# Product Measures

Consider the case $\Omega = \mathbb R^6 , F= B(\mathbb R^6)$ Then the projections $\ X_i(\omega) = x_i ,[ \omega=(x_1,x_2,\ldots,x_6) \in \Omega$ are random variables $i=1,\ldots,6$. Fix $\ S_n = S_0$ $\ u^{\Sigma X_i(\omega)}d^{n-\Sigma X_i(\omega)} \omega \in \Omega$, $\ n=1,\ldots,6$.

Choose the measure P = $\bigotimes_{i=1}^6 Q$ on ($\Omega,F$) where $Q$ denotes the measure $p\delta_1 + q\delta_0$ on $(\mathbb R, B(\mathbb R))$ for some $p,q>0$ such that $p+q = 1$. Show that the projections $\ X_i(\omega), i=1,\ldots,6$ are mutually independent.

Since $\ X_i(\omega)$ is a random variable then am I correct in saying that to show their independence I must show that their sigma algebras $\sigma(\ X_i(\omega))$ are independent how would I go about doing this?

Thanks very much!

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Yes, that is correct. You have to show that $\sigma(X_i)$ and $\sigma(X_j)$ are independent, when $j\neq i$ (note that I have omitted the $\omega$ in $\sigma(X_i(\omega))$, because that is not what you want). Now, recall that $$\sigma(X_i)=\sigma(\{X_i^{-1}(A)\mid A\in \mathcal{B}(\mathbb{R})\}),$$ and hence it is enough to show that $\{X_i^{-1}(A)\mid A\in \mathcal{B}(\mathbb{R})\}$ and $\{X_j^{-1}(A)\mid A\in \mathcal{B}(\mathbb{R})\}$ are independent when $i\neq j$. Now, if $A\in\mathcal{B}(\mathbb{R})$ then $$X_i^{-1}(A)=\{(x_1,\ldots,x_6)\in\mathbb{R}\mid x_i\in A\}=\mathbb{R}\times\cdots \times A\times\cdots\times\mathbb{R},$$ where $A$ is on the $i$'th place. If $j\neq i$, then $$X_i^{-1}(A)\cap X_j^{-1}(B)=\mathbb{R}\times\cdots \times A\times B\times\cdots\times\mathbb{R},$$ where $A$ is on the $i$'th place and $B$ is on the $j$'th place. Now $$P(X_i^{-1}(A)\cap X_j^{-1}(B))=Q(\mathbb{R})^{4}Q(A)Q(B)=Q(A)Q(B)=P(X_i^{-1}(A))P(X_j^{-1}(B)),$$ and hence the events are independent for every choice of $A,B\in\mathcal{B}(\mathbb{R})$.
I actually asked this question on a different username which I can't get back onto, so Im responding like this because I cant seem to add a comment. Thanks very much for you answer , it now asks me to show that $\ S_n/S_{n-1}$and $\ S_{n-1}$ are independent, but how would you divide a random variable by another!? – Rosie Jun 1 '12 at 20:15
Probably by $S_n/S_{n-1}(\omega)=S_n(\omega)/S_{n-1}(\omega)$. So you divide pointwise. – Michael Greinecker Jun 1 '12 at 21:01
@MichaelGreinecker Thanks, I did this and got $\ u^{X_n(\ omega)}d^{1-X_n(\omega)}$. So clearly this contains only $\ X_n$ whereas $\ S_{n-1}$ contains only $\ X_1..X_{n-1}$ is this enough to say that theyre independent (using the fact that the Xi's are the only random element) – Rosie Jun 1 '12 at 21:14
I'm not completely sure whaat your formula implies, but if you have two independent random variables $X$ and $Y$ and any measurable functions $f$ and $g$ (they could be the same), then $f\circ Y$ and $g\circ X$ are independent. – Michael Greinecker Jun 1 '12 at 21:18