Separate Transformations Preserve Independence

Why Transformations Matter

Random variables are often replaced by summaries that are easier to interpret: an indicator records whether a threshold was crossed, an absolute value records magnitude without sign, and a nonlinear function may put a measurement on a more useful scale. A key question is whether these operations alter the relationship between variables. When two variables are independent, applying a measurable function to each one separately preserves that independence.

The word separately is essential: the first transformed variable is a function of the first input, and the second transformed variable is a function of the second input. The result follows from the event-based definition of independence. An event involving a transformed variable can be translated back into an event involving its original variable.

Measurable Transformations and the Main Theorem

Definition: A function $g:\mathbb{R}\to\mathbb{R}$ is Borel measurable if $g^{-1}(C)=\{x:g(x)\in C\}$ is a Borel set in $\mathbb{R}$ for every Borel set $C$. If $X$ is a random variable, then $g(X)$ is a random variable whenever $g$ is Borel measurable.

Continuity is a sufficient condition for Borel measurability, but it is not necessary. Measurability matters because probabilities such as $P(g(X)\in C)$ are defined only when the corresponding event is measurable.

Theorem (Separate measurable transformations preserve independence): Let $X$ and $Y$ be independent real-valued random variables. If $g,h:\mathbb{R}\to\mathbb{R}$ are Borel measurable, then $g(X)$ and $h(Y)$ are independent. Equivalently, for every pair of Borel sets $C,D\subseteq\mathbb{R}$, $$ P\bigl(g(X)\in C,\ h(Y)\in D\bigr) = P\bigl(g(X)\in C\bigr)P\bigl(h(Y)\in D\bigr). $$

Proof. Fix arbitrary Borel sets $C,D\subseteq\mathbb{R}$. Measurability of $g$ and $h$ ensures that $g^{-1}(C)$ and $h^{-1}(D)$ are Borel sets. By the definition of inverse image, $$ \{g(X)\in C\}=\{X\in g^{-1}(C)\},\qquad \{h(Y)\in D\}=\{Y\in h^{-1}(D)\}. $$ Since $X$ and $Y$ are independent, the probabilities of the events on the right factor: $$ \begin{aligned} P\bigl(g(X)\in C,\ h(Y)\in D\bigr) &=P\bigl(X\in g^{-1}(C),\ Y\in h^{-1}(D)\bigr)\\ &=P\bigl(X\in g^{-1}(C)\bigr)P\bigl(Y\in h^{-1}(D)\bigr)\\ &=P\bigl(g(X)\in C\bigr)P\bigl(h(Y)\in D\bigr). \end{aligned} $$ This holds for every Borel $C,D$, which is precisely independence of $g(X)$ and $h(Y)$. $\square$

The proof also gives a useful way to calculate transformed probabilities: $$ P\bigl(g(X)\in C\bigr)=P\bigl(X\in g^{-1}(C)\bigr). $$ Rather than trying to identify a transformed distribution immediately, one can first translate the event through the inverse image.

Worked Examples

Worked Example: Threshold Indicators of Independent Normals

Let $X,Y$ be independent standard normal random variables, and define $I=\mathbf{1}\{X>1\}$ and $J=\mathbf{1}\{Y\leq 0\}$. Show that $I,J$ are independent and calculate their joint success probability.

The indicator functions are Borel measurable: each is the indicator of a Borel subset of the real line. Applying the theorem with $g(x)=\mathbf{1}\{x>1\}$ and $h(y)=\mathbf{1}\{y\leq 0\}$ gives independence of $I$ and $J$. Their success probabilities are $$ P(I=1)=P(X>1)=1-\Phi(1)\approx 0.15865525,\qquad P(J=1)=P(Y\leq 0)=\Phi(0)=0.5, $$ where $\Phi$ is the standard normal distribution function. Therefore $$ P(I=1,J=1)=P(I=1)P(J=1) =(1-\Phi(1))(0.5)\approx 0.07932763. $$ Rounded to five decimal places, the joint probability is $0.07933$. In particular, the product was calculated from the unrounded normal probability before rounding the result.

The variables $I,J$ are Bernoulli indicators. The theorem establishes their independence, not merely the factorization for the one pair of success events: its conclusion applies to every pair of subsets of their possible values.

Worked Example: Absolute Values of Independent Normals

Let $X,Y$ be independent standard normal random variables. Set $U=|X|$ and $V=|Y|$. Find their marginal distribution functions for nonnegative arguments and calculate $P(U\leq 1,V\leq 2)$.

The absolute-value function is continuous, hence Borel measurable. Thus $U,V$ are independent. For $t\geq 0$, $$ \begin{aligned} P(U\leq t) &=P(-t\leq X\leq t)\\ &=\Phi(t)-\Phi(-t)\\ &=2\Phi(t)-1, \end{aligned} $$ using the symmetry $\Phi(-t)=1-\Phi(t)$ of the standard normal distribution. The same formula holds for $V$. For $t<0$, either absolute value is at most $t$ with probability zero.

The joint event is $\{|X|\leq 1,\ |Y|\leq 2\}$. Independence gives $$ \begin{aligned} P(U\leq 1,V\leq 2) &=P(-1\leq X\leq 1)P(-2\leq Y\leq 2)\\ &=[2\Phi(1)-1][2\Phi(2)-1]. \end{aligned} $$ This expression uses the actual marginal probabilities of the two intervals; no assumption that the absolute values retain normal distributions is needed.

