Suppose we have a function $g$ and two random variables $\tilde{X} = (\tilde{X}_1, \tilde{X}_2, \tilde{X}_3)$ and $X = (X_1, X_2, X_3)$ which are iid. Furthermore, $\tilde{X}_1, \tilde{X}_2, \tilde{X}_3$ and $X_1, X_2, X_3$ are independent random vectors. I am interested in the expectation $\mathbb{E}(g(\tilde{X}_1, X_2, X_3)) $.

According to Serfling, Robert J. (1980), I could use the V-statistic (or U-statistic) by defining the kernel$$ h(\tilde{X}, X) = h((\tilde{X}_1, \tilde{X}_2, \tilde{X}_3), (X_1, X_2, X_3)) := g(\tilde{X}_1, X_2, X_3)$$ and using $\mathbb{E}( h(\tilde{X}, X))$.

But I would like to do this differently, for example: \begin{align*} \mathbb{E}(g(\tilde{X}_1, X_2, X_3)) &= \mathbb{E}_{(X_2, X_3)}(\mathbb{E}_{\tilde{X_1} \mid (X_2, X_3)} (g(\tilde{X}_1, X_2, X_3)) \\ &= \mathbb{E}_{(X_2, X_3)}(\mathbb{E}_{\tilde{X_1}} (g(\tilde{X}_1, X_2, X_3)) \\ &= \mathbb{E}_{(X_2, X_3)}(\mathbb{E}_{X_1} (g(X_1, X_2, X_3)) \\ \end{align*}

My idea was to 1) use the law of total expectation (first line), then remove the condition due to independence (second line) and finally use$$ \mathbb{E}_{X_1} (g(X_1, X_2, X_3) = \mathbb{E}_{\tilde{X}_1} (g(\tilde{X}_1, X_2, X_3)$$ since $X_1$ and $\tilde{X}_1$ are iid.

Except that I get a biased estimation analogous to the V-statistic, does someone see a problem in my approach? Is it mathematically correct to do this?


This step is incorrect:$$ E(g(\widetilde{X}_1, X_2, X_3) \mid (X_2, X_3)) = E(g(\widetilde{X}_1, X_2, X_3)), $$ because $(\widetilde{X}_1, X_2, X_3)$ is not independent from $(X_2, X_3)$, although $\widetilde{X}_1$ is independent from $(X_2, X_3)$.

  • $\begingroup$ So the only way to solve this properly is using the V-Statistic (or U-statistic) ? $\endgroup$ – Giuseppe Apr 5 '18 at 13:02
  • 1
    $\begingroup$ @Giuseppe No need to be so complicated. Actually since $\widetilde X_1,X_2,X_3$ are independent and $\widetilde X_1\stackrel{\mathrm d}=X_1$, the product measure induced by $\widetilde X_1,X_2,X_3$ is$$μ=μ_{\widetilde X_1}×μ_{X_2}×μ_{X_3}=μ_{X_1}×μ_{X_2}×μ_{X_3},$$which implies\begin{align*}E(g(\widetilde X_1,X_2,X_3))&=\iiint g(\widetilde x_1,x_2,x_3)\,μ_{\widetilde X_1}(\mathrm d\widetilde x_1)μ_{X_2}(\mathrm dx_2)μ_{X_3}(\mathrm dx_3)\\&=\iiint g(x_1,x_2,x_3)\,μ_{X_1}(\mathrm dx_1)μ_{X_2}(\mathrm dx_2)μ_{X_3}(\mathrm dx_3)=E(g(X_1,X_2,X_3)).\end{align*} $\endgroup$ – Saad Apr 5 '18 at 13:15
  • $\begingroup$ So I can do \begin{array} \mathbb{E}(g(\tilde{X}_1, X_2, X_3)) &=& \int \int \int g(\tilde{x}_1, x_2, x_3) d\mu_{\tilde{X}_1}(\tilde{x}_1) d\mu_{X_2}(x_2) d\mu_{X_3}(x_3) \\ &=& \int \int g(\tilde{x}_1, x_2, x_3) d\mu_{\tilde{X}_1}(\tilde{x}_1) d\mu_{X_2 \times X_3}\\ &=& \mathbb{E}_{(X_2, X_3)}(\mathbb{E}_{\tilde{X}_1} (g(\tilde{X}_1, X_2, X_3)) = \mathbb{E}_{(X_2, X_3)}(\mathbb{E}_{X_1} (g(X_1, X_2, X_3)) \end{array} and use the estimate $\frac 1 n \sum_{k=1}^n \frac 1 n \sum_{i=1}^n g(x_{i,1}, x_{k,2}, x_{k,3})$ (= V-statistic Estimator) if I don't have samples for $\tilde{X}_1$. $\endgroup$ – Giuseppe Apr 5 '18 at 14:10

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