Let X1,…,Xn be i.i.d. random variable with pdf fθ defined as follows:
fθ(x)=θxθ−11(0≤x≤1)
where θ is some positive number.
(a)
1 punto posible (calificable, resultado oculto)
Is the parameter θ identifiable?
Yes
No
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(b)
2.0 puntos posibles (calificables, resultados ocultos)
Compute the maximum likelihood estimator
ˆ
θ
of θ.
(Enter Sigma_i(g(X_i)) for the sum ∑
n
i=1
g(Xi), e.g. enter Sigma_i(X_i^2) for ∑
n
i=1
X
2
i
, enter Sigma_i(ln(X_i)) for ∑
n
i=1
ln(Xi). Do not forget any necessary n in your answer, e.g.
ˉ
X
n will need to be entered as Sigma_i(X_i)/n . Do not worry about the parser not rendering correctly, as the grader will still work independently. If you would like proper rendering, enclose Σi(g(Xi)) in parentheses i.e. use (Σi(g(Xi))).)
Maximum likelihood estimator
ˆ
θ
=
To determine if the parameter θ is identifiable, we need to check if different values of θ can produce different probability density functions (pdfs) for the random variable X.
In this case, the pdf fθ(x) is defined as:
fθ(x) = θx^(θ-1) * 1(0 ≤ x ≤ 1),
where 1(0 ≤ x ≤ 1) is the indicator function that is 1 if 0 ≤ x ≤ 1 and 0 otherwise.
To check if θ is identifiable, we can compare the pdfs for two different values of θ and see if they can be equal for all possible values of X.
Let's consider two different values of θ, θ₁ and θ₂.
For θ₁, the pdf is given by fθ₁(x) = θ₁x^(θ₁-1) * 1(0 ≤ x ≤ 1).
For θ₂, the pdf is given by fθ₂(x) = θ₂x^(θ₂-1) * 1(0 ≤ x ≤ 1).
To check if fθ₁(x) = fθ₂(x) for all x in the domain [0, 1], we can equate the two expressions:
θ₁x^(θ₁-1) * 1(0 ≤ x ≤ 1) = θ₂x^(θ₂-1) * 1(0 ≤ x ≤ 1).
Since the indicator function 1(0 ≤ x ≤ 1) is equal to 1 for all x in the interval [0, 1], we can cancel it out:
θ₁x^(θ₁-1) = θ₂x^(θ₂-1).
We can simplify this equation further by dividing both sides by x^(θ₁-1):
θ₁ = θ₂ * (x^(θ₂-1) / x^(θ₁-1)).
Since x is positive and in the domain [0, 1], x^(θ₂-1) / x^(θ₁-1) will always be positive. Therefore, the only way for the equation to hold for all x is if θ₁ = θ₂.
Based on this analysis, we can conclude that the parameter θ is identifiable since different values of θ will produce different pdfs.
For part (b) of the question, we need to compute the maximum likelihood estimator (MLE) of θ.
The likelihood function for a set of i.i.d. observations X₁, X₂, ..., Xₙ, with pdf fθ(x), is given by:
L(θ) = ∏(i=1 to n) [fθ(Xᵢ)],
where Xᵢ is the ith observation.
To find the MLE of θ, we need to maximize the likelihood function L(θ) with respect to θ.
Maximizing the product is equivalent to maximizing the logarithm of the product. Taking the natural logarithm of the likelihood function, we get:
ln(L(θ)) = ∑(i=1 to n) [ln(fθ(Xᵢ))].
Now, substitute the expression for fθ(x):
ln(L(θ)) = ∑(i=1 to n) [ln(θXᵢ^(θ-1))].
Using properties of logarithms, we can simplify further:
ln(L(θ)) = ∑(i=1 to n) [ln(θ) + (θ-1)ln(Xᵢ)].
Using the property of sums, we can split the summation:
ln(L(θ)) = nln(θ) + (θ-1)∑(i=1 to n) [ln(Xᵢ)].
Now, to find the maximum likelihood estimator ˆθ of θ, we need to maximize ln(L(θ)) with respect to θ. This can be done by taking the derivative of ln(L(θ)) with respect to θ, setting it equal to zero, and solving for θ.
d/ dθ [ln(L(θ))] = 0.
Differentiating, we get:
0 = n/θ - ∑(i=1 to n) [ln(Xᵢ)].
To solve for θ, we can rearrange the equation:
n/θ = ∑(i=1 to n) [ln(Xᵢ)].
Now, isolate θ:
θ = n / ∑(i=1 to n) [ln(Xᵢ)].
Therefore, the maximum likelihood estimator of θ, denoted as ˆθ, is given by:
ˆθ = n / ∑(i=1 to n) [ln(Xᵢ)].