Expectation of Functions of Several Random Variables

In statistics, computing the mean of a test statistic that is a nonlinear function of several sample values (e.g., a sample range, a ratio of sums of squares, an $F$-statistic numerator) is done via multivariate LOTUS applied to the joint density of the order statistics or sufficient statistics, avoiding the need to derive their full sampling distributions. In finance and actuarial science, expected payoffs of options on multiple correlated assets (basket options, rainbow options paying $\max$ of several stock prices) are computed as integrals of the payoff function against the joint (often multivariate lognormal) density of asset prices — this is the entire content of risk-neutral pricing for path-independent multi-asset derivatives. In machine learning, the expected loss (risk) of a model is $E[\ell(Y, \hat f(X))]$, a direct application of multivariate LOTUS to the joint distribution of features $X$ and label $Y$; empirical risk minimization approximates this integral by an average over data. In statistical physics, expectation values of observables that depend jointly on several particle coordinates or momenta (e.g., pair correlation functions, virial expressions) are computed by integrating the observable against the joint Boltzmann density — a direct physical instance of the theorem. In information theory, quantities like mutual information involve $E\!\left[\log \frac{p(X,Y)}{p_X(X)p_Y(Y)}\right]$, computed by LOTUS applied to the joint law of $(X,Y)$. In queueing theory and reliability engineering, expected system lifetimes for series/parallel configurations use $g=\min$ or $g=\max$ of component lifetimes, evaluated by joint-density LOTUS rather than by first deriving the lifetime distribution of the whole system. And linearity of expectation, needing no independence, is the workhorse behind combinatorial "expected number of X" arguments throughout discrete mathematics and randomized algorithm analysis (e.g., expected number of fixed points of a random permutation, expected number of inversions), where indicator variables are summed even though they are highly dependent.

:::quiz[If $X,Y$ have joint pmf and $g(x,y)=x-y$, which formula computes $E[g(X,Y)]$ using 2-D LOTUS?]

• Find the distribution of $Z=X-Y$ first, then apply 1-D LOTUS to $Z$
• $\sum_x\sum_y (x-y)\,p(x,y)$, summing directly over the joint pmf ✓
• $E[X]\cdot E[-Y]$ regardless of dependence
• $\max(x,y) - \min(x,y)$ averaged over the marginals only

:::quiz[Linearity of expectation $E[X+Y]=E[X]+E[Y]$ requires which of the following?]

• $X$ and $Y$ must be independent
• $X$ and $Y$ must be uncorrelated
• $X$ and $Y$ must have the same distribution
• $E[X]$ and $E[Y]$ must both exist (be finite); no independence needed ✓

:::quiz[In Example 1 of this article, $E[\max(X,Y)]$ was computed to be]

• 4.10, the same as $E[X+Y]$
• 2.90, obtained directly from the joint pmf via LOTUS ✓
• $\max(E[X],E[Y]) = \max(2.70,1.40) = 2.70$
• Undefined, since $\max$ is not a linear function

:::quiz[For the density $f(x,y)=6xy^2$ on the unit square, $E[X^2Y]$ was found to equal]

• $\frac{1}{2}$
• $\frac{3}{4}$
• $\frac{3}{8}$, matching both direct LOTUS integration and the independence shortcut $E[X^2]E[Y]$ ✓
• $1$

:::faq[Why doesn't $E[\max(X,Y)] = \max(E[X], E[Y])$?] Because $\max$ is not a linear function of its arguments, and LOTUS only reduces to the sum of marginal expectations when $g$ is linear (i.e., $g(x,y)=ax+by+c$). For nonlinear $g$ — max, min, product, ratio, square — the correct value must be computed by weighting $g(x,y)$ against the full joint distribution, as shown in Example 1, where $E[\max(X,Y)]=2.90$ while $\max(E[X],E[Y])=\max(2.70,1.40)=2.70$: close in that example but not equal, and in general the gap can be arbitrarily large depending on the joint distribution's shape.

Ask the tutor: How does multivariate LOTUS generalize to computing $E[g(X,Y)]$ when $(X,Y)$ has a joint distribution with a singular component, or when $g$ is only piecewise continuous?

The ordinary law of the unconscious statistician for a single variable says $E[g(X)] = \sum_x g(x)p(x)$: you weight each output value $g(x)$ by the probability of the input $x$ that produced it, without ever building the distribution of $g(X)$ itself. The two-dimensional version is the same idea lifted to a plane. Picture the joint mass function $p(x,y)$ as a collection of point-masses scattered over a lattice of points in the plane, with the mass at $(x,y)$ equal to $p(x,y)$. To find $E[g(X,Y)]$, do not project this mass down onto the (harder-to-describe) axis of $z = g(x,y)$; instead simply weight $g(x,y)$ by the mass sitting at $(x,y)$ and sum. This is exactly analogous to finding the center of mass of a physical system: if masses $m_i$ sit at positions $(x_i,y_i)$ in the plane, and you want the average value of some quantity $g(x_i,y_i)$ (say, gravitational potential) over the system, you do not need to know the distribution of potential values — you sum $g(x_i,y_i)\, m_i$ directly over the known positions and masses. Linearity of expectation is the special case $g(x,y)=x+y$: the weighted sum $\sum_{x,y}(x+y)p(x,y)$ splits into $\sum_x x\left(\sum_y p(x,y)\right) + \sum_y y \left(\sum_x p(x,y)\right) = \sum_x x\, p_X(x) + \sum_y y\, p_Y(y) = E[X]+E[Y]$, using only the marginal distributions — dependence between $X$ and $Y$ never enters. That is the structural reason linearity needs no independence: the marginals alone determine $E[X]+E[Y]$, no matter how $X$ and $Y$ are correlated.

