Why transform a conditional law?
A conditional law describes how a random element is distributed after information about another random element has been specified. Often the quantity of interest is not the original random element itself, but a measurable function of it: a square, a threshold indicator, a clipped measurement, or a statistic formed from several coordinates.
The central operation is direct: apply the function to the conditional law. This preserves the conditioning information and transforms only the random element whose law is being described. The resulting formula works for discrete, continuous, and mixed distributions, without requiring a density.
Pushforward of a probability measure
This definition asks how much \(\mu\)-probability is assigned to the set of inputs that the map sends into \(B\). The preimage \(f^{-1}(B)\) is measurable because \(f\) is measurable. The definition applies whether or not \(f\) is one-to-one: when several inputs map to the same output, their probabilities are combined.
Transforming a conditional kernel
Let \(Z\) be a random element of a measurable space \((S,\mathcal S)\), and let \(X\) be a random element of \((E,\mathcal E)\). Suppose \(K\) is a conditional law of \(X\) given \(Z\). This means that \(K\) is a probability kernel from \(S\) to \(E\), and, for every \(A\in\mathcal E\), $$ K(Z,A)=\mathbb E[\mathbf 1_{\{X\in A\}}\mid\sigma(Z)] \quad\text{almost surely}. $$ Assume such a kernel is available; standard Borel state spaces are the usual setting in which regular conditional laws are available.
The formula keeps the conditioning variable \(Z\) unchanged. Only the values of \(X\) are mapped through \(f\). For each fixed measurable event \(B\), the conditional-law identity holds almost surely; as usual, one should not infer that a chosen version has a uniquely determined value at every possible conditioning state.
Conditional integrals of transformed quantities
This integral form is useful when the desired answer is a conditional mean, rather than a whole conditional distribution. For example, choosing \(g(y)=y\), when integrable, gives the conditional expectation of \(f(X)\) by integrating \(f\) against the original conditional kernel.
Worked Example: A discrete map merges outcomes
Worked Example: A discrete map merges outcomes
Suppose \(Z\) takes values \(0\) and \(1\), each with positive probability. Given \(Z=z\), let \(X\) take values \(-2,2,1\) with the following probabilities:
| Condition | \(P(X=-2\mid Z=z)\) | \(P(X=2\mid Z=z)\) | \(P(X=1\mid Z=z)\) |
|---|---|---|---|
| \(z=0\) | \(1/2\) | \(1/3\) | \(1/6\) |
| \(z=1\) | \(1/4\) | \(1/4\) | \(1/2\) |
Take \(f(x)=x^2\), so \(Y=f(X)\) takes values \(4\) and \(1\). The event \(Y=4\) has preimage \(\{-2,2\}\), while \(Y=1\) has preimage \(\{1\}\). Therefore, when \(Z=0\), $$ P(Y=4\mid Z=0)=\frac12+\frac13=\frac56, \qquad P(Y=1\mid Z=0)=\frac16. $$ When \(Z=1\), $$ P(Y=4\mid Z=1)=\frac14+\frac14=\frac12, \qquad P(Y=1\mid Z=1)=\frac12. $$ Each row sums to one: \(5/6+1/6=1\) and \(1/2+1/2=1\). The two distinct inputs \(-2\) and \(2\) contribute to the same transformed outcome. This is why probabilities are added over the entire preimage, rather than calculated from just one input.
The conditional means also illustrate the integral formula. Given \(Z=0\), \(\mathbb E[Y\mid Z=0]=4(5/6)+1(1/6)=21/6=7/2\). Given \(Z=1\), \(\mathbb E[Y\mid Z=1]=4(1/2)+1(1/2)=5/2\).
Worked Example: Squaring a conditional Gaussian
Worked Example: Squaring a conditional Gaussian
Suppose \(Z\) is real-valued and, given \(Z=z\), \(X\) has a normal distribution with mean \(m(z)\) and standard deviation \(\sigma(z)>0\). Assume \(m\) and \(\sigma\) are measurable. For \(Y=X^2\), the conditional distribution function is zero when \(t<0\). For \(t\geq 0\), the event \(Y\leq t\) is the interval event \(-\sqrt t\leq X\leq\sqrt t\). Writing \(\Phi\) for the standard normal distribution function gives $$ P(Y\leq t\mid Z=z) = \Phi\left(\frac{\sqrt t-m(z)}{\sigma(z)}\right) - \Phi\left(\frac{-\sqrt t-m(z)}{\sigma(z)}\right), \qquad t\geq 0. $$ In particular, \(P(Y=0\mid Z=z)=0\), since a normal distribution with positive variance assigns zero probability to the singleton \(\{0\}\).
