The random variables Θ and X are described by a joint PDF which is uniform on the triangular set defined by the constraints 0≤x≤1 , 0≤θ≤x . Find the LMS estimate of Θ given that X=x , for x in the range [0,1] . Express your answer in terms of x using standard notation.
The conditional PDF of Θ given that X=x is uniform on the set [0,x] . Thus, the conditional expectation of Θ given that X=x is equal to x/2 .
Thank you. It was my thoughts too but I wanted to confirm.
Well, the LMS estimate requires finding the conditional mean of Θ given X=x. In other words, we want to find E[Θ|X=x].
Since the joint PDF is uniform on the triangular set defined by the constraints 0≤x≤1 and 0≤θ≤x, the probability distribution is a triangle with base length 1 and height 1.
To find E[Θ|X=x], we need to find the centroid of the triangle.
Since the triangle has a base length of 1, its centroid is located at x=1/2.
Therefore, the LMS estimate of Θ given X=x, for x in the range [0,1], is Θ = 1/2.
To find the LMS (Least Mean Squares) estimate of Θ given that X = x, we need to find the conditional mean of Θ given X = x.
Given that the joint PDF is uniform on the triangular set defined by the constraints 0 ≤ x ≤ 1 and 0 ≤ θ ≤ x, we can write the joint PDF as:
f(Θ, X) = 2, for 0 ≤ θ ≤ x ≤ 1
0, otherwise
To find the conditional PDF of Θ given X = x, we need to normalize the joint PDF. The normalization constant can be found by integrating the joint PDF over the feasible region:
c = ∫∫ f(Θ, X) dθ dx
= ∫∫ 2 dθ dx
= 2∫[θ=0]^x ∫[x=θ]^1 dθ dx
= 2∫[θ=0]^1 ∫[x=θ]^1 dθ dx
= 2∫[θ=0]^1 (1-θ) dθ
= 2[(θ-θ^2/2)]|[θ=0]^1
= 2(1-1/2)
= 1
Therefore, the normalized conditional PDF is:
f(Θ | X = x) = f(Θ, X)/c
= 2/1
= 2, for 0 ≤ θ ≤ x ≤ 1
0, otherwise
The conditional mean of Θ given X = x is the expected value of Θ conditioned on X = x. So, we can calculate the conditional mean by integrating Θ multiplied by the conditional PDF over the feasible range:
E[Θ | X = x] = ∫ Θ * f(Θ | X = x) dθ
= ∫ Θ * 2 dθ, for 0 ≤ θ ≤ x ≤ 1
E[Θ | X = x] = 2 * ∫[θ=0]^x Θ dθ
= 2 * [(θ^2/2)]|[θ=0]^x
= 2 * (x^2/2 - 0)
= x^2
Therefore, the LMS estimate of Θ given that X = x is x^2.
To find the LMS (Least Mean Squares) estimate of Θ given that X=x for x in the range [0,1], we need to compute the conditional expectation of Θ given X=x.
First, let's write down the joint PDF of Θ and X, denoted as f(Θ, X):
f(Θ, X) = 2 (1 - Θ)
The joint PDF is uniform on the triangular set defined by the constraints 0 ≤ x ≤ 1 and 0 ≤ θ ≤ x.
Next, let's compute the conditional PDF of Θ given X=x, denoted as f(Θ|X=x):
f(Θ|X=x) = f(Θ, X=x) / f(X=x)
To compute f(X=x), we integrate the joint PDF over the entire range of Θ, since X=x is fixed:
f(X=x) = ∫[0,x] f(Θ, X=x) dΘ
Substituting the joint PDF expression:
f(X=x) = ∫[0,x] 2 (1 - Θ) dΘ
Integrating, we get:
f(X=x) = 2 (x - x^2/2) = 2x - x^2
Now, let's substitute the expressions for f(Θ, X=x) and f(X=x) into f(Θ|X=x):
f(Θ|X=x) = (1 - Θ) / (2x - x^2)
Finally, to find the LMS estimate of Θ given X=x, we need to compute the conditional expectation of Θ given X=x:
E(Θ|X=x) = ∫[0,x] Θ f(Θ|X=x) dΘ
Substituting the conditional PDF expression:
E(Θ|X=x) = ∫[0,x] Θ (1 - Θ) / (2x - x^2) dΘ
Integrating, we get:
E(Θ|X=x) = (1 / (2x - x^2)) * (x^3 / 6)
Simplifying, we find:
E(Θ|X=x) = x^3 / (6 (2x - x^2))
So, the LMS estimate of Θ given that X=x, for x in the range [0,1], is expressed as:
E(Θ|X=x) = x^3 / (6 (2x - x^2))