Example 1 — expected maximum of two dependent discrete variables, via 2-D LOTUS directly (no distribution of $Z$ built). Let $(X,Y)$ have the joint pmf below, deliberately asymmetric so no shortcut hides an error:

Check this is a valid pmf: summing all nine entries,

We want $E[\max(X,Y)]$. By 2-D LOTUS with $g(x,y)=\max(x,y)$, we compute $\max(x,y)$ at each of the nine cells and weight by the cell probability — no need to find $P(\max(X,Y)=z)$ for each $z$ first (though we will cross-check by doing exactly that). Cell-by-cell values of $g(x,y) = \max(x,y)$:

Now apply LOTUS:

Compute term by term:

Sum: $0.10+0.20=0.30$; $+0.15=0.45$; $+0.30=0.75$; $+0.10=0.85$; $+0.30=1.15$; $+0.25=1.40$; $+0.50=1.90$; $+1.00=2.90$.

Cross-check by building the distribution of $Z=\max(X,Y)$ the long way. Group cells by $z$-value: $z=1$ occurs at $(1,0),(1,1)$ with probability $0.10+0.20=0.30$; $z=2$ occurs at $(2,0),(2,1)$ with probability $0.15+0.05=0.20$; $z=3$ occurs at $(1,3),(2,3)$ with probability $0.05+0.10=0.15$; $z=5$ occurs at $(5,0),(5,1),(5,3)$ with probability $0.05+0.10+0.20=0.35$. Check: $0.30+0.20+0.15+0.35=1.00.\checkmark$ Then

matching the direct LOTUS computation, as the theorem guarantees. This confirms LOTUS saves the intermediate step: we got $2.90$ without ever tabulating $P(Z=z)$. As a bonus, linearity gives the marginals for free: $p_X(1)=0.35, p_X(2)=0.30, p_X(5)=0.35$ (each row sum), so

and $p_Y(0)=0.30, p_Y(1)=0.35, p_Y(3)=0.35$ (each column sum), so

Then $E[X]+E[Y] = 2.70+1.40=4.10$, which is not equal to $E[\max(X,Y)]=2.90$ — as expected, since $\max$ is nonlinear. But by linearity we do get for free, e.g., $E[X+Y] = 4.10$ directly, matching a LOTUS check with $g(x,y)=x+y$: $\sum(x+y)p(x,y) = 1(0.10)+2(0.20)+4(0.05)+2(0.15)+3(0.05)+5(0.10)+5(0.05)+6(0.10)+8(0.20) = 0.10+0.40+0.20+0.30+0.15+0.50+0.25+0.60+1.60 = 4.10.\checkmark$

Example 2 — a continuous, non-uniform joint density with a genuinely 2-D integral for $E[g(X,Y)]$. Let $(X,Y)$ have joint density

First confirm this is a valid density:

Note $X$ and $Y$ are independent here since $f(x,y) = (2x)(3y^2)$ factors into valid marginal densities $f_X(x)=2x$ on $(0,1)$ and $f_Y(y) = 3y^2$ on $(0,1)$ — chosen deliberately asymmetric (different exponents) so the calculation is non-trivial, though the LOTUS machinery below does not use independence at all.

Task: find $E[X^2 Y]$ directly via 2-D LOTUS, i.e. $g(x,y) = x^2y$, without first finding the distribution of the product $X^2Y$ (which would be considerably harder — that ratio-type distribution has no elementary closed form on its own). By the continuous LOTUS,

Integrate over $y$ first:

Then integrate over $x$:

So

Cross-check using independence (a shortcut only available because $X\perp Y$ here): since $X$ and $Y$ are independent, so are $X^2$ and $Y$ (functions of independent variables are independent), hence $E[X^2Y] = E[X^2]\,E[Y]$. Compute each marginal moment directly:

Then $E[X^2]E[Y] = \frac12\cdot\frac34 = \frac38 = 0.375$, matching the direct 2-D LOTUS computation exactly. This illustrates that the product shortcut $E[X^2Y]=E[X^2]E[Y]$ is a consequence of independence and LOTUS together — it is not automatic and would fail without independence, unlike linearity of sums which needs no such assumption.