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I want to know if the change of variables formula in probability is a special case of the change of variables in calculus, or something different.

Trying to think for myself, first write the CoV for probability within informal derivation, $$ p(y)dy = p(x)dx \quad\Rightarrow\quad p(y) = p(x) \left| \frac{dx}{dy} \right| $$ and I believe we could add integration to make it look more like the calculus case, though the answer is one $$ \int p(y) \, dy = \int p(x) \left|\frac{dx}{dy}\right| \, dx = 1 $$

Now writing the CoV for calculus, mulitidimensional form, $$ \iint_\Omega f(g(p)) |J_g| \, dp = \iint_{g(\Omega)} f(p) \, dp $$ and rewrite this to look like the probability case, $p\rightarrow x$ $$ \iint_\Omega f(y=g(x)) |J| \, dx = \iint_{g(\Omega)} f(x) \, dx $$ move the $|J|$ to the other side, and name $f \rightarrow p$ $$ \iint_\Omega p(y) \, dx = \iint_{g(\Omega)} p(x) |J|^{-1} \, dx $$

This does not quite look like the probability case, because on the left side it should be $$ \iint_\Omega p(y) \, \mathbf{d}y $$ rather than $$ \iint_\Omega p(y) \, \mathbf{d}x $$

Probably I made a mistake. Can anyone explain what the mistake is?

self studying

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    The change of variables formula of probability *is* that of Calculus: there is no difference whatsoever. One rigorous way of approaching it is explained at https://stats.stackexchange.com/a/154298/919 using differential forms. – whuber Dec 28 '19 at 21:32
  • That link looks almost understandable, but I do not know differential forms. Is it possible to comment on what is wrong with the informal derivation I gave above? It tries to put the probability and calculus CoV in the same form, but fails. – stochasticthoughtprocess Jan 01 '20 at 04:43

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