I can't figure out a specific step in the solution for the following problem: Let $X \sim U(-1,3)\,$ and $\,Y=X^4 $, find the PDF of $Y$

Question

I am trying to study from pretty old class notes and encountered the following problem:

$\text{Let } X \ \sim U(-1,3)\text{ and } Y=X^4 $, find the PDF of $Y$.

In the solution they used the following drawing of $Y$:

And then, in order to compute $F_Y(t)$, they divided $t$ to the following ranges: $ t<0; 0\leq t < 1; 1 \leq t < 81; t\geq 81 $, informally stating that:

When looking for the CDF of a transformation, it changes its range whenever the graph of y changes its shape

Well, this is at least the best translation I could come up with. Can anyone please help me to figure out how those ranges where chosen? I encounter the same dilemma over and over and I can't see the logic behind it.

Any help would be appreciated, thanks!

Answer 1

You are performing a variable transformation. The boundary ranges were clearly chosen to indicate those regions of the codomain hyperplane which display different behaviors. Indeed, $X^4$ is a non-negative random variable (and therefore you have a density equal to $0$ for all $t<0$). $X$ can at most be $3$ (not including $3$) and thus you also have a density of $0$ for $t>81=3^4$.

As for the intermediate regions, you have that values of $X\in[1,3)$ are mapped to $1\leq t<81$, while both values of $X\in(-1,0]$ and $X\in[0,1)$ are mapped to $0<t<1$. Densities in these regions behave differently from one another, and therefore different ranges must be outlined in order to study the density of $Y$.

Clarification

It's true that the domain of $X$ is $(-1,3)$, but values from different areas of the domain are mapped in different ways to $Y$.

Values from $X\in[1,3)$ are mapped one-to-one on $Y\in [1,81)$.

However, from $X\in(-1,1)$ the mapping is not bijective. This is due to the fact that $4$ is an even integer.

This is why these different areas of the domain of $Y$ need to be studied separately. You will find that the cumulative distribution function in the ranges you have indicated don't behave in the same way (they follow a different law).

Answer 2

Make sure you understand the following to start: let's consider the interval $(-1, 3)$. If I take every number in $(-1, 3)$ to the fourth power, I end up getting the interval $(1, 81)$.

However, the transformation $x \mapsto x^4$ is not a one-to-one transformation when looking over $(-1, 3)$. Transformations that are one-to-one (or "injective") are the easiest ones to deal with. You may have not taken real analysis yet, so I'll demonstrate what "one-to-one" means.

What is happening when you take every number in $(-1, 3)$ to the fourth power is the following:

To describe the image above, here's what it is saying: the interval $(-1, 0]$ turns into $[0, 1)$ when we take each value in $(-1, 0]$ to the fourth power. Similarly, $[0, 1) \mapsto [0, 1)$ and $[1, 3) \mapsto [1, 81)$ when we take each value in their respective intervals to the fourth power.

$Y = X^4$ is not a one-to-one transformation when looking at values of $X$ in $(-1, 3)$. What a one-to-one transformation means is that for each transformed value $y = x^4$, there is only one $x$ which is transformed to $y$. To see why this isn't a one-to-one transformation, simply take any $y$-value in $(0, 1)$; say, for example, $1/2$. If $1/2 = x^4$, then there are two $x$-values which satisfy this equation, namely $1/\sqrt[4]{2}$ and $-1/\sqrt[4]{2}$, both of which are in the interval $[-1, 1]$.

More generally, we have to specifically call out every $y$-value in $[0, 1)$ because there are two corresponding $x$-values which are transformed to a $y$-value in the interval $(0, 1)$, with one of the $x$-values in $(-1, 0)$ and the other $x$-value in $(0, 1)$.

Next, this problem isn't the case when $y$ is in the interval $[1, 81)$, because there is only one value of $x$ in $[1, 3)$ which is transformed to any value in $[1, 81)$, so the transformation is one-to-one when we look at the outputted interval $[1, 81)$; this is NOT the case for $(0, 1)$. That is, if instead you were told that $X \sim U[1, 3)$, the only values that $Y$ could take are in $[1, 81)$, and it would be a one-to-one transformation and therefore you would not have to split $[1, 81)$ into special cases, but that is not the case in this problem.

It is also hopefully clear why $(-\infty, 0)$ and $[81, \infty)$ are called out as well; these are intervals over which the density of $Y$ has to be zero, because there is no value in $(-1, 3)$ which maps to any value in $(-\infty, 0)$ or $[81, \infty)$.