# TransformedDistribution with respect to new variable

Given a custom pdf $f_x(x)$, I'm trying to find it's transformation $f_y(y)$ where $$y=x^2$$ and $$f_x(x)=30*x^2 (1 - x)^2, 0<x<1$$ I tried to using the following commands:

y = x^2

PDF[TransformedDistribution[y, x \[Distributed]
ProbabilityDistribution[30*x^2 (1 - x)^2, {x, 0, 1}]]]


However, the answer comes up in terms of x, not y:

$\text{Function}\left[x, \begin{array}{cc} \{ & \begin{array}{cc} 15. \left(\sqrt{x}-1.\right)^2 \sqrt{x} & 0.\leq x\leq 1.\land 0.<\sqrt{x}<1. \\ 0. & \text{True} \\ \end{array} \\ \end{array} ,\text{Listable}\right]$

Could you please point me in the right direction?

## 1 Answer

What looks like x is actually \[FormalX] (a dummy argument of the pure function). If you name that function (e.g., pdF= PDF[Transformed....]) you can use it as pdF[z] or pdF[y] etc.

y = x^2;
pdF = PDF[TransformedDistribution[y,
x \[Distributed] ProbabilityDistribution[30*x^2 (1 - x)^2, {x, 0, 1}]]]


pdF[z]


pdF[1/2] // N
(* 0.909903 *)


If you want to use y as the argument, you need to Clear the values and definition of y first:

ClearAll[y];
pdF[y]


Plot[pdF[z], {z, 0, 1}]


Integrate[PiecewiseExpand@pdF[z], {z, 0, 1}]
(* 1 *)
Integrate[PiecewiseExpand@pdF[z], {z, 0, #}] & /@ Range[0, 1, .1]
(* {0, 0.185201, 0.401758, 0.588938, 0.736979, 0.846194,
0.920709, 0.966401, 0.990018, 0.998751, 1.} *)