0
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

Outline

(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.

$\endgroup$
  • $\begingroup$ part a) seems asking $\Pr\{X \leq - \frac {1} {2}\}$ instead. $\endgroup$ – BGM Feb 25 '16 at 3:02
  • $\begingroup$ @BGM Oh! Thanks! $\endgroup$ – Em. Feb 25 '16 at 3:09
  • $\begingroup$ @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. $\endgroup$ – Em. Feb 25 '16 at 18:08
  • $\begingroup$ @probablyme Awesome! I got that. Part (c) would be $\frac{dy}{dx}(\frac{x-3}{5})=\frac{1}{5}=f(x)$, right? $\endgroup$ – E. Otero Feb 25 '16 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.