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 density function (PDF) factorizes into the product of their individual PDFs.

First, let's find the individual probability density functions (PDFs) for z1, z2, and z3.

Given that y1 < y2 < y3 are the order statistics from a random sample of size a distribution, we know that y1, y2, and y3 are all independent and identically distributed random variables.

Let's find the PDF for each of them:

1. For y3:
The probability density function (PDF) of y3 can be determined using the order statistics. Since y3 is the maximum value, its PDF is given by the cumulative distribution function (CDF) of the original distribution raised to the power of the sample size (a):

F(y3) = P(y3 ≤ t) = [F(t)]^a

Differentiating both sides with respect to t, we get the PDF of y3:

f(y3) = a*[F(t)]^(a-1)*f(t)

where f(t) is the PDF of the original distribution.

2. For y2:
The PDF of y2 can be found by considering the event that y2 ≤ t. For this event to happen, both y1 and y2 must be less than or equal to t. Therefore, we write the probability as:

P(y2 ≤ t) = P(y1 ≤ t and y2 ≤ t)

Since y1 and y2 are independent, we can write:

P(y2 ≤ t) = P(y1 ≤ t) * P(y2 ≤ t)

The PDF of y2 can be derived by differentiating both sides of the above equation with respect to t:

f(y2) = f(t) * F(t)

where f(t) is the PDF of the original distribution and F(t) is the CDF.

3. For z1:
We are given that z1 = y1/y3. To find the PDF of z1, we need to consider the joint distribution of y1 and y3. Since y1 and y3 are independent, the joint PDF can be expressed as the product of their individual PDFs:

f(y1, y3) = f(y1) * f(y3)

To find the PDF of z1, we need to perform a change of variables. Let's define a new variable, u = y1/y3. Solving for y1 and y3 in terms of u, we have:

y1 = u * y3

Taking the derivative with respect to u, we get:

dy1/du = y3

Rearranging, we get:

y3 = (1/u) * y1

Now, we need to find the joint PDF of y1 and y3, and use the change of variables to find the PDF of z1. Substituting the expressions for y1 and y3 into the joint PDF, we obtain:

f(y1, y3) = f(u * y3) * f((1/u) * y1)

Since y1 and y3 are independent, their PDFs multiply:

f(y1, y3) = f(u * y3) * f((1/u) * y1) = f(u * y3) * f(y1/u)

Multiplying both sides by u, we get:

u * f(y1, y3) = f(u * y3) * f(y1/u)

Taking the derivative with respect to u, we find the joint PDF of z1 and y3:

d(u * f(y1, y3))/du = d(f(u * y3))/du * [d(f(y1/u))/du * (-y1/u^2)]

Simplifying, we get:

f(z1, y3) = f(y3) * f(z1/y3) * (-y1/y3^2)

Since z1 = y1/y3, we can replace y1/y3 with z1, and we have:

f(z1, y3) = f(y3) * f(z1)

Therefore, the joint PDF of z1 and y3 factors into the product of their individual PDFs.

Similarly, you can follow the above steps to find the PDFs for z2 and z3. Once you have the PDFs for z1, z2, and z3, you can check if they factorize into the product of their individual PDFs. If they do, it shows that z1, z2, and z3 are mutually stochastically independent.