# Random variable on product space as product of random variables

Let $\left(\Omega_i,\mathcal{A}_i,P_i\right)$ for $i=1,2$ a probability space and assume that $\mathcal{A}_1$ is independent of $\mathcal{A}_2$. Define the probability space $\left(\Omega,\mathcal{A},P\right)$ where $\Omega=\Omega_1\times\Omega_2$, $\mathcal{A}=\mathcal{A}_1\otimes\mathcal{A}_2$ and $P=P_1\times P_2$. Consider a random variable $X$ on $\left(\Omega,\mathcal{A},P\right)$. Why can we find r.v. $X_i$ defined on $\left(\Omega_i,\mathcal{A}_i,P_i\right)$ s.t. $X(\omega_1,\omega_2)=X_1(\omega_1)\cdot X_2(\omega_2)$ holds.

Furthermore i am interested, if the martingale property is transfered to $X_1$ and $X_2$: $X$ is a martingale $\Rightarrow$ $X_i$ is a $(\mathcal{A}_i,P_i)$-martingale.

• What is the meaning of the assumption that $\mathcal{A}_1$ is independent of $\mathcal{A}_2$? Please be very precise here since the current statement seems absurd. – Did Jan 23 '17 at 17:01
• @Did That's a good question. It is basically assumption 4.1. of this paper and the claim is stated in remark 4.2: arxiv.org/pdf/1406.6902v1.pdf – peer Jan 23 '17 at 17:06
• The sigma-algebra $\mathcal{A}=\mathcal{A}_1\otimes\mathcal{A}_2$ refers to $\mathcal{A}=\sigma(\mathcal C)$ where $$\mathcal C=\{A_1\times A_2\mid A_1\in\mathcal{A}_1,A_2\in\mathcal{A}_2\}$$ The independence assumption is absurd (is it in the paper?), as already mentioned. The claim is obviously false. If the paper you linked to really interests you, you might want to ask the authors how they proceed to find $Z_1$ and $Z_2$ when $\left(\Omega_i,\mathcal{A}_i,P_i\right)=([0,1],\mathcal B([0,1]),\mathrm{Leb})$ and $$Z(\omega_1,\omega_2)=\omega_1+\omega_2$$ – Did Jan 23 '17 at 17:16
• @Did The paper constructs an insurance market and a financial market, which gernerates time dependent informations given by filtrations $\mathbb{G}$ and $\mathbb{F}$. The assumption states that the insurance market is independent of the financial market (they don't affect each other). I don't know exactly how this is transfered into mathematical terms. So maybe the statement is only true because of the specific definition of the sigma-algebra $\mathcal{F}$ and $\mathcal{G}$. – peer Jan 23 '17 at 17:42
• Even knowing nothing about nothing, if remark 4.2 was true, we would have $Z$ and $T$ random variables with $Z(\omega_1,\omega_2)=Z_1(\omega_1)Z_2(\omega_2)$ and $T(\omega_1,\omega_2)=T_1(\omega_1)T_2(\omega_2)$ but $U=Z+T$ not a random variable? To the bin! – Did Jan 23 '17 at 17:45