How can I find this distribution function?

Question

The random variable $X$ has the probability density function $f_X(x)$ represented in the figure

How can I find the $F_X(x)$ (distribution function)? And how can I find the value of $C$ when $P(|X|\le 0.5) \ge C$?

Answer 1

Integrating the triangular density you will easy find

$$ F_X(x) = \begin{cases} 0, & \text{if $x<-1$} \\ x+\frac{x^2}{2}+\frac{1}{2}, & \text{if $-1\leq x <0$} \\ x-\frac{x^2}{2}+\frac{1}{2}, & \text{if $0 \leq x <1$} \\ 1, & \text{if $x \geq 1$} \end{cases}$$

To calculate the requested probability

$$\mathbb{P}[|X|>0.5]=0.25$$

it is immediate to calculate the area of the two triangles in the tails (you will do it by mind, without any complicate calculation)

Answer 2

In general, to find the CDF $F_X(x)$ given the PDF $f_X(x)$, you should integrate: $$ F_X(x) = \int_{-\infty}^x f_X(t) \,dt. $$ In this case, $f_X$ is a piecewise function defined separately for $x<-1$, $-1 \le x < 0$, $0 \le x < 1$, and $x \ge -1$. So you have to integrate separately on those intervals:

• For $x < -1$, $F_X(x) = \int_{-\infty}^x f_X(t) \,dt = \int_{-\infty}^x 0 \,dt = 0$.
• For $-1 \le x < 0$, $F_X(x)$ can be broken up as $\int_{-\infty}^{-1} f_X(t)\,dt + \int_{-1}^x f_X(t)\,dt$, which becomes $0 + \int_{-1}^x (t+1)\,dt$.
• For $0 \le x < 1$, $F_X(x)$ can be broken up as $\int_{-\infty}^{-1} f_X(t)\,dt + \int_{-1}^0 f_X(t)\,dt + \int_0^x f_X(t)\,dt$, which becomes $0 + \int_{-1}^0 (t+1)\,dt + \int_0^x (1-t)\,dt$.
• Finally, for $x \ge 1$, we have to integrate over all four parts, but we expect that at this point $F_X(x) = 1$.

Another way to phrase it is that if we have intervals $(-\infty,-1) \cup [-1,0) \cup [0,1) \cup [1,\infty)$ on which $f_X$ has different definitions, we:

1. Set $F_X(t) = \int_{-\infty}^x 0\,dt$ for $x \in (-\infty,-1)$ and use this to compute $F_X(-1)$. (We'll get $0$.)
2. Set $F_X(t) = F_X(-1) + \int_{-1}^x (t+1)\,dt$ for $x \in [-1,0)$ and use this to compute $F_X(0)$.
3. Set $F_X(t) = F_X(0) + \int_0^x (1-t)\,dt$ for $x \in [0,1)$ and use this to compute $F_X(1)$. (We'll get $1$.)
4. Set $F_X(t) = F_X(1) + \int_1^x 0\,dt$ for $x \in (1,\infty)$.

The idea is that on every interval, $F_X(x)$ should be the whole probability of the previous intervals, plus some portion of the probability of the current interval.

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Finally, to find $\Pr[|X| \le 0.5]$ or $\Pr[-0.5 \le X \le 0.5]$, we can take $F_X(0.5) - F_X(-0.5)$.