Conditioning on a Function of a Variable

What Does the Transformation Reveal?

Suppose we observe $T=g(X)$, rather than $X$ itself. Knowing $T$ tells us which values of $X$ are possible, but it may not distinguish between them. For example, if $T=|X|$, then observing $T=2$ tells us that $X$ is either $2$ or $-2$, if both values are possible. The conditional mean of $X$ given $T=2$ is therefore an average over those possibilities.

The average is generally weighted: values of $X$ that are more probable contribute more. The weights come from the distribution of $X$, not just from counting how many preimages a transformation has. This principle applies to discrete variables, continuous variables, and coarsened categories.

Definition: Let $X$ be a random variable, let $g$ be a measurable function, and set $T=g(X)$. For an integrable random variable $h(X)$, $E[h(X)\mid T]$ denotes the conditional expectation of $h(X)$ given the information in $T$, that is, given the sigma-field generated by $T$. It is a random variable that can be written as a measurable function of $T$, up to almost-sure equality. A value such as $E[h(X)\mid T=t]$ refers to a version of that function evaluated at $t$; for continuous $T$, this value is specified only for almost every $t$ under the distribution of $T$.

The Discrete Averaging Formula

First suppose $X$ takes values in a finite or countable set $\{x_i\}$, with $P(X=x_i)=p_i$. For an output $t$ of the transformation, the values compatible with $T=t$ are exactly those $x_i$ for which $g(x_i)=t$. Conditional expectation averages $h(x_i)$ over that set, using the conditional probabilities of those values.

Theorem: Suppose $E[|h(X)|]<\infty$, and let $T=g(X)$. For any $t$ such that $P(T=t)>0$, $$ E[h(X)\mid T=t] = \frac{\displaystyle\sum_{i:g(x_i)=t}h(x_i)p_i} {\displaystyle\sum_{i:g(x_i)=t}p_i}. $$ Consequently, $E[h(X)\mid T]$ is the function of $T$ given by this ratio on outputs with positive probability. Its values on outputs with probability zero do not affect the conditional expectation.

To verify the formula, fix an output $t$ with positive probability. The event $\{T=t\}$ consists of the disjoint events $\{X=x_i\}$ for which $g(x_i)=t$, so $$ P(T=t)=\sum_{i:g(x_i)=t}p_i. $$ For each such $i$, $P(X=x_i\mid T=t)=p_i/P(T=t)$. Multiplying $h(x_i)$ by this conditional probability and summing gives the stated ratio. The sums involving $h$ are well-defined because $E[|h(X)|]<\infty$. This also shows that the resulting function of $T$ has the defining conditional-average property on each event $\{T=t\}$. $\square$

Worked Example: Conditioning on Absolute Value in a Discrete Model

Let $X$ take values $-2,-1,1,2$ with probabilities $0.1,0.2,0.3,0.4$, respectively, and let $T=|X|$. Find $E[X\mid T]$.

If $T=1$, the compatible values are $-1$ and $1$, and their total probability is $0.2+0.3=0.5$. Thus $$ E[X\mid T=1] =\frac{(-1)(0.2)+(1)(0.3)}{0.5} =\frac{0.1}{0.5} =0.2. $$

If $T=2$, the compatible values are $-2$ and $2$, again with total probability $0.1+0.4=0.5$. Therefore $$ E[X\mid T=2] =\frac{(-2)(0.1)+(2)(0.4)}{0.5} =\frac{0.6}{0.5} =1.2. $$

Since $T$ takes only the values $1$ and $2$, the answer can be written as $E[X\mid T]=0.2$ when $T=1$, and $E[X\mid T]=1.2$ when $T=2$. The positive answers reflect the greater probability assigned to the positive values in each pair.

Worked Example: Conditioning on an Indicator

Let $X$ take values $-2,0,1,3$ with probabilities $0.2,0.1,0.4,0.3$. Define $T=\mathbf{1}_{\{X>0\}}$, which records whether $X$ is positive. Find $E[X\mid T]$.

When $T=1$, $X$ is either $1$ or $3$, and $P(T=1)=0.4+0.3=0.7$. Hence $$ E[X\mid T=1] =\frac{1(0.4)+3(0.3)}{0.7} =\frac{1.3}{0.7} =\frac{13}{7}. $$

When $T=0$, $X$ is either $-2$ or $0$, and $P(T=0)=0.2+0.1=0.3$. Thus $$ E[X\mid T=0] =\frac{(-2)(0.2)+0(0.1)}{0.3} =\frac{-0.4}{0.3} =-\frac{4}{3}. $$

The indicator retains only the information “positive” or “not positive.” Within each group, the conditional mean is the probability-weighted average of the original values.

Worked Example: Replacing Values by Category Labels

Suppose $X$ takes values $0,1,2,3$ with probabilities $0.1,0.2,0.3,0.4$. Let $C$ be the category “low” when $X\leq 1$ and “high” when $X\geq 2$. Find $E[X\mid C]$ in each category.

The low category has probability $0.1+0.2=0.3$. Its conditional mean is $$ E[X\mid C=\text{low}] =\frac{0(0.1)+1(0.2)}{0.3} =\frac{2}{3}. $$ The high category has probability $0.3+0.4=0.7$, so $$ E[X\mid C=\text{high}] =\frac{2(0.3)+3(0.4)}{0.7} =\frac{1.8}{0.7} =\frac{18}{7}. $$

A category label is simply another transformation: it groups original values into sets. The conditional mean averages within each set using the probabilities of its members.

