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Is $f(x_1, x_2, x_3) = f(x_1| x_2, x_3)f(x_2 |x_3)f(x_3)$ equivalent to the standard chain rule of probability $f(x_1, x_2, x_3) = f(x_3 |x_2, x_1)f(x_2|x_1)f(x_1)$?

I ask because $f(x_1, f_2) = f(x_1|x_2)f(x_2) = f(x_2|x_1)f(x_1)$ and thought this may apply to high order cases.

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  • $\begingroup$ can you correct your first equation? $\endgroup$ – gunes Feb 15 at 22:51
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Based on your second paragraph, I assume you ask if the following is true or not: $$p(x_1|x_2,x_3)p(x_2|x_3)p(x_3)=p(x_3|x_2,x_1)p(x_2|x_1)p(x_1)$$

Yes, it holds. The indices doesn't matter. Call them $x_1=a,x_2=b,x_3=c$ and you won't have a standard form. To be specific, you'll have $n!$ different factorizations if you have $n$ RVs.

The intuition is you can always group RVs as if they're single, e.g. $x_1=a, (x_2,x_3)=b$. When you apply your standard chain rule formula you can end up in any possible factorization.

Note: $p(x),f(x)$ are all abuse of notations. A better notation is to use something like $p_X(x)$, to indicate both the RV and the specific value.

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  • $\begingroup$ Thank you very much. I have also corrected the question so it is clearer. $\endgroup$ – InvestingScientist Feb 16 at 9:12

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