Cumulative distribution (uniform distribution/continuous)

probability

Let $X$ be a continuous uniform random variable on the interval $[0,2]$.

How to determine the cumulative distribution function of $Y=X(2-X)$?

I got:

Since $X$ is uniformly distributed, the probability density function is $f_X(x)=\frac{1}{2}\boldsymbol{1}_{0<x<2}$

Now I tried to find the cumulative distribution function:

$F_Y(y)=\mathbb{P}(Y \leq x)=\mathbb{P}(X(2-X) \leq x)=\mathbb{P}((2X-X^2) \leq x)$

How to continue the transformation such that I get $\mathbb{P}(X \leq …)$?

I don't see how to rewrite $2X-X^2$ here.

Best Answer

Add $-1$ to both sides,$$2X-X^2-1=-(X-1)^2\le x-1\implies(X-1)^2\ge1-x$$When $x\ge1$, the inequality is always true and the c.d.f. of $Y$ is $1$. For $x<1$,$$P[(X-1)^2\ge1-x]\iff P(|X-1|\ge\sqrt{1-x})\\\iff P(X\le1-\sqrt{1-x})+P(X\ge1+\sqrt{1-x})$$For $x<0$, the first and second terms are both $0$ (Why?). So you only have to talk about $x\in[0,1)$. Can you take this forward?

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Probability Distributions – Finding the Distribution of the Reciprocal of a Random Variable

Observe that $$\left(-\infty, \dfrac{1}{x}\right) \cup \left\{\dfrac{1}{x} \right\}=\left(-\infty, \dfrac{1}{x} \right]$$ and that the sets $\left(-\infty, \dfrac{1}{x}\right)$, $\left\{\dfrac{1}{x} \right\}$ are disjoint.

Therefore, on the left-hand side, $$\mathbb{P}\left[\left(-\infty, \dfrac{1}{x}\right) \cup \left\{\dfrac{1}{x} \right\}\right]=\mathbb{P}\left[\left(-\infty, \dfrac{1}{x}\right) \right] + \mathbb{P}\left(\left\{\dfrac{1}{x} \right\}\right) = \mathbb{P}\left(X < \dfrac{1}{x}\right)+0$$ because $$\mathbb{P}\left(\left\{\dfrac{1}{x} \right\}\right) = 0$$ due to $X$ being a continuous random variable, and on the right-hand side, we simply have $$\mathbb{P}\left[\left(-\infty, \dfrac{1}{x} \right] \right] = \mathbb{P}\left(X \leq \dfrac{1}{x}\right)$$ Hence, equating the two sides, $$\mathbb{P}\left(X \leq \dfrac{1}{x}\right) = \mathbb{P}\left(X < \dfrac{1}{x}\right)$$

Cumulative distribution of mixed random variable function

You can rely on basic principles. Note that $Y$ is of the form

$$ Y = g(X), \qquad \text{where} \quad g(x) = \begin{cases} 2x, & \text{if $x < \frac{1}{2}$,} \\ 0, & \text{if $x \geq \frac{1}{2}$.} \end{cases} $$

From this and using the definition of CDF,

\begin{align*} F_Y(y) &= \mathbf{P}(Y \leq y) = \mathbf{P}(g(X) \leq y). \end{align*}

So the problem boils down to solving the inequality $g(x) \leq y$ for each given $y$. To this end, you may study the behavior of the function $g(x)$. Note that the graph of $g(x)$ looks like:

From this, we can solve the inequality $g(x) \leq y$ for $x$ for each given value of $y$:

Range of $y$	Sol. of $g(x) \leq y$	$F_Y(y)$
$y < 0$	$x \leq y/2$	$\begin{aligned} \mathbf{P}(X \leq y/2) = 0 \end{aligned}$
$0 \leq y < 1$	$x \leq y/2$ or $x \geq \frac{1}{2}$	$\begin{aligned} &\mathbf{P}(X \leq y/2) + \mathbf{P}(X \geq \tfrac{1}{2}) \\ &= 1 - e^{-y} + e^{-1} \end{aligned}$
$y \geq 1$	$x \in \mathbb{R}$	$1$

Summarizing, the CDF of $Y$ is given by

$$ F_Y(y) = \begin{cases} 0, & y < 0, \\ 1 - e^{-y} + e^{-1}, & 0 \leq y < 1, \\ 1, & y \geq 1. \end{cases} $$

For fun, I also included the graph of $F_Y$ below:

Best Answer

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Cumulative distribution of mixed random variable function

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