I know that $E(X_1) = E[E(X_1 | X_2)]$, but I’m wondering if I can generalize this to $E[g(X_1,…,X_n)] = E[E(g(X_1,…,X_n) | X_2,…X_n)]$ based on the following:

$$E(g(X_1,…,X_n)) = \int_{-\infty}^{\infty}\dots\int_{-\infty}^{\infty} g(x_1,…,x_n)f(x_1,\dots,x_n)dx_1\dots dx_n$$

$$= \int_{-\infty}^{\infty}\dots\int_{-\infty}^{\infty} g(x_1,…,x_n)f(x_1 | x_2,\dots,x_n)f(x_2,\dots,x_n)dx_1\dots dx_n$$

$$= \int_{-\infty}^{\infty}\dots\int_{-\infty}^{\infty} g(x_1,…,x_n)f(x_1 | x_2,\dots,x_n)dx_1 f(x_2,\dots,x_n)dx_2\dots dx_n$$

$$= \int_{-\infty}^{\infty}\dots\int_{-\infty}^{\infty} E(g(X_1,…,X_n) | X_2 = x_2, \dots, X_n = x_n) f(x_2,\dots,x_n)dx_2\dots dx_n$$

$$= E[E(g(X_1,…,X_n) | X_2, \dots X_n)]$$

Is this correct?


1 Answer 1


Yes, that equation holds: The law of iterated expectation works for functions of random vectors, as well as random variables. Thus, you can let $\mathbf{X}_* = (X_2,...,X_n)$ and you then have:

$$\begin{equation} \begin{aligned} \mathbb{E}(g(X_1,...,X_n)) &= \mathbb{E}(g(X_1,\mathbf{X}_*)) \\[6pt] &= \mathbb{E}(\mathbb{E}(g(X_1,\mathbf{X}_*)|\mathbf{X}_*)) \\[6pt] &= \mathbb{E}(\mathbb{E}(g(X_1,...,X_n)|X_2,...,X_n)). \\[6pt] \end{aligned} \end{equation}$$


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