Conditional Expectations of Functions

Why Evaluate a Function Under a Conditional Law?

A conditional distribution describes how $X$ behaves once the value of $Y$ is known. Often the quantity of interest is not $X$ itself, but a function $g(X)$: for example, a deductible-adjusted payment, a rounded measurement, or a nonlinear score. The direct approach is to evaluate $g$ at each possible value of $X$, then average those results using the conditional probabilities or density of $X$ given $Y=y$.

This approach avoids an unnecessary intermediate step. One need not first derive the distribution of $g(X)$. The key is to use the conditional law of $X$, including its support and its weights, rather than the marginal law of $X$.

Definition: For a specified value $y$, the conditional expectation of $g(X)$ given $Y=y$ is the average of $g(X)$ under the conditional distribution of $X$ given $Y=y$. If $X$ is discrete, with conditional probability mass function $p_{X\mid Y}(x\mid y)$, then

E[g(X)\mid Y=y]=\sum_x g(x)p_{X\mid Y}(x\mid y).

If $X$ is continuous with conditional density $f_{X\mid Y}(x\mid y)$, then

E[g(X)\mid Y=y]=\int_{-\infty}^{\infty}g(x)f_{X\mid Y}(x\mid y)\,dx.

The sum or integral is over the conditional support. The expectation must be finite; a sufficient condition is that the corresponding conditional average of $|g(X)|$ is finite.

For a discrete conditioning variable, the conditional mass function is defined when $P(Y=y)>0$, by $p_{X\mid Y}(x\mid y)=P(X=x,Y=y)/P(Y=y)$. For continuous variables, conditional densities are obtained from the joint and marginal densities wherever the marginal density of $Y$ is positive. A conditional density at a particular $y$ is understood through the chosen conditional-density model; values on zero-probability conditioning points cannot be determined from ordinary conditioning on an event of positive probability alone.

Why the Weighted-Sum and Integral Formulas Hold

The formula is the ordinary definition of an average applied to the conditional law. Conditioning changes the weights assigned to the possible values of $X$, but it does not change how a function is evaluated: first calculate $g(x)$, then average those values with the conditional weights.

Discrete case.
Given $Y=y$, the mutually exclusive outcomes $X=x$ have conditional probabilities $p_{X\mid Y}(x\mid y)$, which sum to 1. The expected value of a quantity taking value $g(x)$ on outcome $X=x$ is therefore the sum of each value multiplied by its probability: $\sum_x g(x)p_{X\mid Y}(x\mid y)$.

Continuous case.
For a small interval of width $dx$ near $x$, the conditional probability is approximately $f_{X\mid Y}(x\mid y)\,dx$. The contribution to the average is approximately $g(x)f_{X\mid Y}(x\mid y)\,dx$. Summing these contributions over the support in the limit gives $\int g(x)f_{X\mid Y}(x\mid y)\,dx$.

Use the conditional law throughout.
The weights in either calculation must describe $X\mid Y=y$. Replacing them with marginal probabilities or a marginal density answers a different question.

The formulas work for functions that are nonlinear or piecewise-defined. The function may change the values being averaged, but there is no need to calculate the probability distribution of those changed values first. This is particularly useful when $g$ has thresholds, caps, or several pieces.

Worked Example: A Discrete Deductible-and-Limit Payment

A hypothetical policy has a deductible of $1,000 and a payment limit of $2,000 per claim. Let $X$ be the claim severity in thousands of dollars. Conditional on a particular risk class $Y=2$, suppose $X$ has the following conditional law: $P(X=0\mid Y=2)=0.25$, $P(X=2\mid Y=2)=0.50$, and $P(X=4\mid Y=2)=0.25$. Find the expected insurer payment in thousands of dollars. Assume there is no interest adjustment; no mortality assumption is relevant to this claim-payment calculation.

Method. Evaluate the payment function for each conditional value of $X$, then take the probability-weighted sum. With amounts measured in thousands, the payment per loss is $g(x)=\min(\max(x-1,0),2)$.

At $X=0$, the loss is below the deductible, so $g(0)=\min(0,2)=0$. At $X=2$, the amount above the deductible is 1, so $g(2)=\min(1,2)=1$. At $X=4$, the amount above the deductible is 3, but the limit caps payment at 2, so $g(4)=\min(3,2)=2$.

The conditional probabilities sum to $0.25+0.50+0.25=1$. Therefore,

\begin{aligned} E[g(X)\mid Y=2] &=0(0.25)+1(0.50)+2(0.25)\\ &=0+0.50+0.50\\ &=1.00. \end{aligned}

The expected payment is $1,000. This result lies between the smallest and largest possible payments, $0 and $2,000, as it should. The common mistake is to apply the deductible but forget the payment limit: that would assign a payment of 3, rather than 2, when $X=4$, producing an incorrect average of 1.25 thousand dollars.

Continuous Conditional Laws

For a continuous conditional law, the integral uses the conditional density and its conditional support. When $g$ is piecewise-defined, split the integral at every point where its formula changes. This is not a transformation-of-variables calculation: the goal is an average, not the density or distribution function of $g(X)$.

Worked Example: A Threshold Function Under a Conditional Uniform Law

Suppose that, for positive $y$, the conditional distribution of $X$ given $Y=y$ is uniform on $(0,y)$. Thus its conditional density is $f_{X\mid Y}(x\mid y)=1/y$ for $0<x<y$, and zero otherwise. Given $Y=4$, calculate $E[(X-1)_+\mid Y=4]$, where $(z)_+=\max(z,0)$.

