1
$\begingroup$

I am currently working on distribution with density function

$$f(x)=\begin{cases} \frac{2}{5}|x-2|,& \text{0 ≤ x ≤ 3} \\\\0 & \text{otherwise}\end{cases}$$

I have found that cumulative distribution function of $F(X)$ for $2 ≤ x ≤ 3$ which is $\frac{1}{5}x^2-\frac{4}{5}x + \frac{8}{5}$ and $F(X)$ for $0≤x≤2$ is $\frac{4x-x^2}{5}$.

I know that to find median you have to set

$$\int_{-\infty}^x f(x)\,dx= 0.5.$$

But what $f(x)$ do I use to find the median of the probability distribution of $X$.

Also how do I find $E(X)$ for this distribution?

Do I just use integral of $xf(x)$? and what $f(x)$ should I use?

I just cannot find any distribution with absolute value thanks!

$\endgroup$
0

2 Answers 2

1
$\begingroup$

A number $m$ is a median of $X$ if $\mathbb P(X\leqslant m)\geqslant\frac12$ and $\mathbb P(X\geqslant m)\leqslant\frac12$. The density of $X$ is $$f(x) = \frac25(2-x)\mathsf 1_{(0,2)}(x) + \frac25(x-2)\mathsf 1_{(2,3)}(x), $$ and so the CDF is obtained by integrating: $$F(x)=\int_0^x f(t)\ \mathsf dt = \left(4x-\frac15x^2\right)\mathsf 1_{(0,2)}(x) + \frac15\left(4 + (t-2)^2 \right)\mathsf 1_{[2,3)}(x) + \mathsf 1_{[3,\infty)}(x). $$ Since $F(2)=\frac45>\frac12$ and $F$ is continuous on $(0,2)$, it follows that the median is the unique number $m$ which satisfies $$\int_0^m \frac25(2-x)\ \mathsf dx = \frac12, $$ that is, $m = 2 - \sqrt{\frac32}$.

$\endgroup$
1
  • 1
    $\begingroup$ thanks alot for the help I really liked your explanation $\endgroup$
    – Danny Yoon
    Feb 25, 2016 at 23:12
1
$\begingroup$

Did you graph $f_X(x)$? It should be clear that the median is in the interval $[0,2]$. Hence you could do $$\int_0^c \frac{2}{5}(-x+2)\,dx = .5$$ from the outset, or, since you found $F_X(x)$, use $$F_X(c) = .5$$ to find $c$, the median.

The expectation can be found from the usual $$E[X] = \int_0^3 f_X(x)\,dx$$

or you could use the tail sum $$E[X] = \int_0^31-F_X(x)\,dx.$$ Of course, there are cases.

$\endgroup$

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.