I was hoping to see. It starts making sense now. $\endgroup$ Commented Mar 21, 2015 at 13:49

If $\{X_i\}$ are iid and $\{Y_i\}$ are iid, then $\{(X_i,Y_i)\}$ are iid. Hence $\operatorname{Cov}(X_i,Y_j) = 0$ when $i \ne j$. Since you are requiring $\operatorname{Cov}(X_i,Y_i) = \operatorname{Cov}(X_i,Y_i)$, you end up concluding that $\operatorname{Cov}(X_i, Y_i) = 0$, and thus $\operatorname{Cov}(X,Y) = 0$.

  • 1
    $\begingroup$ I did not say anywhere that they are independent. I only said that they are identically distributed. And then I assumed that their covariances are not zero (eq. 1), so certainly not independent. $\endgroup$ Commented Mar 21, 2015 at 0:33
  • $\begingroup$ Ah, in that case, ${\rm\hat Cov}(X,Y)$ is almost certainly not an unbiased estimate of $\operatorname{Cov}(X,Y)$ (and definitely not in this case). The unbiased nature of ${\rm\hat Cov}$ depends on having an iid sample (or possible some very specific types of violations of iid). $\endgroup$
    – Jay S
    Commented Mar 21, 2015 at 0:55
  • $\begingroup$ But this is not the point of the question. I couldn't care less about unbiasedness here. The issue is how a standard sample-moment covariance expression ends up having zero expected value when so much non-zero covariability exists in the sample. $\endgroup$ Commented Mar 21, 2015 at 0:59
  • $\begingroup$ Sorry, early submission. When we have an iid sample $\{(X,Y)\}$, then ${\rm\hat Cov}(X,Y)$ is unbiased because $\operatorname{E}(\bar{X},\bar{Y}) = \frac{1}{n}\operatorname{E}(XY)+\frac{n-1}{n}\operatorname{E}(X)\operatorname{E}(Y)$. When the sample is not independent, and $\operatorname{Cov}(X_i,Y_j) \ne 0$ for some $i\ne j$, this equality does not necessarily hold, and thus ${\rm \hat Cov}(X,Y)$ may be biased. $\endgroup$
    – Jay S
    Commented Mar 21, 2015 at 2:06
  • 1
    $\begingroup$ Maybe it would be worth looking into whether $\{X_i,Y_i\}$ is sufficient for $\operatorname{Cov}(X,Y)$ (equivalently, sufficient for $\operatorname{E}[\operatorname{E}(X)\operatorname{E}(Y)]$). Constructing an unbiased estimator for $\operatorname{E}[\operatorname{E}(X)\operatorname{E}(Y)]$ (or showing that it is not possible) would help illuminate what exactly is happening. I'll think some more on it. $\endgroup$
    – Jay S
    Commented Mar 21, 2015 at 2:52

I understand from the comments that this is not what you are looking for; I leave this answer anyway in case another reader find may find it useful.

We are given a setting and a derivation of a bias in the sample covariance. I understand from the comments that the fact that the bias makes the expectation zero in this case is what interests the OP. I argue here that it is of no particular interest and try to address the different points in OP's text.

We have "seriously entangled" (and "linearly" so) random variables, and the unbiased sample covariance, an almost "automatic" metric to calculate when getting to know the data, has expected value zero...

The sample covariance is not a measure of some overall linear dependence structure in a dataset generated by 2n random variables, and hence it is not surprising that insisting on calculating it in such a setting yields an expected value of zero. For this to be surprising, there must be some other value or range of values that makes more sense and I can see no such value.

But I feel this is not just a "theoretical curiosity" but it may be telling us something useful about the limitations of our tools... something that may be already well-known -but since it is not well-known to me, I decided to post it as a question

I would agree that a bias is a limitation, and a bias is indeed what you have derived, nothing else. I do not, however, see any argument why a bias making the expectation zero is more of a limitation than any other bias. For example, I would argue it's often worse to estimate the wrong sign than to estimate zero.

Why a familiar and widely used estimator of sample covariance has expected value zero, in a situation where the variables involved are characterized by non-zero and equal pair-wise covariance?

Because you have forced the estimator on such a setting, which it is not meant to handle. In particular the estimator has, perhaps a bit strongly stated, nothing to do with the pairwise covariance among 2n variables. In essence what you have done here is take a measure (not in the mathematical sense) designed for one situation, applied it to another and in a resulting equation such as $a+b=c$ forced $b=-a$.

This example shows nicely how dependence can cause bias. Disregarding bias, it tells us nothing about the usefulness of sample covariance and it's application. That being said, it may still be interesting to examine what a certain equation tells us about the nature of the data, as whuber's very nice answer and simulation shows. Again though, the particular example in the question does not have bearing on the usefulness or properties of sample covariance if bias is deemed uninteresting.

  • $\begingroup$ Thanks for the contribution. I see that you chose to focus on the issue of unbiasedness. The fact that the specific estimator would lose its unbiasedness property in such a set up was obvious from the beginning. What struck me as odd and worthy of some more investigation and thought was that its expected value becomes exactly zero. And I consider zero to be a special value, since in our number system is the only one that indicates non-existence, while all other numbers quantify what exists. $\endgroup$ Commented Mar 22, 2015 at 3:28
  • $\begingroup$ @AlecosPapadopoulos I will try to re-word and see if I can get my points across. That fact that there is a bias is not my point at all. My point is that I don't see anything interesting or surprising in that, whereas your question indicates that you do. What other number would have been more intuitive to see, and why? Maybe I can salvage this into something useful if I understand more what surprises you about 0, and why you think it may be a limitation of the tool in addition to bias, which you claim is obvious. $\endgroup$
    – KOE
    Commented Mar 22, 2015 at 11:04

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