Let x1,x2,x3 be random sample of f(x)=exp(-x),x>0,zero elsewhere,if y1=x1/x2,y2=x3/(x1+x2),y3=x1+x2 find g(y1,y2,y3),g(y1),g(y2,y3)
To find the probability density function (pdf) of a function of multiple random variables, we need to use the technique called the method of transformations. Let's start by finding the joint pdf of y1, y2, and y3.
Step 1: Finding the joint pdf of y1, y2, and y3
Since we know the distribution of x1, x2, and x3, we can find the cumulative distribution functions (CDFs) of these random variables, and then use the method of transformations to find the joint pdf of y1, y2, and y3.
1. Finding the CDF of x1:
The random variable x1 follows an exponential distribution with parameter λ=1. Therefore, the CDF of x1 can be derived as:
F(x1) = 1 - exp(-x1)
2. Finding the CDF of x2:
Similar to x1, x2 also follows an exponential distribution with parameter λ=1. Thus, the CDF of x2 is:
F(x2) = 1 - exp(-x2)
3. Finding the CDF of x3:
Since x3 is the sum of x1 and x2, we can use the convolution property of the CDF to find its CDF:
F(x3) = ∫[0,∞] F(x3 | x1, x2) * f(x1, x2) dx1 dx2
Here, F(x3 | x1, x2) is the conditional CDF of x3 given x1 and x2, and f(x1, x2) is the joint pdf of x1 and x2.
We can take advantage of the independence assumption between x1, x2, and x3 to simplify the above expression. Since x1, x2, and x3 are independent, the joint pdf can be written as the product of individual pdfs:
f(x1, x2) = f(x1) * f(x2) * f(x3)
4. Finding the joint pdf of y1, y2, and y3:
To find the joint pdf of y1, y2, and y3, we need to express them in terms of x1, x2, and x3.
y1 = x1/x2
y2 = x3/(x1+x2)
y3 = x1 + x2
We can solve these equations to express x1, x2, and x3 in terms of y1, y2, and y3, and then find their Jacobian.
x1 = y3/(y1 + 1)
x2 = y3 - x1 = y3 - y3/(y1 + 1) = y3 * y1/(y1 + 1)
x3 = y2(y1 + 1)/(1 - y2)
Next, we can use the transformations x1 = g1(y1, y2, y3), x2 = g2(y1, y2, y3), and x3 = g3(y1, y2, y3) to find the joint pdf of the transformed variables.
Now, let's derive the joint pdf of y1, y2, and y3.
Step 2: Finding the joint pdf of y1, y2, and y3
To find the joint pdf of y1, y2, and y3, we need to find the Jacobian determinant of the transformation from x1, x2, x3 to y1, y2, y3.
Jacobian determinant (J) = |∂(x1, x2, x3)/∂(y1, y2, y3)|
To find the Jacobian determinant, we need to compute the partial derivatives ∂x1/∂y1, ∂x1/∂y2, ∂x1/∂y3, ∂x2/∂y1, ∂x2/∂y2, ∂x2/∂y3, ∂x3/∂y1, ∂x3/∂y2, ∂x3/∂y3.
∂x1/∂y1 = y3/(y1+1)^2
∂x1/∂y2 = 0
∂x1/∂y3 = 1/(y1+1)
∂x2/∂y1 = y3/(y1+1) - y3 * y1/(y1+1)^2 = y3/(y1+1)^2
∂x2/∂y2 = 0
∂x2/∂y3 = y1/(y1+1)
∂x3/∂y1 = y2/(1-y2)
∂x3/∂y2 = (y1+1)/(1-y2) + y2 * (y1+1)/(1-y2)^2
∂x3/∂y3 = 0
Now, we can compute the Jacobian determinant:
J = |∂(x1, x2, x3)/∂(y1, y2, y3)|
= |[y3/(y1+1)^2 0 1/(y1+1) ]
[y3/(y1+1)^2 0 y1/(y1+1) ]
[y2/(1-y2) (y1+1)/(1-y2) 0 ]|
= y3 * y2/(y1+1)^2
Step 3: Finding the joint pdf of y1, y2, and y3
The joint pdf of y1, y2, and y3 can be derived using the joint pdf of x1, x2, and x3, and the Jacobian determinant:
g(y1, y2, y3) = f(x1, x2, x3) * |J|
Since the joint pdf of x1, x2, and x3 is the product of individual pdfs, we can substitute the expressions we found earlier:
f(x1) = exp(-x1), f(x2) = exp(-x2), f(x3) = exp(-x3)
g(y1, y2, y3) = exp(-y3/(y1 + 1)) * exp(-y3 * y1/(y1 + 1)) * exp(-y2(y1 + 1)/(1 - y2)) * y3 * y2/(y1 + 1)^2
This is the joint pdf of y1, y2, and y3.
Step 4: Finding marginal pdfs
To find the marginal pdfs, we need to integrate the joint pdf over the other variables.
1. Finding the marginal pdf of y1 (g(y1)):
To find the marginal pdf of y1, we integrate the joint pdf over y2 and y3:
g(y1) = ∫∫ g(y1, y2, y3) dy2 dy3
2. Finding the marginal pdf of y2 given y3 (g(y2 | y3)):
To find the marginal pdf of y2 given y3, we integrate the joint pdf over y1:
g(y2 | y3) = ∫ g(y1, y2, y3) dy1
3. Finding the marginal pdf of y3 (g(y3)):
To find the marginal pdf of y3, we integrate the joint pdf over y1 and y2:
g(y3) = ∫∫ g(y1, y2, y3) dy1 dy2
These integrations need to be done numerically, as the resulting expressions can be complex.