[This question is cross-posted on math SE here ]

Suppose I have two iid streams of data that are independent of each other: $X = (X_1, X_2, \ldots)$ and $Y = (Y_1, Y_2, \ldots)$. I want to estimate the difference in means between the two groups. Between the two streams, I want to sample a total of $n$ points. For notation's sake, say I'm sampling one point per unit of time for $T$ total time units.

Now consider the following sampling scheme which divides up the $T$ time period into two halves:

  • Up until time $t = T/2$, sample from $X$ and $Y$ with equal probability.
  • From $t = (T/2+1)$ until $T$, sample from $X$ with probability $p$ and from $Y$ with probability $1-p$, where $p$ is some function of the data I observed in the first half.

Now consider $\hat{\theta}_1 := \bar{X}_1 - \bar{Y}_1$, the difference in sample means calculated from only the data collected up until time $t=T/2$ and $\hat{\theta}_2 := \bar{X}_2 - \bar{Y}_2$ calculated from only the data collected from time $t=(T/2+1)$ to $t=T$.

Question: Without knowing more about how $p$ depends on the data in the first half, can we tell whether $\hat{\theta}_1$ and $\hat{\theta}_2$ are correlated?

Obviously, $\hat{\theta}_1$ and $\hat{\theta}_2$ are not independent, but nevertheless I thought they would be uncorrelated. My reasoning was that the dependence of $p$ only affects the allocation between $X$ and $Y$, and doesn't introduce any bias as far as the expected value of $\bar{X} - \bar{Y}$. I feel like I oversimplified this, but I'm a bit stuck as to how to work this out rigorously.

  • $\begingroup$ This is not very rigorous, but under the assumptions that $E[Y]$ and $E[X]$ are stationary, and * $E[X] - E[Y] \neq 0$, then it should hold that $\hat{\theta}_1 = c \cdot \hat{\theta}_2$, where $c = p \frac{E[X] + E[Y]}{E[X] - E[Y]} + E[Y]$. Note that $c$ is a linear relationship, hence suggesting that there can be a non-zero correlation. $\endgroup$ – NofP Oct 21 '18 at 23:36
  • $\begingroup$ Did you prove that "Obviously, $\hat{\theta}_1$ and $\hat{\theta}_2$ are not independent"? $\endgroup$ – user158565 Oct 22 '18 at 2:25
  • 1
    $\begingroup$ @a_statistician I don't know if this is a proof, but the variance of $\hat{\theta}_2$ depends on $p$, so if $p$ depends in turn on $\hat{\theta}_1$ (or the data used to generate it) then that clearly implies that they are dependent, no? $\endgroup$ – gogurt Oct 22 '18 at 4:09
  • $\begingroup$ How $Var(\hat \theta_2)$ depends on $p$? $\endgroup$ – user158565 Oct 22 '18 at 14:39

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