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suppose g is a deterministic function, and X a rv. Then we define Y=g(X) and

$f_Y(y) = f_X(g^{-1}(y))/|g'(g^{-1}(y))|$

But I don't understand at all why the $/|g'(g^{-1}(y))|$ is there. Seems like it would make perfect sense without it.

violin timbre
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    Your "definition" of $f_Y(y)$ is not applicable in all cases of deterministic $g$; you need to restrict $g$ to be a strictly monotone function. But instead of just pulling definitions out of thin air, you tried to understand the _derivation_ of the result, the reason for why the denominator is there (and why we need $g$ to be strictly monotone) will be perfectly obvious. – Dilip Sarwate Sep 16 '17 at 12:22
  • http://stla.github.io/stlapblog/posts/ChangeOfVariables.html – Stéphane Laurent Sep 16 '17 at 12:41
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    You need to remember that a PDF $f(x)$ is always (implicitly) accompanied by an infinitesimal term $\mathrm{d}x$. See https://stats.stackexchange.com/questions/4220 for an elementary explanation, https://stats.stackexchange.com/a/154298/919 for a modern (dating to the late 19th century) approach, or search this site for [Jacobian](https://stats.stackexchange.com/search?tab=votes&q=jacobian). – whuber Sep 17 '17 at 15:36

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From comments, whuber writes

You need to remember that a PDF $f(x)$ is always (implicitly) accompanied by an infinitesimal term $dx$. See Can a probability distribution value exceeding 1 be OK? for an elementary explanation, https://stats.stackexchange.com/a/154298/919 for a modern (dating to the late 19th century) approach, or search this site for Jacobian.

Sycorax
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