2
$\begingroup$

Suppose we got 2 non indepenendent random variables $X_1:(\Omega,\mathcal{A})\rightarrow (\mathcal{X}_1,\mathcal{B}_1)$ and $X_2:(\Omega,\mathcal{A})\rightarrow (\mathcal{X}_2,\mathcal{B}_2)$ with densities with respect to $\sigma$-finite measures $\mu_1$ and $\mu_2$, respectively. Is it true that random vector $X=(X_1,X_2)$ has joint density with respect to product measure $\mu_1\times\mu_2$ even if $\mathbb{P}_X\neq\mathbb{P}_{X_1}\times\mathbb{P}_{X_2}$?

Edit: Due to misunderstanding of my question, let me explain it a little bit.

Suppose we got 2 probability spaces $(\Omega_1,\mathcal{A}_1,\mathbb{P}_1)$ and $(\Omega_2,\mathcal{A}_2,\mathbb{P}_2)$. Let $X_1:(\Omega,\mathcal{A})\rightarrow (\mathcal{X}_1,\mathcal{B}_1)$ and $X_2:(\Omega,\mathcal{A})\rightarrow (\mathcal{X}_2,\mathcal{B}_2)$ be two random variables and let $\mathbb{P}_{X_1}$ and $\mathbb{P}_{X_2}$ be their distributions (measures induced by $X_1$ and $X_2$ from $\mathbb{P}_1$ and $\mathbb{P}_2$ respectively).

Suppose $X_1$ has density $f_{X_1}$ with respect to $\mu_1$, so $$\mathbb{P}_{X_1}(B)=\int_B f_{X_1}(x)\;d\mu_1\quad \forall B\in\mathcal{B}_1 $$ and $X_2$ has density $f_{X_2}$ with respect to $\mu_2$, so $$\mathbb{P}_{X_2}(B)=\int_B f_{X_2}(x)\;d\mu_2\quad \forall B\in\mathcal{B}_2 $$.

Now consider joint probability space $(\Omega_1\times\Omega_2,\mathcal{A}_1\otimes\mathcal{A}_2,\mathbb{P})$, where $\mathbb{P}$ is a probability measure of joint experiment and a random vector $(X_1,X_2)=X:(\Omega_1\times\Omega_2,\mathcal{A}_1\otimes\mathcal{A}_2)\rightarrow(\mathcal{X}_1\times\mathcal{X}_2,\mathcal{B}_1\otimes\mathcal{B}_2)$ with it's distribution $\mathbb{P}_X$.

Is it true, that random vector $X$ has density $f_X$ with respect to $\mu_1\times\mu_2$ even though $X_1$ and $X_2$ could be non indenpendent, so $$\mathbb{P}_X(B)=\int_B f_X(x) \; d\mu_1\times\mu_2 \quad \forall B \in \mathcal{B}_1\otimes\mathcal{B}_2?$$

$\endgroup$
7
  • $\begingroup$ Could you please explain what you mean by "$\mathbb{P}_{X_i}$"? In your notation is it implicit that $(\Omega, \mathcal A)$ has a fixed probability measure for both the $X_i.$ Are you trying to say instead that $X_i:(\Omega,\mathcal{A},\mathbb{P}_{X_i})\rightarrow (\mathcal{X}_i,\mathcal{B}_i)$ for $i=1,2$? $\endgroup$
    – whuber
    Aug 2, 2022 at 16:47
  • $\begingroup$ What I meant by $\mathbb{P}_{X_i}$ is the distribution of $X_i$, the measure induced by $X_i$. $\endgroup$
    – MatEZ
    Aug 2, 2022 at 18:52
  • $\begingroup$ A random variable does not induce a measure without something more involved. Perhaps you are assuming measures on the codomains $(\mathcal{X}_i, \mathcal{B}_i)$? $\endgroup$
    – whuber
    Aug 2, 2022 at 19:34
  • $\begingroup$ There is a probability measure $\mathbb{P}$ on $\mathcal{A}$ and the measure induced from $\mathbb{P}$ by a map $X:(\Omega,\mathcal{A})\rightarrow (\mathcal{X},\mathcal{B})$ I denote by $\mathbb{P}_X$ - it's called the distribution of $X$ (measure on $\mathcal{B}$. $\endgroup$
    – MatEZ
    Aug 2, 2022 at 19:48
  • $\begingroup$ So: are you trying to talk about the push forward measures on the spaces $(\mathcal X_1, \mathcal B_i)$? $\endgroup$
    – whuber
    Aug 2, 2022 at 21:11

1 Answer 1

3
$\begingroup$

For a counterexample let

  • $\Omega_1 = \Omega_2 = [0,1]$
  • $\mathcal A_1 = \mathcal A_2 = \mathcal B_1 = \mathcal B_2 = \mathcal B([0,1]) \equiv \mathcal B(\mathbb R)\big|_{[0,1]}$
  • $\mathbb P_1 = \mathbb P_2 = \mathop{\mathrm{Unif}}([0,1]) \equiv \lambda\big|_{[0,1]}$
  • $X_1 = X_2 = \mathrm{id}_{[0,1]}$.

Clearly, $(X_1, X_2)$ takes values in $N \mathrel{:=} \{(x_1, x_2) \in [0,1]^2 : x_1 = x_2\}$.
But $(\mathbb P_1 \times \mathbb P_2)(N) = 0$, so $(X_1, X_2)$ can't have a density w.r.t to $\mathbb P_1 \times \mathbb P_2.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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