0
$\begingroup$

For a random variable given as:

$X=\begin{cases}3 , -3<x<5 \\ x,5 \leq x \leq 7 \\ 8, 7<x<9\end{cases}$

I am not sure if this describes a discrete random variable or continuous, as if it is discrete I would just list all the values $X$ can take and their probability:

$X={3,5,6,7,8}$

$P(X=3)=\frac{7}{11}$

$P(X=5)=P(X=6)=P(X=7)=P(X=8)=\frac{1}{11}$

$F(x)=\begin{cases}0 , x\le -3 \\ \frac{7}{11},-3<x<5 \\ \frac{10}{11}, 5 \leq x \leq 7 \\ 1, x> 7\end{cases}$

I am not sure how to do it if X was a continuous random variable, as other examples I have seen already have a pdf given so cdf is just the integral.

UPDATE: as brought up by Bungo, I am also given that the random variable is sampled uniformly.

So I guess using the uniform distribution

$f(x)=\begin{cases}0 , otherwise \\ \frac{1}{8},-3<x<5 \\ \frac{1}{2}, 5 \leq x \leq 7 \\ \frac{1}{2}, 7<x<9 \end{cases},F(x)=\begin{cases}0 , otherwise \\ \frac{1}{8},-3<x<5 \\ \frac{5}{8}, 5 \leq x \leq 7 \\ \frac{9}{8}, 7<x<9 \end{cases} $

But I think this is incorrect as $\frac{9}{8}$ should be 1

$\endgroup$
  • 1
    $\begingroup$ What is the probability distribution of $x$? Without this, there's not enough information to answer. $\endgroup$ – Bungo Aug 27 '18 at 6:40
  • $\begingroup$ @Bungo , i am not given it. but if it helps. i am also told that the value of the random variable is chosen uniformly at random from the interval (-3,9). Does this suggest that the pdf is of the uniform distribution? $\endgroup$ – glockm15 Aug 27 '18 at 6:47
1
$\begingroup$

The random variable is neither discrete nor continuous. It takes values other than $3,5,6,7,8$. It is defined on the space $(-3,9)$ with normalized Lebesgue measure as the basic probability measure. The correct values of $F(x)$ are $0$ for $x <3$, $2/3$ for $3 \leq x \leq 5$, $\frac {x+5} {12}$ for $5 \leq x \leq 7$, $1$ for $x \geq 8$.

$\endgroup$
  • $\begingroup$ Is there another method to find F(x) instead of normalized Lebesgue measure? So what I have for f(x) is correct? $\endgroup$ – glockm15 Aug 27 '18 at 7:46
  • 1
    $\begingroup$ By the very definition this random variable takes uncountable number of values: it takes all values between $5$ and $7$. So your answer is not correct. Answering this question requires some familiarity with general probability spaces because $X$ is neither discrete nor continuous. Is it a mixture of the two types. $\endgroup$ – Kavi Rama Murthy Aug 27 '18 at 7:53
  • $\begingroup$ The formula for $5 \leq x \leq 7$ isn't right. In that interval, the cdf should climb linearly from $2/3$ to $5/6$. Also should mention that the cdf equals the constant $5/6$ for $7 \leq x < 8$. $\endgroup$ – Bungo Aug 27 '18 at 15:31

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.