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 $(X_2, X_3)$.