Given two not independent r.v. $X,Y$ such that $\mathbf{E}[X]=1$, ¿Is it true that $\mathbf{E}[XY]\approx \mathbf{E}[Y]$? And in that case, ¿can we stablish the rate of approximation?

I think that result is used by Efron and Tibshirani in section 9 of the article Bootstrap Methods for Standard Errors, Confidence Intervals, and Other Measures of Statistical Accuracy, when deriving an approximation of the coefficient of variation of the Bootstrap estimate of the standard error based on Monte Carlo simulation. I'll explain why. The formula is: $$ \mathbf{Cv}[\hat{\mathbf{se}}_{B}] \approx \sqrt{\mathbf{Cv}\left[\hat{\mathbf{se}}_{\infty}\right]^2+ \frac{1}{4B}(\mathbf{E}\left[\Delta^*\right]+2) } $$ Where $\hat{\mathbf{se}}_{B}$ is the Bootstrap estimate of the standard error of an statistic $\hat{\theta}$ computed from B Bootstrap samples, and $\hat{\mathbf{se}}_{\infty}$ is the ideal Bootstrap estimate of the standard error, which turns out to be the plug-in estimate of the standard error. $\Delta^*$ is the excess kurtosis of the Bootstrap distribution of the statistic $\hat{\theta}$. The randomness comes from two sources. One is given by the variability of a Bootstrap sample computed from a fixed sample and the other source comes from the variability of that original sample in the population. In other words, the original sample is not fixed.

After aplying law of total variance to $\mathbf{Var}[\hat{\mathbf{se}}_{B}]$ it is possible to get: $$ \mathbf{Var}[\hat{\mathbf{se}}_{B}] \approx \mathbf{Var}\left[\hat{\mathbf{se}}_{\infty}\right] + \mathbf{E}\left[\frac{\hat{\mathbf{se}}_{\infty}^2}{4B}(\Delta^*+2)\right] $$ Now, if we divide by $\mathbf{E}[\hat{\mathbf{se}}_{B}]^2$ and use the following approximation given by the law of total expecation: $\mathbf{E}[\hat{\mathbf{se}}_{B}]=\mathbf{E}[\mathbf{E}[\hat{\mathbf{se}}_{B}|\mathbf{X}=\mathbf{x}]] \approx \mathbf{E}[\hat{\mathbf{se}}_{\infty}]$ we obtain: $$ \mathbf{Cv}[\hat{\mathbf{se}}_{B}]^2 \approx \mathbf{Cv}\left[\hat{\mathbf{se}}_{\infty}\right]^2+ \frac{1}{4B}\mathbf{E}\left[\frac{\hat{\mathbf{se}}_{\infty}}{\mathbf{E}[\hat{\mathbf{se}}_{\infty}]}\cdot \frac{\hat{\mathbf{se}}_{\infty}}{\mathbf{E}[\hat{\mathbf{se}}_{\infty}]} \cdot (\Delta^*+2)\right] $$ So, in order to obtain the formula given in the article we must assume: $$ \mathbf{E}\left[\frac{\hat{\mathbf{se}}_{\infty}}{\mathbf{E}[\hat{\mathbf{se}}_{\infty}]}\cdot \frac{\hat{\mathbf{se}}_{\infty}}{\mathbf{E}[\hat{\mathbf{se}}_{\infty}]} \cdot (\Delta^*+2)\right] \approx \mathbf{E}\left[\Delta^*+2\right] $$ ¿Why is this assumption possible?

Thanks in advance.

Pd: $\hat{\mathbf{se}}_{\infty}$ is the plug-in estimate of the standard error of the sampling distribution of the statistic $\hat{\theta}$, so as plug-in estimate is a consistent estimator of the true standard error of the statistic, and hence its variance over all samples of size $n$ must ¿converge to zero?.

  • 2
    $\begingroup$ there's nothing special about 1, you can scale and shift any variable to make its expectation equal to 1 $\endgroup$
    – Aksakal
    Jun 23 at 1:26
  • $\begingroup$ There's nothing here to suggest a "rate" would make any sense. Given any value of $E[Y]$ and any value of $E[X],$ $E[XY]$ can literally be anything (even infinite). After all, the covariance of $(X,Y)$ is the difference $E[XY]-E[X]E[Y]$ and it's easy to make the covariance anything you like by suitably scaling one or both variables, whereas you can maintain their expectations with compensatory shifts of location. $\endgroup$
    – whuber
    Jun 23 at 20:31
  • $\begingroup$ Re the edit: I strongly, strongly doubt E&T are invoking any such general "result." Most likely in their setting they are supposing almost all the probability of $X$ is concentrated in a small neighborhood of $1.$ $\endgroup$
    – whuber
    Jun 23 at 20:37
  • $\begingroup$ Thanks for your comment. I re-edited again the question and explained the part of the bootstrap CV. Pd: I'm trying to derive that formula for my final degree proyect. $\endgroup$ Jun 23 at 21:07
  • 1
    $\begingroup$ That final expectation (and the cv) corresponds to the whole population. The inicial sample from which we construct the bootstrap samples is not fixed, so $\Delta^*$ is a random variable over al samples of sized $n$ of the population. In this conditions, $X=\frac{\hat{\mathbf{se}}_{\infty}^2}{\mathbf{E}[\hat{\mathbf{se}}_{\infty}]^2}, Y = \Delta^*+2$. Precisely, using first-order taylor i saw that $\mathbf{E}[X]\approx 1$. I know how to use taylor theorem over expectations but not over r.v. which is what (i think) you are suggesting. In fact, what you told in the last lines is the key. $\endgroup$ Jun 23 at 21:58


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