# Probability Density Function of a certain random variable

I am currently trying to understand probability to the continuous probability space. I am at a lost of where to begin with regards to tackling a practice problem that I found at the end of text book. I understand that the sample space would be $[-1,1]$ and that for the first part we would like to find the probability of $[-1,\frac{1}{2})$; however, how do I use the pdf to find this value? I understand it when using uniform probability on a unit interval, but I am not sure how to apply that to this situation.

In addition, how would I go about then finding the cumulative distribution function for this?

I will list the entire question in case any context is needed.

(a) A certain random variable, $X$, takes real values between $-1$ and $1$, and its pdf is given by the following expression for $-1 \le x \le 1$: $$f(x)=1-\lvert x \rvert$$ Compute the probability $P(X < -1/2)$.

(b) Compute the cdf of the random variable described in the preceding part.

(c) The cumulative distribution of a certain real-valued random variable, $X$, is given by: $$F(x) = \begin{cases} 0, & \text{if n\le 3} \\ \frac{x-3}{5}, & \text{if 3 < x \le 8} \\ 1, & \text{if x > 8} \end{cases}$$ Find the PDF of this random variable.

I would appreciate any help or guidance to point me in the right direction!

(a) We are told $$f_X(x) = 1-|x|$$ where $-1\leq x\leq 1$. Hence $$P(X\leq -1/2) = \int_{-1}^{1/2} f_X(x)\,dx = \int_{-1}^{1/2} 1-|x|\,dx.$$
(b)Essentially, you are being asked $$P(X\leq x) = \int_{-1}^x f_X(x)\,dx = \begin{cases} 0,& x<-1\\ \_\_\_\_,& -1\leq x<0\\ \_\_\_\_,& 0\leq x \leq 1\\ 1,& x>1\end{cases}$$ where again there are cases since there is an issue at $x = 0$.
(c) Recall that the pdf is the derivative of the cdf. Furthermore, notice that if $x<3$, then $F_X(x) = 0$. Intuitively, this means there is no density when $x<3$; hence no 'accumulation of area' under the curve up to $3$. When $3\leq x\leq 8$, there is density; hence you are accumulating area under the curve. Finally, when $x> 8$, you're done! There is no more area to accumulate.
• part a) seems asking $\Pr\{X \leq - \frac {1} {2}\}$ instead. – BGM Feb 25 '16 at 3:02
• @E.Otero No problem. One small note. In part (b), the second $\_\_\_\_$ should be $-\frac{1}{2}x^2+x+\frac{1}{2}$. Students tend to make a mistake here. – Em. Feb 25 '16 at 18:08
• @probablyme Awesome! I got that. Part (c) would be $\frac{dy}{dx}(\frac{x-3}{5})=\frac{1}{5}=f(x)$, right? – E. Otero Feb 25 '16 at 18:12