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

  • $\begingroup$ 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
    Commented 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
    Commented 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
    Commented 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
    Commented 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
    Commented Aug 2, 2022 at 21:11

1 Answer 1


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 $\mathbb P_1 \times \mathbb P_2.$


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