Expected value of a random variable drawn from joint distribution, separated into parts

Question

Suppose we have three random variables $X$, $Y$, and $Z$ that are drawn from a joint distribution $F(X,Y,Z)$ with joint density $f(X,Y,Z)$. I would like to write out the expected value of $X$ separated into three terms corresponding to when $X$ is largest, when $Y$ is largest, and when $Z$ is largest. Would the following be correct?

$$ E[X] = $$ $$\int_{-\infty}^{\infty}\int_{-\infty}^{X}\int_{-\infty}^{X}Xf(X,Y,Z)dZdYdX$$ $$+ \int_{-\infty}^{\infty}\int_{-\infty}^{Y}\int_{-\infty}^{Y}Xf(X,Y,Z)dZdXdY$$ $$+\int_{-\infty}^{\infty}\int_{-\infty}^{Z}\int_{-\infty}^{Z}Xf(X,Y,Z)dYdXdZ$$

If it is correct, could you suggest how I would go about proving it?

Answer 1

Assuming that the tri-variate joint density function $f_{X,Y,Z}(x,y,z)$ is integrable, integrate over both $x$ and $y$ (from $-\infty$ to $\infty$) to get the marginal density $f_X(x)$ of $X$. then the expectation of $X$ is $E(X) = \int_{-\infty}^\infty xf_X(x)\,dx.$

Answer 2

Yes, it's (almost*) correct. What you have is the iterated expectation:

$$\begin{align} \mathsf E(X) = & \quad\; \mathsf E(X\mid Y<X>Z)\,\mathsf P(Y<X>Z) \\[0ex] & + \mathsf E(X\mid X<Y>Z)\,\mathsf P(X<Y>Z) \\[0ex] & + \mathsf E(X\mid X<Z>Y)\,\mathsf P(X<Z>Y) \\[2ex] = & \quad\; \int_{-\infty}^{\infty}\int_{-\infty}^x\int_{-\infty}^x x\;f(x,y,z)\operatorname d z\operatorname d y\operatorname d x \\[0ex] & + \int_{-\infty}^{\infty}\int_{-\infty}^y\int_{-\infty}^y x\;f(x,y,z)\operatorname d z\operatorname d x\operatorname d y \\[0ex] & + \int_{-\infty}^{\infty}\int_{-\infty}^z\int_{-\infty}^z x\;f(x,y,z)\operatorname d x\operatorname d y\operatorname d z \end{align}$$

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(* Don't use capital letters for integration variables, reserve them for random variables.)