Measurable Images Preserve Independence

Why Measurable Transformations Matter

Independence is often easiest to establish for the original random elements, while the quantities of interest are transformed versions of them. An analysis may use absolute values, threshold indicators, or selected coordinates rather than the original variables. The key fact is that a measurable transformation applied to each of several independent inputs preserves their independence.

The reason is structural: an event about a transformed value pulls back to an event about its original input. Independence of the original inputs already gives the required factorization for those preimage events. No assumption that the transformations are one-to-one is needed.

Definitions

Definition: Let $(S_i,\mathcal S_i)$, $i=1,\ldots,n$, be measurable spaces, and let $X_i:\Omega\to S_i$ be measurable random elements on a probability space $(\Omega,\mathcal F,P)$. The random elements $X_1,\ldots,X_n$ are independent if, for every choice of sets $A_i\in\mathcal S_i$, $$ P(X_1\in A_1,\ldots,X_n\in A_n) = \prod_{i=1}^n P(X_i\in A_i). $$

Here $P(X_i\in A_i)$ denotes $P(\{\omega\in\Omega:X_i(\omega)\in A_i\})$. The sets in the definition range over the measurable sets of the respective spaces; this allows the definition to apply to random elements more general than real-valued random variables.

Definition: A function $f_i:S_i\to T_i$, where $(T_i,\mathcal T_i)$ is a measurable space, is measurable if $f_i^{-1}(B)\in\mathcal S_i$ for every $B\in\mathcal T_i$. The transformed random element $f_i(X_i)$ is the composition $f_i\circ X_i:\Omega\to T_i$.

Measurability ensures that events concerning $f_i(X_i)$ are events in $\mathcal F$, so their probabilities are defined. More specifically, for any $B\in\mathcal T_i$, $$ \{f_i(X_i)\in B\}=\{X_i\in f_i^{-1}(B)\}. $$ This identity is the essential step in the preservation result.

The Independence-Preservation Theorem

Theorem: Suppose $X_1,\ldots,X_n$ are independent random elements, and for each $i$, $f_i:S_i\to T_i$ is measurable. Then $f_1(X_1),\ldots,f_n(X_n)$ are independent.

Proof. Choose arbitrary $B_i\in\mathcal T_i$, for $i=1,\ldots,n$. By the preimage identity and the independence of the $X_i$, $$ \begin{aligned} P(f_1(X_1)\in B_1,\ldots,f_n(X_n)\in B_n) &=P(X_1\in f_1^{-1}(B_1),\ldots,X_n\in f_n^{-1}(B_n))\\ &=\prod_{i=1}^n P(X_i\in f_i^{-1}(B_i))\\ &=\prod_{i=1}^n P(f_i(X_i)\in B_i). \end{aligned} $$ Each $f_i^{-1}(B_i)$ is in $\mathcal S_i$ by measurability, so the independence assumption applies. Since the factorization holds for every choice of measurable $B_i$, it is exactly the definition of independence of the transformed random elements. $\square$

The theorem permits each function to be different, and it permits the functions to discard information. For example, a threshold indicator keeps only whether a value lies above a specified level. Independence is preserved because the event that an indicator equals a particular value still corresponds to a measurable event about its own input.

A View Through Joint Laws

The same fact can be expressed in terms of distributions. The law of a random element $X$ with values in $S$ is the probability measure $\mu_X$ on $\mathcal S$ defined by $\mu_X(A)=P(X\in A)$. If $f:S\to T$ is measurable, the law of $f(X)$ assigns to $B\in\mathcal T$ the probability $\mu_X(f^{-1}(B))$. This is the pushforward of $\mu_X$ by $f$.

Theorem: Under the assumptions of the independence-preservation theorem, the joint law of $(f_1(X_1),\ldots,f_n(X_n))$ is the product of the individual laws of $f_1(X_1),\ldots,f_n(X_n)$.

Proof. Write $Y_i=f_i(X_i)$, and let $\mu_i$ be the law of $Y_i$. For measurable sets $B_i\in\mathcal T_i$, the theorem just proved gives $$ P(Y_1\in B_1,\ldots,Y_n\in B_n) = \prod_{i=1}^n \mu_i(B_i). $$ The right-hand side is the value of the product measure $\mu_1\otimes\cdots\otimes\mu_n$ on the measurable rectangle $B_1\times\cdots\times B_n$. Thus the joint law and this product measure agree on every measurable rectangle. Rectangles form a generating pi-system for the product sigma-algebra, and probability measures agreeing on that pi-system agree on the sigma-algebra it generates. Therefore the joint law is $\mu_1\otimes\cdots\otimes\mu_n$. $\square$

This formulation says that transforming independent inputs does not merely preserve event factorization: the entire joint distribution of the outputs is the product of their marginal distributions. It is useful when the aim is to identify a joint law, not only to answer whether the outputs are independent.

Worked Example: Absolute Values of Independent Normal Variables

Let $X$ and $Y$ be independent standard normal random variables. The function $f:\mathbb R\to[0,\infty)$ given by $f(x)=|x|$ is Borel measurable. The theorem therefore implies that $|X|$ and $|Y|$ are independent.