Worked Example: Nonlinear Functions of Independent Uniform Variables

Let $X,Y$ be independent Uniform$(0,1)$ random variables. Define $U=X^2$ and $V=\log(1+Y)$ almost surely. Verify measurability carefully, find the marginal distribution functions, and calculate $P(U\leq 1/4,V\leq \log(3/2))$.

The function $g(x)=x^2$ is continuous on $\mathbb{R}$. The expression $\log(1+y)$, by contrast, is not real-valued for every $y\in\mathbb{R}$. To use the theorem with a function defined on all of $\mathbb{R}$, take the Borel-measurable extension $$ h(y)= \begin{cases} \log(1+y),&y>0,\\ 0,&y\leq 0. \end{cases} $$ It is continuous on each of the Borel sets $(-\infty,0]$ and $(0,\infty)$, and is Borel measurable on their union $\mathbb{R}$. Since $P(Y>0)=1$, $h(Y)=\log(1+Y)$ almost surely. Consequently $U=g(X)$ and $V=h(Y)$ are independent.

For $0\leq u<1$, monotonicity of squaring on $(0,1)$ gives $$ P(U\leq u)=P(X\leq \sqrt{u})=\sqrt{u}. $$ The distribution function is $0$ for $u<0$ and $1$ for $u\geq 1$. For $0\leq v<\log 2$, monotonicity of the logarithm gives $$ P(V\leq v)=P\bigl(Y\leq e^v-1\bigr)=e^v-1. $$ For $v<0$, this probability is $0$; for $v\geq\log 2$, it is $1$, since $0<Y<1$ almost surely. In particular, $$ P(U\leq 1/4)=1/2,\qquad P(V\leq \log(3/2))=e^{\log(3/2)}-1=1/2. $$ By independence, the requested joint probability is $$ P(U\leq 1/4,V\leq \log(3/2))=(1/2)(1/2)=1/4. $$

A Finite-Family Extension

The same argument applies when each member of a finite independent family is transformed separately. This extension is useful when several measurements are recoded or summarized at once.

Theorem: Let $X_1,\ldots,X_m$ be mutually independent real-valued random variables, where $m$ is a positive integer. For each $i$, let $g_i:\mathbb{R}\to\mathbb{R}$ be Borel measurable. Then $g_1(X_1),\ldots,g_m(X_m)$ are mutually independent.

Proof. By mutual independence, for any Borel sets $B_1,\ldots,B_m$, $$ P\left(\bigcap_{i=1}^{m}\{X_i\in B_i\}\right) =\prod_{i=1}^{m}P(X_i\in B_i). $$ Now choose arbitrary Borel sets $C_1,\ldots,C_m$. Each $g_i^{-1}(C_i)$ is Borel by measurability. Substituting $B_i=g_i^{-1}(C_i)$ into the independence identity and using $\{g_i(X_i)\in C_i\}=\{X_i\in g_i^{-1}(C_i)\}$ yields $$ P\left(\bigcap_{i=1}^{m}\{g_i(X_i)\in C_i\}\right) =\prod_{i=1}^{m}P\bigl(g_i(X_i)\in C_i\bigr). $$ This is the definition of mutual independence of the transformed variables. $\square$

Common Pitfalls and Why the Result Matters

Using functions that are not measurable. Without measurability, inverse images of Borel sets need not be events to which the probability argument applies. Check the domain and measurability of each function.
Checking only continuity on the support. If a theorem is stated for Borel functions on $\mathbb{R}$, continuity on a smaller interval alone does not make the proposed function a map $\mathbb{R}\to\mathbb{R}$. Give a measurable extension when needed, as in the logarithm example.
Thinking every transformation preserves independence. The theorem concerns separate functions of independent inputs. Its hypotheses do not say that arbitrary combinations of inputs, or transformations of variables that were not independent, become independent.
Assuming transformed distributions are unchanged. Independence is preserved, but the marginal distributions can change substantially. For instance, $|X|$ is nonnegative even when $X$ is standard normal.

The inverse-image proof explains the practical value of the result: once independence is known, separate recoding does not require a new independence calculation from scratch. One can focus on the transformed marginal distributions and compute probabilities by translating the relevant events back to the original variables.

Check Your Understanding

Answer each question using the event-based definition of independence.

State the measurability condition on $g$ needed to ensure that $g(X)$ is a random variable when $X$ is real-valued.
If $X,Y$ are independent, why does $P(g(X)\in C,h(Y)\in D)$ factor for Borel sets $C,D$? Identify the two inverse-image events.
Suppose $X,Y$ are independent and $I=\mathbf{1}\{X\leq 2\}$, $J=\mathbf{1}\{Y>0\}$. Which theorem establishes independence of $I,J$, and what are its function measurability conditions?
Why is a measurable extension useful when applying $h(y)=\log(1+y)$ to a Uniform$(0,1)$ variable?
If $X_1,\ldots,X_m$ are mutually independent, what additional condition on the functions $g_i$ ensures their separate transformations remain mutually independent?