# Conditioning on random sets?

Let $$( X_i)_{i=1}^n$$ be a random vector of (possibly dependent) real random variables.

Let $$A_i$$ be any measurable set such that $$P(X_i \in A_i \,|\, X_{i-1} = x_{i-1},\ldots,X_1 = x_1) = \beta_i$$. What I want to work out is the probability $$P(X_n \in A_n,\ldots, X_1\in A_1).$$ The answer (I think) is just $$P(X_n \in A_n,\ldots, X_1\in A_1) = \prod_{i=1}^n\beta_i$$, since $$P(X_n \in A_n,\ldots, X_1\in A_1) = P(X_n \in A_n|X_{n-1}\in A_{n-1},\ldots,X_1\in A_1)\ldots P(X_1\in A_1),$$ but I am unsure how to make this formal? Is the following proof valid, where $$p(x_n,\ldots,x_1)$$ is the pdf of $$X_n,\ldots,X_1$$: \begin{align} P(X_n \in A_n,\ldots, X_1\in A_1) &= \int_{A_1}\ldots\int_{A_n} p(x_n,\ldots,x_1) \mathrm{d}x_n\ldots \mathrm{d}x_1 \\ &=\int_{A_1}p(x_1)\ldots\int_{A_{n-1}}p(x_{n-1}\,|\,x_{n-2}\ldots,x_1)\int_{A_n} p(x_n\,|\,x_{n-1}\ldots,x_1)\, \mathrm{d}x_n\,\mathrm{d}x_{n-1}\ldots \mathrm{d}x_1 \\ &=\beta_n\int_{A_1}p(x_1)\ldots\int_{A_{n-1}}p(x_{n-1}\,|\,x_{n-2}\ldots,x_1)\,\mathrm{d}x_{n-1}\ldots \mathrm{d}x_1 \\ &= \prod_{i=1}^n\beta_i. \end{align}

Maybe this is still not well formulated, since the $$A_i$$ are really any measurable set with constant probability for any $$x_1,\ldots,x_{i-1}$$. Is the correct mathematical formulation that the $$A_i$$ are random sets? I am quite new to probability theory so am unsure...

• Intuitively, no. In fact, if you know all about $Y_1 = X_1,$ so you know $Y_2 = X_2 \mid X_1.$ – Will M. Nov 28 '18 at 0:53
• What does $Y_2 = X_2 | X_1$ mean? There is no such thing defined in probability theory. Do you mean $Y_2 = E[X_2|X_1]$? – Michael Nov 28 '18 at 1:09
• Thanks Michael for the comment. Yes, I now realise that this didn't make sense. I have tried to rewrite the question so it is closer to what I really wanted to know. – Timothy Hedgeworth Nov 28 '18 at 14:46