Check the conditional law. At $y=4$, the density is $1/4$ on $0<x<4$, and its integral is $\int_0^4(1/4)\,dx=1$. The function is zero for $x\leq1$ and equals $x-1$ for $x>1$. Hence

\begin{aligned} E[(X-1)_+\mid Y=4] &=\int_0^4 (x-1)_+\frac14\,dx\\ &=\int_0^1 0\cdot\frac14\,dx+\int_1^4 (x-1)\frac14\,dx\\ &=\frac14\left[\frac{(x-1)^2}{2}\right]_{1}^{4}\\ &=\frac14\left(\frac{9}{2}-0\right)\\ &=\frac98=1.125. \end{aligned}

The conditional expectation is 1.125. The common mistake is to integrate $x-1$ over the full interval from 0 to 4. On $0<x<1$, the positive-part function is zero, not negative; the integral must respect that piece of the definition.

The conditional density may be nonconstant, so values of $X$ need not receive equal weight. A density is not itself a probability at a point; the probability of a region is found by integrating the density over that region. In an expectation integral, the density weights the function values in precisely this way.

Worked Example: A Nonlinear Function Under a Triangular Conditional Density

For $y>0$, suppose $X\mid Y=y$ has density $f_{X\mid Y}(x\mid y)=2x/y^2$ for $0<x<y$, and zero otherwise. Given $Y=2$, find $E[\sqrt{X}\mid Y=2]$. Also compare it with $\sqrt{E[X\mid Y=2]}$. There is no interest-rate or mortality component in this probability model.

Check the density and substitute the conditioning value. Its integral over the conditional support is $\int_0^y 2x/y^2\,dx=[x^2/y^2]_0^y=1$. At $y=2$, the density is $x/2$ for $0<x<2$. Apply the expectation integral directly:

\begin{aligned} E[\sqrt{X}\mid Y=2] &=\int_0^2 \sqrt{x}\frac{x}{2}\,dx\\ &=\frac12\int_0^2 x^{3/2}\,dx\\ &=\frac12\left[\frac{2}{5}x^{5/2}\right]_0^2\\ &=\frac{2^{5/2}}{5} =\frac{4\sqrt{2}}{5}\\ &\approx 1.1314. \end{aligned}

The comparison requires calculating the conditional mean of $X$ first:

\begin{aligned} E[X\mid Y=2] &=\int_0^2 x\frac{x}{2}\,dx\\ &=\frac12\left[\frac{x^3}{3}\right]_0^2 =\frac43,\\ \sqrt{E[X\mid Y=2]} &=\sqrt{\frac43} =\frac{2}{\sqrt{3}} \approx 1.1547. \end{aligned}

Rounded to four decimal places, the two quantities are 1.1314 and 1.1547, respectively, so they are not equal. The wrong calculation would replace $E[\sqrt{X}\mid Y=2]$ with $\sqrt{E[X\mid Y=2]}$. The correction is to apply the square-root function to each possible $X$ inside the expectation integral, as in the first calculation.

Checks and Common Exam Traps

Before finalizing a calculation, verify that the conditional law is the one requested, that its support is correct, and that the conditional probabilities sum to 1 or the conditional density integrates to 1. Then check that the function has been evaluated on every part of that support. In particular, thresholds and caps can change the formula for $g(x)$ within the support.

Using a marginal law. $E[g(X)\mid Y=y]$ uses the law of $X\mid Y=y$, not the unconditional law of $X$.
Applying the function after averaging. In general, $E[g(X)\mid Y=y]$ is not $g(E[X\mid Y=y])$. The worked nonlinear example shows this difference numerically.
Ignoring part of a piecewise function. Write down the intervals on which each formula applies before integrating.
Forgetting the conditional support. A density formula only applies on its stated support; integrating outside it can introduce incorrect contributions.
Failing to check existence. If the weighted sum or integral of $|g(X)|$ is infinite, the expectation need not be a finite number.

Key takeaway: To evaluate $E[g(X)\mid Y=y]$, obtain the conditional law of $X$ given $Y=y$, evaluate $g$ on that law's support, and take the conditional weighted sum or density integral. No distribution for $g(X)$ is needed.

Check Your Understanding

For each problem, calculate the requested conditional expectation directly from the stated conditional law.

Given $Y=1$, $X$ takes values $-1,0,2$ with conditional probabilities $0.20,0.30,0.50$, respectively. Find $E[(X+1)^2\mid Y=1]$. (Answer: 4.8)
A hypothetical claim count $N$, given a risk category $Y=3$, takes values $0,1,3$ with probabilities $0.40,0.40,0.20$. A payment is $g(N)=\max(N-1,0)$. Find the expected payment count given $Y=3$. (Answer: 0.4)
Given $Y=6$, $X$ is uniform on $(0,6)$. Find $E[(X-2)_+\mid Y=6]$. (Answer: $4/3$)
Given $Y=2$, $X$ has density $3x^2/8$ on $0<x<2$. Find $E[1/(1+X)\mid Y=2]$, to four decimal places. (Answer: $3\ln(3)/8\approx0.4120$)
Given $Y=5$, $X$ has density $2x/25$ on $0<x<5$. A hypothetical payment in thousands of dollars is $g(X)=\min(\max(X-1,0),2)$. Find its conditional expectation, in thousands of dollars. (Answer: $124/75\approx1.6533$ thousand dollars)

What You'll Learn