# Quick question about density with respect to product measure

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 $$\mathbb{P}_X(B)=\int_B f_X(x) \; d\mu_1\times\mu_2 \quad \forall B \in \mathcal{B}_1\otimes\mathcal{B}_2?$$

• 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 $i=1,2$?
– whuber
Aug 2, 2022 at 16:47
• What I meant by $\mathbb{P}_{X_i}$ is the distribution of $X_i$, the measure induced by $X_i$. Aug 2, 2022 at 18:52
• A random variable does not induce a measure without something more involved. Perhaps you are assuming measures on the codomains $(\mathcal{X}_i, \mathcal{B}_i)$?
– whuber
Aug 2, 2022 at 19:34
• 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 $\mathcal{B}$. Aug 2, 2022 at 19:48
• So: are you trying to talk about the push forward measures on the spaces $(\mathcal X_1, \mathcal B_i)$?
– whuber
Aug 2, 2022 at 21:11

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