$$f(x) = \begin{cases} 2(1-\frac{1}{x^2}) & \text{if }1\le x\le2 \\ 0 & \text{otherwise} \end{cases} $$ Compute the CDF of X:
$$ \int^X_12(1-\frac{1}{y^2})dy = 2x+\frac{2}{X}-4 $$
So I got $$ F(X) = \begin{cases} 0 & \text{if }x\lt1 \\ 2X+\frac{2}{X}-4 & \text{if }1\le X\le2 \\ 1 & \text{if }x\gt2 \end{cases} $$ My problem starts here:
It asks to obtain an expression for the (100p)th percentile and the value of the median.
Isn't the given expression $$f(x) = 2(1-\frac{1}{x^2}) $$ (100p)th percentile?
In order to find median I guess I will have to do $$ \int^c_12(1-\frac{1}{x^2})dx = 0.5 $$ and then solve for c.
And the expected value which I think: $$ E(x) = \int^2_1x2(1-\frac{1}{x^2})dx $$
How do the mean and median of this distribution compare?
Can you help me out with this?
Thank you!
The CDF is indeed: $$F_X(x) = \begin{cases} 0 & : x\lt 1 \\ 2x+\frac{2}{x}-4 & : 1\le x < 2 \\ 1 & : 2\le x \end{cases}$$ Note: Be careful with your capitalisations. $X$ should be your random variable. $x$ should be a value or variable of integration. They are not the same thing. Don't mix and match.
The $100p^{\rm th}$ percentile is the inverse of the CDF:$$ F_X^{-1}(p)=\begin{cases} 1 & ; p<0 \\ \tfrac 1 4(p+\sqrt{~p+8~}\sqrt{~p~}+4) & : 0\leq p < 1\\ 2 & : 1\leq p \end{cases}$$
The median is, of course, the 50th percentile, so $m=F_X^{-1}(0.50)$
The mean is, as you had, $\mathsf E(X)=2\,\int_1^2 x\,(1-\tfrac 1 {x^2})\operatorname d x = 3-\ln 4$