Continuous Variables: Weighting the Inverse Branches

For a continuous $X$, individual values usually have probability zero, so the discrete formula cannot be applied by assigning probabilities to points. Instead, density determines the relative weight of each possible preimage. When $g$ is many-to-one, a given output $t$ may have several inverse branches $x_j(t)$. A branch receives weight proportional to the density of $X$ there and to the change-of-variables factor $1/|g'(x_j(t))|$.

Theorem: Suppose $X$ has density $f_X$, $E[|h(X)|]<\infty$, and the support of $X$, apart from a set of probability zero, can be divided into finitely many intervals on each of which $g$ is continuously differentiable and strictly monotone, with $g'(x)\ne 0$. For almost every $t$, let $x_1(t),\ldots,x_k(t)$ be the inverse-branch values in the support for which $g(x_j(t))=t$, and define $$ w_j(t)=\frac{f_X(x_j(t))}{|g'(x_j(t))|}. $$ At outputs where $\sum_{j=1}^k w_j(t)>0$, a version of the conditional mean is $$ E[h(X)\mid T=t] = \frac{\displaystyle\sum_{j=1}^k h(x_j(t))w_j(t)} {\displaystyle\sum_{j=1}^k w_j(t)}, \qquad T=g(X). $$ This expression is defined for almost every $t$ with respect to the distribution of $T$.

To justify the formula, apply change of variables separately on each monotone interval. For any bounded measurable function $a$, $$ E[h(X)a(T)] =\int h(x)a(g(x))f_X(x)\,dx =\int a(t)\sum_{j=1}^k h(x_j(t))w_j(t)\,dt. $$ The same change of variables with $h(x)=1$ shows that $T$ has density $f_T(t)=\sum_{j=1}^k w_j(t)$ almost everywhere. Define $m(t)$ by the displayed ratio wherever $f_T(t)>0$, and choose its value arbitrarily where $f_T(t)=0$. Then $$ E[m(T)a(T)] =\int a(t)m(t)f_T(t)\,dt =\int a(t)\sum_{j=1}^k h(x_j(t))w_j(t)\,dt =E[h(X)a(T)]. $$ These identities hold for all bounded measurable $a$, and $E[|h(X)|]<\infty$ ensures the relevant integrals exist. Thus $m(T)$ has the defining conditional-expectation property given $T$, which proves the formula. $\square$

Worked Example: Conditioning on the Absolute Value of a Continuous Variable

Let $X$ have density $f_X(x)=2(x+1)/9$ for $-1<x<2$, and zero elsewhere. This is a density because $$ \int_{-1}^{2}\frac{2(x+1)}{9}\,dx =\frac{[(x+1)^2]_{-1}^{2}}{9} =\frac{9}{9} =1. $$ Set $T=|X|$ and find $E[X\mid T=t]$.

For $0<t<1$, both inverse branches, $x=t$ and $x=-t$, lie in the support. Since the absolute-value transformation has derivative with absolute value $1$ on each branch, their weights are $$ w_+(t)=f_X(t)=\frac{2(t+1)}{9}, \qquad w_-(t)=f_X(-t)=\frac{2(1-t)}{9}. $$ Therefore $$ E[X\mid T=t] =\frac{t\,w_+(t)+(-t)\,w_-(t)}{w_+(t)+w_-(t)} =t\frac{(t+1)-(1-t)}{(t+1)+(1-t)} =t^2. $$

For $1<t<2$, the negative branch $-t$ is outside the support, so only $x=t$ is possible. Consequently $E[X\mid T=t]=t$. The cases $t=1$ and the endpoints do not affect the result, since they have probability zero under the continuous distribution of $T$. The complete answer, almost everywhere, is $t^2$ for $0<t<1$ and $t$ for $1<t<2$.

Common Pitfall: Averaging Preimages Without Weights

A tempting shortcut is to average the possible values of $X$ equally whenever they map to the same $T$. That is correct only when those possibilities have equal conditional probabilities. In the discrete absolute-value example, the two values mapping to $T=1$ are $-1$ and $1$, but their probabilities are $0.2$ and $0.3$. Their unweighted average is $0$, whereas the correct conditional mean is $0.2$.

The same issue appears for continuous transformations. The weights depend not only on $f_X$ at the inverse points but also on the derivative factor $1/|g'|$. A second mistake is to assume that conditioning on $|X|$ always gives a mean of zero. That conclusion holds in symmetric settings where the positive and negative possibilities balance; it does not follow from the transformation alone. In the continuous example above, the conditional mean is positive because the density favors positive values.

Key takeaway: To condition on $T=g(X)$, identify all original values compatible with the observed output, then average using their conditional probabilities. In discrete problems the weights are point probabilities; in smooth continuous problems they are density and change-of-variables weights. The resulting conditional mean is a function of $T$, determined up to almost-sure equality.

Check Your Understanding

Work through these questions by identifying the values or inverse branches compatible with each observed transformed value.

If $X$ takes values $0,1,2$ with probabilities $0.2,0.5,0.3$ and $T=\mathbf{1}_{\{X=2\}}$, what is $E[X\mid T=0]$?
For a discrete $X$, why must the conditional mean given $g(X)=t$ use the probabilities of the values in that preimage?
If $X$ has a symmetric distribution about zero, under what conditions would symmetry imply $E[X\mid |X|]=0$?
For a smooth transformation with two inverse branches, what two factors determine each branch’s weight in the continuous formula?
Why is a conditional mean at a single output value not uniquely determined when that output has probability zero?