Distribution of $Y=X(1-X)$ when $X\sim U(0,L)$

Question

I was wondering if someone could provide a hint of how to solve the following problem. I am working on some of my own research and can't seem to remember how to solve a problem like this:

Let $X\sim U(0,L)$ where $0 < L \le 1$. We wish to solve for the distribution $Y$ when

\begin{equation} Y = X(1-X) \end{equation}

I would provide my work but I honestly have not made it very far. Any assistance would be great. Thank you.

Answer 1

Comment: The first step is to determine the support of $Y.$ Then the so-called 'PDF method' or 'transformation method' should work nicely. Maybe you can start on this, edit results into your Question, and let us know of any specific difficulties you encounter.

It seems worthwhile to look at two kinds of examples:

The following brief simulation in R (for $L = 0.3 < .5)$ should help you decide on the support and give you a hint about the answer.

x = runif(10^6, 0, .3)
y = x*(1-x)
hist(y, prob=T, br = 50, col="skyblue2")

And for $L = .8 > .5,$

x = runif(10^6, 0, .8)
y = x*(1-x)
hist(y, prob=T, br = 50, col="skyblue2")

Plots of $y$ vs. $x$ for the two cases simulated above:

Answer 2

This is essentially a change of variable. Since $X$ takes values between $0$ and $L$, the random variable $Y=X(1-X)$ takes values between $0$ and $1/4$ (if $L \geq 1/2$) or between $0$ and $L(1-L)$ (if $L \leq 1/2$).

I'll deal with the second cas. For any bounded measurable function $f$,

$$\mathbb{E} (f(Y)) = \mathbb{E} (f(X(1-X))) = \frac{1}{L} \int_0^L f(x(1-x)) \ dx.$$

Since $L \leq 1/2$, the function $x \mapsto x(1-x)$ is bijective from $[0,L]$ to $[0,1-L]$. Making the change of variables $y = x(1-x)$, you get the density of $Y$.

If $L \geq 1/2$, the function $x \mapsto x(1-x)$ is no longer bijective, and you will have to split the interval $[0,L]$ on two sub-intervals on which it will be bijective.