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: graph

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!


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 $Y$.


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).

  • $\begingroup$ thanks, but what do you mean by "behave differently"? $\endgroup$ – superuser123 Jan 16 at 15:27
  • $\begingroup$ Sorry, poor word choice. I meant that they behave differently from one another. In substance, they are different from each other, and when calculating the density function you must consider these regions separately, because they "take inputs" from different regions of the domain of $X$. $\endgroup$ – Easymode44 Jan 16 at 16:06
  • $\begingroup$ I actually did understand that you meant "different from each other" but I still can't understand what is the difference, I mean, the domain of X is $ [-1, 3] $, so what's the division about? $\endgroup$ – superuser123 Jan 16 at 16:20
  • $\begingroup$ Added some clarification, hope it helps $\endgroup$ – Easymode44 Jan 17 at 12:36

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:

enter image description here

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 $[81, \infty)$.


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