Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Discrete Case:

Random variable $X, Y$ are independent if $P(X \mid Y=y)$ is a function (which can be deemed as pmf for random variable $X \mid Y$) without $y$.

Continuous Case:

Random variable $X, Y$ are independent if $f(X \mid Y=y)$ is a function (which can be deemed as pdf for random variable $X \mid Y$) without $y$.

It seems to be true and I came across the use of that in many context. However, I cannot find any reference of the formalized theorem about it?

Could anyone direct me any reliable reference?

Update: Are there any non-Measure reference on such topic?

share|cite|improve this question
what is $P(X|Y=y)$, a probability? density function? – Sasha Feb 22 '13 at 13:17
I guess, a random variable, and perhaps its meant that their density functions altogether are independent from $y$. – Berci Feb 22 '13 at 13:56
@Sasha Hopefully I made it clearer after edit. – colinfang Feb 22 '13 at 14:57
up vote 1 down vote accepted

I suppose you mean the following: Let $(\Omega, \Sigma, P)$ be a probability space and for simplicity assume that $X,Y$ take their values in $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ (any other measurable spaces, even different ones for each variable would be fine). Let $P(\cdot | Y=-):\: \Sigma \times \mathbb{R} \rightarrow [0,1]$ be a measurable function for fixed first variable such that $P(\cdot | Y=-) \circ (\mathrm{id_\Sigma},Y):\: \Sigma \times \Omega \rightarrow [0,1]$ is a regular conditional probability given $Y$. Then $P(X \in \cdot | Y=-):\: \mathcal{B}(\mathbb{R}) \times \mathbb{R} \rightarrow [0,1], P(X \in A | Y=y) := P(\{X \in A\} | Y=y)$ defines the conditional distribution of $X$ given $Y=-$. And your claim is that $X,Y$ are independent iff this map is constant in the second variable.

If $X,Y$ are independent, then $$\int_{\{Y \in B\}} {P(X \in A)} \mathrm{d}P =P(X \in A) P(Y \in B) = P(X \in A, Y \in B) = \int_{\{Y \in B\}} {P(\{X \in A\} | \sigma(Y)) \mathrm{d}P}$$ for all $A,B \in \mathcal{B}(\mathbb{R})$, so by definition of the conditional probability, we have for all $A \in \mathcal{B}(\mathbb{R})$: $P(\{X \in A\} | \sigma(Y)) = P(X \in A)$ P-a.s. But then $P(\{X \in A\}|Y=-)$ can be chosen to be constant $=P(X \in A)$.

If $P(X \in \cdot | Y=-)$ is constant for fixed first variable, we must have $$P(X \in A | Y=y_0) \cdot P(Y \in B) = \int_B {P(X \in A | Y =y)} \mathrm{d}P_Y(y) = \int_{\{Y \in B\}} {P(X \in A | \sigma(Y))} \mathrm{d}P = P(X\in A, Y \in B)$$ for all $y_0 \in \mathbb{R}$, $A,B \in \mathcal{B}(\mathbb{R})$. But then choosing $B=\mathbb{R}$, we see that $P(X \in A | Y=y_0) = P(X \in A)$, and the independence follows.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.