Finding the CDF of a transformation of a random variable without first finding its PDF This is in the context of random variables and their transformations.
Given a random variable $X$, its probability density function (PDF) $f_{X}$, and another random variable $Y$, which is a a function of $X$, how do I calculate the cumulative density function (CDF) of $Y$ (without first finding the PDF of $Y$)?
Below is a question and my solution:
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Question:
The PDF of a random variable $X$ is $f_{X}(x) =\begin{cases}\dfrac{1}{3}&,& -2 < x < 1\\ 0&,&\text{ elsewhere }\end{cases}$.
Find the CDF of $Y$ where $Y=X^{4}$.
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My solution:
From the PDF of $X$, we get the CDF of $X$ by using $F_X(x)=\int_{-\infty}^{x} f_X(t)\space dt$.
This comes out to be: $F_{X}(x)=\begin{cases}0&,& x <-2\\ \dfrac{x+2}{3}&,& -2\leq x <1\\ 1&,& x\geq 1\end{cases}$.
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Since $X\in (-2, 1)$ here, and $Y=X^{4}$, hence $Y\in [0, 16)$.
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Finding the CDF of $Y=X^{4}$, that is, $F_{Y}(y)$:
$\begin{align}F_{Y}(y) &= P(Y \leq y) \\ &= P(X^{4} \leq y) \end{align}$
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Firstly, for$\space $ $-2 < x < 0 \space$ ($\equiv \space 0 < y < 16$), we have $X=-Y^{1/4}$ (since $X$ is negative for these values of $Y$).
So,
$\begin{align} F_{Y}(y) &= P(-y^{1/4}\leq X <0) \\ &=\int ^{0}_{-y^{1/4}}f_{x}(x)\space dx \\ &= \dfrac{x}{3}\Bigg|_{x\space =\space -y^{1/4}}^{x\space =\space 0} \\ &= \dfrac{1}{3}y^{1/4}\end{align}$
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Now, for $\space $ $0 \le x < 1 \space$ ($\equiv \space 0 \le y < 1$), we have $X=Y^{1/4}$ (since $X$ is non-negative for these values of $Y$).
So,
$\begin{align}F_{Y}(y) &= P(0 \leq X <y^{1/4}) \\ &=\int ^{y^{1/4}}_{0}f_{x}(x)\space dx \\ &= \dfrac{x}{3}\Bigg|_{x\space =\space 0}^{x\space =\space y^{1/4}} \\ &= \dfrac{1}{3}y^{1/4}\end{align}$
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Combining the above two results, I am getting:
(1). For $0 \le y < 1$: $\space F_{Y}(y) = \dfrac{2}{3}y^{1/4}$, and,
(2). For $1 \le y < 16$: $\space F_{Y}(y) = \dfrac{1}{3}y^{1/4}$.
The second one above is clearly wrong since it is $\textbf{not}$ giving $\space F_{Y}(16^-) = 1$, while the first one is correct (as confirmed by the answer that I have).
What have I missed here while finding the CDF of $Y$?
I know we can first find the PDF of $Y=X^{4}$ using a transformation formula and then find its CDF from its PDF, but I do not want to solve this using that formula.
 A: Your problem is with the supposed equivalences
$$
-2<x<0\ \equiv\ 0<y<16\\
$$
and
$$
0\le x<1\ \equiv\ 0\le y<1
$$
neither of which is a true equivalence, since
$\ 0<y<16\Leftrightarrow$$\,\{-2<x<0\}\color{red}{\vee\{0<x<2\}}\ $ and $\ 0\le y<1\Leftrightarrow$$\,\color{red}{-1<}x<1\ $.
For any $\ y\ge0 $ you have
\begin{align}
F_Y(y)&=P(Y\le y)\\
&=P\left(-y^{-\frac{1}{4}}\le X\le y^\frac{1}{4}\right)\\
&=P\left(\max\left(-2,-y^{-\frac{1}{4}}\right)\le X\le \min\left(1,y^\frac{1}{4}\right)\right)\\
&=\cases{P\left(-y^{-\frac{1}{4}}\le X\le y^\frac{1}{4}\right)&if $\ 0\le y\le1$\\
P\left(-y^{-\frac{1}{4}}\le X\le1\right)&if $\ 1< y\le1$6\\
1&if $\ 16<y$}\\
&=\cases{\frac{2y^\frac{1}{4}}{3}&if $\ 0\le y\le1$\\
\frac{y^\frac{1}{4}}{3}+\frac{1}{3}&if $\ 1< y\le16$\\
1&if $\ 16<y$}\ .
\end{align}
While you did end up with the correct result for $\ 0\le y\le1 $, your derivation of it was not fully coherent.
A: For $0 \le y \le 1$ you have that $x$ spans $-y^{1/4} \le x \le y^{1/4}$
instead for $ 1 < y \le 16$ $x$ spans $ -y^{1/4} \le x < -1$
