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Let $X_1$ and $X_2$ be independent random variable of the describe type with joint pdf $f_1(x_1)$, $f_2(x_2)$, $x_1$, $x_2$ belongs to $A$. Let $y_1=u_1(x_1)$ and $y_2=u_2(x_2)$ the note a $1-1$ transformation that maps $A$ onto $B$. Show that $y_1$ and $y_2$ are independent.

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If considering the joint pdf, it should be one function of 2 variables. Aand... What is the question? – Berci Oct 8 '12 at 11:51
And what is "the describe type"? – joriki Oct 8 '12 at 11:52
up vote 0 down vote accepted

OK. So, we have $u_1,u_2:A\to B$ maps (suppose for simplicity that $A,B\subseteq\mathbb R$), and $X_1,X_2$ random variables, and these are mapped to $u_1(X_1)$ and $u_2(X_2)$ random variables with values in $B$. Let $F_i$ be the distr.function of $X_i$ (the integral of the density function), that is: $$P(X_i<t) = F_i(t) = \int_{-\infty}^t f_i(x)dx$$ Being independent means that their joint distr.function is the product of the distr.functions (same holds for density functions). That is, for any $t_1,t_2\in A$, $$P(X_1<t_1,\ X_2<t_2)= P(X_1<t_1)\cdot P(X_2<t_2)$$ In intuition, it's something like: $X_1$ and $X_2$ 'live in separate dimensions', that's why probabilities (viewed as 'area' or 'length' or 'volume') are multiplied.

For general (more abstract) $A, B$ measure spaces, and $S_1,S_2\subseteq A$, by def. of density function, we have $$P(X_i\in S_i) = \int_{S_i} f_i$$ but, in this exercise not much need for this integral, because being independent is also equivalent to $$P(X_1\in S_1,\, X_2\in S_2) = P(X_1\in S_1)\cdot P(X_2\in S_2)$$ for all measurable sets $S_1,S_2\subseteq A$.

Now, if we want to prove the independencies of $Y_1=u_1(X_1)$ and $Y_2=u_2(X_2)$, consider $H_1,H_2\subseteq B$ measurable subsets. Then $$P(Y_i\in H_i) =P(u(X_i)\in H_i) = P\left(X_i\in u^{-1}(H_i)\right)$$ $$\begin{align*} P(Y_1\in H_1,\, Y_2\in H_2) &= P\left(X_1\in u^{-1}(H_i),\, X_2\in u^{-1}(H_2)\right) = \\ &= P\left(X_1\in u_1^{-1}(H_1)\right)\cdot P\left(X_2\in u_2^{-1}(H_2)\right) = \\ &= P(Y_1\in H_1)\cdot P(Y_2\in H_2) \end{align*} $$ QED.

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