let y1<y2<y3 be the order statistics of random sample of size a distribution
f(x)=2x , 0<x<1 show that z1=y1/y3, z2=y2/y3 ,z3=y3 are mutually stochastically independent
To show that z1, z2, and z3 are mutually stochastically independent, we need to show that their joint probability distribution can be written as the product of their marginal probability distributions.
Let's start by finding the marginal probability distribution of each of the variables z1, z2, and z3.
For z1 = y1/y3, we can write its cumulative distribution function (CDF) as follows:
Fz1(z1) = P(z1 ≤ z) = P(y1/y3 ≤ z) = P(y1 ≤ zy3) = [P(y1 ≤ zy3, y2 ≤ y3)] / P(y2 ≤ y3)
Since the order statistics y1, y2, and y3 are assumed to be mutually independent, we can write this as:
= [P(y1 ≤ zy3) * P(y2 ≤ y3)] / P(y2 ≤ y3)
Now, let's focus on each of the terms separately.
P(y1 ≤ zy3) can be calculated as:
∫(0 to 1) ∫(0 to zy3) 2x dy1 dx
= ∫(0 to 1) zy3(2x) dx
= 2z ∫(0 to 1) xy3 dx
= zy3
P(y2 ≤ y3) can be calculated as:
∫(0 to 1) ∫(y2 to 1) 2x dy3 dx
= ∫(0 to 1) (1-y2)(2x) dx
= 2 ∫(0 to 1) x - xy2 dx
= 2[(1/2) - (1/3)y2]
= 1 - (2/3)y2
P(y2 ≤ y3) is simply 1 since y2 ≤ y3.
Now, plugging in these values back into the expression for Fz1(z1), we get:
Fz1(z1) = [zy3 * (1 - (2/3)y2)] / (1 - (2/3)y2)
= zy3
Since Fz1(z1) simplifies to zy3, we can conclude that z1 follows a uniform distribution on the interval [0, 1].
Similarly, we can show that z2 and z3 also follow uniform distributions on the intervals [0, 1] and [0, 1] respectively.
Now, since z1, z2, and z3 each follow a uniform distribution on the interval [0, 1], we can conclude that they are mutually stochastically independent.
Thus, we have shown that z1 = y1/y3, z2 = y2/y3, and z3 = y3 are mutually stochastically independent.