What is the LMS Estimate of Theta for a Joint PDF on a Triangular Set?

[Sitingbull]

I have the following question and I am struggling to find the right answer.

The random variables and are described by a joint PDF which is uniform on the triangular set defined by the constraints 0 <= x <= 1, 0<= theta <= x Find the LMS estimate of theta given that X = x , for in the range [0,1] . Express your answer in terms x.

I started by calculating the joint pdf by first calculating the area of the triangle which is 1/2 * x * 1 = x /2 . The joint pdf will be 1 / (x/2) = 2 / x

Then I integrate over integral over (x to 1) of theta times 2 / x. I got an integral of \theta^ 2 / x which gives me a final answer of (1-X^2) / x

Does that look good or I am missing something ...

Sorry for my notation which lacks LATEX, I am new here.Thank youSB

 

[HOI]

I have the following question and I am struggling to find the right answer.

The random variables and are described by a joint PDF which is uniform on the triangular set defined by the constraints 0 <= x <= 1, 0<= theta <= x Find the LMS estimate of theta given that X = x , for in the range [0,1] . Express your answer in terms x.

I started by calculating the joint pdf by first calculating the area of the triangle which is 1/2 * x * 1 = x /2 .

What triangle are you talking about? I assumed you meant the triangle given here, 0<= x<= 1, 0<= theta<= x but the area of that does not depend on x! The area is 1/2.

The joint pdf will be 1 / (x/2) = 2 / x

Further, The probability density function is exactly what is said above- a constant since this is a "uniform distribution": 2 for all x, theta in the regon.

Then I integrate over integral over (x to 1) of theta times 2 / x. I got an integral of \theta^ 2 / x which gives me a final answer of (1-X^2) / x

Does that look good or I am missing something ...

Sorry for my notation which lacks LATEX, I am new here.Thank youSB

Given that x= X, a fixed value, theta can rang from 0 up to X. Take the square root of the integral of Theta^2 from 0 to X.

 

[Sitingbull]

Thank you Country Boy you are totally right, I managed at the end to get it. Its simply the expecation of theta which is the midpoint between 0 and x. Thank you a lot