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Given the conditional probability density $P(x)$, the following is correct for any change of variables $x\rightarrow z$:

Eq.1: $P(x)dx=P(z)dz$

Eq.2: So $P(z)=P(x)[dx/dz]$

Is this also valid for the conditional distribution? Or otherwise, what is the general relation in this case?

Eq.3: Is $P(x|y)dx=P(z|y)dz$ ?

mccurcio
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user1611107
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  • I think you have answered your own question by using the simple re-arrangement of the first equation to find equation 2. X may go to Z BUT dx does not necessarily go to dz. correct? – mccurcio Mar 19 '21 at 16:04
  • (1) and (3) can be understood as a mere change of notation from "$x$" to "$z.$" Why would introducing "$\mid y$" in (3) create any difference compared to (1)? – whuber Mar 19 '21 at 16:38
  • @ oaxacamatt and whuber: I am pretty sure that the Jakobian changes somehow, but don't know how.. I definitely don't understand the answer yet... Can't there be a case, for example, where x is dependent on y, but after change of variables it no longer true? In this case for example, it should come out in the correct version of Eq. (3)... I think – user1611107 Mar 19 '21 at 16:45
  • I think you should be a little more rigorous with your definitions. i.e by $x\to z$ do you mean $z = f(x)$ for some function $f$? By $P(x)$ and $P(z)$ do you mean the probability density function with respect to $x$ and $z$ respectively? In general the pdf of transformed r.v. is not the same as the original r.v. [see transformed r.v.s](https://en.wikipedia.org/wiki/Random_variable#Functions_of_random_variables) – bdeonovic Mar 19 '21 at 18:48
  • In a post at https://stats.stackexchange.com/a/154298/919 I explain where the Jacobian comes from. – whuber Mar 19 '21 at 19:02
  • @bedeonovic From the standpoint of [Differential algebra](https://en.wikipedia.org/wiki/Differential_algebra) as developed by Ritt, Thomas, and others between 1930 and 1950, this is perfectly rigorous notation! – whuber Mar 19 '21 at 19:11

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