Consider the following problem:
Let $X,Y$ be random variables with values in $[0,+\infty)$ and $V$ a random variable with values in $(0,+\infty)$. Suppose that $X$ and $Y$ are independent from $V$ and that $\phi_{\log V}$ (characteristic function of $\log V$) has isolated zeroes. Then
- If $X \neq 0$ a.c. and $Y \neq 0$ a.c. and $XV$ and $YV$ have the same distribution, prove that $X$ and $Y$ have the same distribution
- Prove point 1. without the hypotesis of $X \neq 0$ a.c. and $Y \neq 0$ a.c.
Point 1. follows from properties of the characteristic function and it is clear to me. The problem is the proof of point 2. that is stated as follows:
Observe that $P(X=0)=P(XV=0)=P(YV=0)=P(Y=0)$ and that, given a borel set $A\subseteq \mathbb{R}$ we have that $$P(X \in A)=P(X \in A \mid X \neq 0)P(X \neq 0) +\mathbf{1}_{\{0 \in A\}}P(X=0)$$ and similarly for $Y$, so it is enough to prove that $$P(X \in A \mid X \neq 0)=P(Y \in A \mid Y \neq 0)$$ but this follows immediately from point 1. since under $P(\cdot \mid X \neq 0)$ we have that $X \neq 0$ a.c. and similarly for $Y$.
I really don't understand how the bolded part should imply the thesis: the first thing that is not clear to me is that changing the probability on the space that we have (switching from $P$ to $P(\cdot \mid X \neq 0)$) can change also the distribution of the variables, and so the hypothesis that we have could not be more valid (in particular $\phi_{\log V}$ can be different), but this doesn't bother me much since I suppose that we are taking for granted that the distributions of the variables remain untouched.
What I really don't understand is that $X \neq 0$ and $Y \neq 0$ a.c. under two different probabilities, so how can we apply the first point since they have to be nonzero a.c. at the same time?
Revisit the proof of 1, you'll notice you don't actually need for $X$ and $Y$ to be on the same probability space if you consider $V_{1},V_{2} \sim V$, $V_{1}X \sim V_{2}Y$ , $V_{1}$ independent of $X$ and $V_{2}$ independent of $Y$.
By independence, $V$ w.r.t. $P(.|X \neq 0)$ and $P(.|Y \neq 0)$ are distributed as $V$ w.r.t. $P$. Also, because $\{XV \neq 0 \} = \{X \neq 0\}$ and $\{YV \neq 0 \} = \{Y \neq 0\}$ you can easily show by definition of conditional probability that $XV$ w.r.t. $P(.|X \neq 0)$ $\sim$ $YV$ w.r.t. $P(.|Y \neq 0)$. Hence you are in the conditions of 1.