# Chain rule of probability equivalence?

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.

• can you correct your first equation? – gunes Feb 15 at 22:51

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.