probabilityprobability distributions
Suppose that the cumulative distribution function of a random variable X is given by
$
F(a) =
\begin{cases}
0,& a < 0 \\
1/5, & 0 \leq a < 2 \\
2/5, & 2 \leq a < 4 \\
1, & a \geq 4
\end{cases}
$
Find the probability mass function of X?
My reasoning is as follows:
The cdf is discontinuous at the points 0, 2, and 4. Between these $F'(a)$ is defined and $=0$, hence the pmf needs definition only at these points. But how do we get the probabilities at a = 0, 2, 4?
Every random variable $X$ has a distribution function $F(x)=\mathbb P(X\leq x)$, by definition. Also, by definition, if we're dealing with continuous random variables, we have: $$ F(x)=\mathbb P(X\leq x)=\int_{-\infty}^xf(t)dt, $$ where $f$ (non-negative, continuous) is our density function. Therefore, $$ f(x)=\frac{d}{dx}F(x), $$ where we're applying the Fundamental Theorem of Calculus.
Since your distribution function is defined in cases, you have to calculate the derivative in cases too. I will show one case: $F(x)=0$ for $x<-1$. Thus we have: $\frac{d}{dx}F(x)=0$ for $x<-1$. You should finish the other two cases, and then write out $f(x)$ in the same way as $F(x)$ has been written out, i.e. by using cases.
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 $g(x) \leq y$ | $F_Y(y)$ | Graph |
|---|---|---|---|
| $y < 0$ | $x \leq y/2$ | $\begin{aligned} \mathbf{P}(X \leq y/2) = 0 \end{aligned}$ |  |
| $0 \leq y < 1$ | $x \leq y/2$ 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
HINT:
$$F'(4)=F(a>4)-F(2\le a<4)$$ $$F'(2)=F(2\le a<4)-F(0\le a<2)$$ $$F'(0)=F(0\le a<2)-F(a<0)$$