IID Uniform probability problem Three students independently attempt to solve a problem. Assume that the times taken by each student to solve the problem are iid according to U(0,30). Find the probability that the student who finishes last takes more than twice as long as the student who finishes first.
What is the answer if there are n students and the distribution is U(0,x)?
 A: Find $\mathsf P(Y_{(3)}>2Y_{(1)})$ , the probability that the largest of three iid rv is more than twice the least.
That is the probability that one of the values is more than twice another, and that the third value lays somewhere between both of them; for any given selection of the three r.v..
$$\begin{align}\mathsf P(Y_{(3)}>2Y_{(1)}) & = \int_0^{\boxed{15}} \boxed{3}\,f(x)\int_{\boxed{2x}}^{30} \boxed{2}\,f(y) \boxed{\big(F(y)-F(x)\big)}\operatorname d y\operatorname d x \\[1ex] & = 6\int_0^{15}\int_{2x}^{30}\dfrac{(y-x)}{30^3}\operatorname d y \operatorname d x \\[1ex] & = \dfrac 3 4\end{align}$$
Fill in the blanks, then extend the same principle to the multitude.


*

*What is the largest least value that another value in the interval can be twice or more?


15


*How many ways are there to pick a which variable is the least order statistic?


3?


*Where does the interval for the largest order statistic begin?


*

*Hint: we want the probability that this is at least twice the least o.s.


*How many ways are there to select a variable to be the most order statistic (given a least o.s.)?


2


*What goes in the brackets to for the middle order statistic(s) (which must lay somewhere between the least and most order statists)?


not sure – William Bernard
  
  
*
  
*Hint: $F(z)=\frac{z}{30}\mathbf 1_{[1;30]}(z)+\mathbf 1_{(30;\infty)}(z)$ is the CDF of the distribution, $f(z)=\frac 1{30}\mathbf 1_{[0;30]}(z)$ is the pdf.
  


$$\begin{align}\mathsf P(Y_{(n)}\geq 2Y_{(1)}) & = \int_0^{1/2} \int_{2 x}^1 n(n-1) (y-x)^{n-2} \operatorname dy \operatorname dx \\[1ex] & = (1-2^{-(n+1)}) \end{align}$$
