Compute the order statistics probability of $P(X_{(3)} < aX_{(2)})$

Question

Suppose $a>1$, $X_1,X_2, X_3,$ and $X_4$ are four iid continuous random variables with $U(0,a)$ random variables so their common density function is $f(x)=\frac{1}{a}, 0<x<a.$

$a.$ joint pdf of $X_{(2)}$ and $X_{(3)}$ $i.e.$ $f_{X_{(1)}X_{(2)}}(x,y), x<y$

I'm fairly sure i have this part solved correctly. I found the CDF's to be $\frac{x}{a}$ and $\frac{y}{a}$ and then plugged into the formula: $$\frac{n!}{(i-j)!(j-i-1)!(n-j)!}F(x)^{i-1}f(x)[F(y)-F(x)]^{j-i-1}f(x)[1-F(y)]^{n-j}$$ which resulted in my solution of: $$\frac{24x(1-\frac{y}{a})}{a^3}$$

$b.$ Compute $P(X_{(3)} < aX_{(2)})$

As for this part im completely unsure what to do and if my result from part a is necessary to solve this part. How do i proceed to solve this part? A detailed solution would be nice, since the chapter in my book hasn't helped me very much in regards to this question.

$Edit:$

I think the integral to solve it may look something like this: $$\int_{0}^1\int_{0}^\frac{y}{a}\frac{24x(1-\frac{y}{a})}{a^3}dxdy$$which i simplified to: $$\frac{-3}{a^6}+\frac{4}{a^5}$$

am i on the right track here?

Answer 1

It is quite easy to check ones work with a computer algebra system.

Given: Parent random variable $X \sim \text{Uniform}(0,a)$, where $a>1$, with pdf $f(x)$:

(a) Then, the joint pdf of the $2^{\text{nd}}$ and $3^{\text{rd}}$ order statistics, in a sample of size 4, is say $g(x_2, x_3)$:

where I am using the OrderStat function from the mathStatica package for Mathematica to automate. This is the same as the solution you obtained, except that you have not specified the constraint that $X_{(2)} < X_{(3)}$.

(b) $P\big(X_{(3)} < a X_{(2)} \big)$, where $a>1$ (above), is:

... so your second part requires some attention.

Notes

1. As disclosure, I should add that I am one of the authors of the software used above.

Answer 2

Notes (too long for a Comment):

(1) @wolfies answer is correct as $a \rightarrow 1$; not so for @After_Sunset.

(2) Also, a simulation in R based on 100,000 runs of the experiment with $a = 2$ gives very nearly 0.75. Similar agreement for several other values of $a.$ That's no substitute for a general formula. It seems worthwhile trying to find what's wrong in the traditional integration, now that we know something is.

> m = 10^5;  n = 4;  a = 2;  u2 = u3 = numeric(m)
> for(i in 1:m) {s = sort(runif(4, 0, a))
+ u2[i] = s[2];  u3[i] = s[3]}
> mean(u3 < a*u2);  1 - a^-2
[1] 0.74869
[1] 0.75