For an explicit check, let $a,b\geq 0$, and let $\Phi$ be the standard normal distribution function. Symmetry of the standard normal law gives $$ P(|X|\leq a)=\Phi(a)-\Phi(-a)=2\Phi(a)-1, $$ and likewise $P(|Y|\leq b)=2\Phi(b)-1$. Since the original variables are independent, $$ P(|X|\leq a,\ |Y|\leq b) =P(-a\leq X\leq a,\ -b\leq Y\leq b) =(2\Phi(a)-1)(2\Phi(b)-1). $$ This is the product of the two marginal probabilities, as required. The absolute-value map is not one-to-one, but injectivity is not needed.

Thresholding and Other Coarsenings

Thresholding turns a variable into a binary-valued random element. For a fixed threshold $c$, the map $x\mapsto \mathbf 1_{\{x>c\}}$ is measurable. The resulting indicator records less information than the original variable, but applying such maps separately to independent inputs still preserves independence.

Worked Example: Independent Threshold Indicators

Let $X$ be uniform on $[0,1]$, and let $Y$ have the exponential distribution with rate $1$. Assume $X$ and $Y$ are independent. Define $$ A=\mathbf 1_{\{X>1/2\}}, \qquad B=\mathbf 1_{\{Y>\log 2\}}. $$ Both are measurable functions of their respective inputs, so $A$ and $B$ are independent.

Their success probabilities are $$ P(A=1)=P(X>1/2)=1/2, \qquad P(B=1)=P(Y>\log 2)=e^{-\log 2}=1/2. $$ Independence of $X$ and $Y$ also gives $P(A=1,B=1)=1/4$. To check the full binary joint distribution, the other three probabilities are $$ \begin{aligned} P(A=1,B=0)&=P(X>1/2,\ Y\leq\log 2)=1/2\cdot1/2=1/4,\\ P(A=0,B=1)&=P(X\leq1/2,\ Y>\log 2)=1/2\cdot1/2=1/4,\\ P(A=0,B=0)&=P(X\leq1/2,\ Y\leq\log 2)=1/2\cdot1/2=1/4. \end{aligned} $$ Each cell has probability $1/4$, which agrees with the product of the corresponding marginal probabilities.

Coordinate Projections

A coordinate projection selects one component of a vector. For example, on $\mathbb R^2$, the maps $\pi_1(x_1,x_2)=x_1$ and $\pi_2(x_1,x_2)=x_2$ are Borel measurable. If two random vectors are independent, applying a coordinate projection to each vector preserves their independence. The vectors need not have independent coordinates internally: the theorem concerns independence between the separate input vectors.

Worked Example: Projections of Independent Random Vectors

On $\Omega=\{0,1\}\times\{0,1\}$ with each point having probability $1/4$, write a point as $(z,w)$, and define $U(z,w)=(z,z)$ and $V(z,w)=(w,w)$. The vector $U$ takes each of $(0,0)$ and $(1,1)$ with probability $1/2$; the same is true of $V$. For $u=(z,z)$ and $v=(w,w)$, $$ P(U=u,V=v)=1/4=P(U=u)P(V=v). $$ If either vector is outside its two-point support, the joint probability and the product of the marginal probabilities are both zero. Hence $U$ and $V$ are independent random elements.

Project onto the first coordinate of $U$ and the second coordinate of $V$. The outputs are $Z=z$ and $W=w$. Directly, $$ P(Z=1)=1/2,\qquad P(W=1)=1/2,\qquad P(Z=1,W=1)=1/4. $$ The other three pairs each also have probability $1/4$. Thus the projected outputs are independent, even though each individual input vector has two identical coordinates.

What the Result Does Not Say

The transformations must be applied separately to independent inputs. Applying two functions to the same random element does not generally produce independent outputs. For instance, let $Z$ be Bernoulli with $P(Z=0)=P(Z=1)=1/2$. The variables $Z$ and $1-Z$ are both measurable functions of $Z$, but $$ P(Z=1,\ 1-Z=1)=0 \quad\text{while}\quad P(Z=1)P(1-Z=1)=1/4. $$ They are not independent.

Likewise, the coordinates of one random vector are not necessarily independent merely because coordinate projections are measurable. Measurability guarantees that the projections are valid random elements; it does not establish independence between them. The preservation theorem starts with independence of the separate inputs and then transforms each input on its own.

This result is useful whenever a problem replaces independent variables by simpler or more relevant summaries: signs, indicators, absolute values, or selected coordinates. It allows independence to be carried through those operations without recalculating all possible joint events. The essential check is always the same: identify the independent inputs, verify that each transformation is measurable and acts on one input, and then apply the theorem.

Check Your Understanding

Answer the following questions using the definitions and results above.

Why does the event $\{f_i(X_i)\in B_i\}$ correspond to an event about $X_i$?
Does a measurable transformation need to be one-to-one for independence to be preserved? Explain.
If $X$ and $Y$ are independent, what can be concluded about $\mathbf 1_{\{X>c\}}$ and $\mathbf 1_{\{Y>d\}}$?
Why does the theorem not imply that two coordinates of the same random vector are independent?
How does agreement on measurable rectangles establish the product-law conclusion?