For \(y>0\), the two inverse branches of \(x\mapsto x^2\) are \(x=\sqrt y\) and \(x=-\sqrt y\). The derivative of each inverse branch has absolute value \(1/(2\sqrt y)\). If \(\phi\) is the standard normal density, change of variables on the positive and negative half-lines yields the conditional density $$ p_{Y\mid Z=z}(y) = \frac{1}{2\sigma(z)\sqrt y} \left[ \phi\left(\frac{\sqrt y-m(z)}{\sigma(z)}\right) + \phi\left(\frac{-\sqrt y-m(z)}{\sigma(z)}\right) \right], \qquad y>0. $$ Both terms are needed because both square roots can produce the same value of \(Y\). The density is supported on the positive half-line, and there is no additional atom at zero.
Worked Example: Clipping a conditional uniform variable
Worked Example: Clipping a conditional uniform variable
Let \(Z\) take values in \((0,\infty)\), and suppose \(X\) given \(Z=z\) is uniform on \([0,z]\). Define \(Y=f(X)=\min(X,1)\). If \(0<z\leq1\), then \(X\leq1\) almost surely and \(Y=X\), so \(Y\) is uniform on \([0,z]\).
If \(z>1\), then every \(x\in[0,1)\) is unchanged, while all \(x\in[1,z]\) map to the single point \(1\). For \(0\leq y<1\), $$ P(Y\leq y\mid Z=z)=P(X\leq y\mid Z=z)=\frac{y}{z}. $$ The remaining probability is concentrated at \(1\): $$ P(Y=1\mid Z=z)=P(X\geq1\mid Z=z)=\frac{z-1}{z}. $$ Thus the transformed conditional law has density \(1/z\) on \([0,1)\) and an atom of mass \((z-1)/z\) at \(1\). The continuous part has total mass \(1/z\), so the total mass is \(1/z+(z-1)/z=1\). This example shows that a deterministic measurable map can turn a continuous conditional distribution into a mixed one.
How to use the transformation rule
Write \(K(s,\cdot)\) for the conditional law of \(X\) at conditioning state \(s\).
For an event \(Y\in B\), express it as \(X\in f^{-1}(B)\).
Compute \(K(s,f^{-1}(B))\), combining the probability of every input that maps into \(B\).
For a conditional mean or another conditional expectation, integrate the transformed function against \(K(s,\cdot)\).
Common pitfalls and why the rule matters
A frequent mistake is to transform a summary of the original conditional law instead of transforming the law itself. In general, the conditional mean of \(f(X)\) is not \(f\) applied to the conditional mean of \(X\). In the discrete example, given \(Z=0\), the conditional mean of \(X\) is $$ (-2)\frac12+2\frac13+1\frac16=-\frac16. $$ Its square is \(1/36\), whereas the conditional mean of \(X^2\) is \(7/2\). The transformed conditional law, or equivalently the integral of \(f\) against the original conditional law, is what gives the correct answer.
A second pitfall is to assume that a transformed law must have the same type as the original one. A many-to-one map can merge discrete outcomes, as squaring does, or create atoms from a continuous distribution, as clipping does. The preimage formula handles both effects without special cases in the general theorem.
The rule is useful whenever a conditional model is specified for a latent or underlying quantity but conclusions are needed about an observable transformation. It lets one obtain the conditional distribution of that observable while retaining exactly the same conditioning information. Its basic scope is a measurable transformation of the random element; it does not, by itself, change what is being conditioned on.
Check Your Understanding
Use the preimage formula to answer these questions.
- If \(L(s,B)=K(s,f^{-1}(B))\), which part of the conditional law is transformed, and which part remains unchanged?
- Why must probabilities of distinct inputs be added when those inputs have the same image under \(f\)?
- For the clipped uniform example with \(z>1\), what fraction of the conditional probability is concentrated at \(1\), and why?
- For a conditional normal \(X\), why does the density of \(X^2\) include contributions from both square roots of a positive value?
- Under what condition does the integral formula for \(\mathbb E[f(X)\mid\sigma(Z)]\) give a finite conditional expectation?