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...

  • $\begingroup$ Intuitively, no. In fact, if you know all about $Y_1 = X_1,$ so you know $Y_2 = X_2 \mid X_1.$ $\endgroup$ – Will M. Nov 28 '18 at 0:53
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    $\begingroup$ What does $Y_2 = X_2 | X_1$ mean? There is no such thing defined in probability theory. Do you mean $Y_2 = E[X_2|X_1]$? $\endgroup$ – Michael Nov 28 '18 at 1:09
  • $\begingroup$ 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. $\endgroup$ – Timothy Hedgeworth Nov 28 '18 at 14:46

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