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
(b)
2.0 puntos posibles (calificables, resultados ocultos)
Compute the maximum likelihood estimator ˆθ of θ.
(Enter Sigma_i(g(X_i)) for the sum ∑ni=1g(Xi), e.g. enter Sigma_i(X_i^2) for ∑ni=1X2i, enter Sigma_i(ln(X_i)) for ∑ni=1ln(Xi). Do not forget any necessary n in your answer, e.g. ˉXn 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 ˆθ=
c)
As in previous ecercise let X1…Xn be iid with pdf where θ>0
ocultos)
Compute the Fisher information.
I(θ)=
d)
What kind of distribution does the distribution of √nˆθ approach as n grows large?
Bernoulli
Poisson
Normal
Exponential
e) Ha realizado 1 de 3 intentosAlgunos problemas tienen opciones como guardar, restablecer, sugerencias o mostrar respuesta. Estas opciones aparecen después de oprimir el botón Enviar.
(e)
1 punto posible (calificable, resultado oculto)
What is the asymptotic variance V(ˆθ) of ˆθ ?
To avoid double jeopardy, you may use I for the Fisher information I(θ) evaluated at θ, or you may enter your answer without using I.
V(ˆθ)
f)
Using the MLE ˆθ, find the shortest confidence interval for θ with asymptotic level 85% using the plug-in method.
To avoid double jeopardy, you may use V for the appropriate estimator of the asymptotic variance V(ˆθ), and/or I for the Fisher information I(ˆθ) evaluated at ˆθ, or you may enter your answer without using V or I.
(Enter hattheta for ˆθ. If applicable, enter Phi(z) for the cdf Φ(z) of a normal variable Z, q(alpha) for the quantile qα for any numerical value α. Recall the convention in this course that P(Z≤qα)=1−α for Z∼N(0,1).)
Iplug-in=[A,B] where
A=
sin responder
B=
(a) Yes, the parameter θ is identifiable.
(b) To compute the maximum likelihood estimator ˆθ of θ, we need to maximize the likelihood function. The likelihood function is given by:
L(θ) = ∏(θx^(θ-1))
Taking the logarithm of the likelihood function, we have:
ln(L(θ)) = ∑(θ-1)ln(x)
To find the maximum likelihood estimator, we differentiate ln(L(θ)) with respect to θ and set it equal to zero:
(∂/∂θ)(ln(L(θ))) = ∑ln(x) - nln(θ) = 0
Solving for θ, we have:
ˆθ = (∑ln(x))/n
(c) The Fisher information is given by the expected value of the square of the score function, which is the derivative of ln(fθ(x)) with respect to θ.
Taking the derivative of ln(fθ(x)) with respect to θ, we have:
(∂/∂θ)(ln(fθ(x))) = x - 1/θ
Squaring the score function, we have:
[(x - 1/θ)^2] = x^2 - 2x/θ + 1/θ^2
The Fisher information is the expected value of the score function squared, which can be calculated as:
I(θ) = E[(x^2 - 2x/θ + 1/θ^2)] = E(x^2) - 2E(x)/θ + 1/θ^2
Since the random variables X1,...,Xn are uniformly distributed on the interval [0,1], we have:
E(x) = 1/2
E(x^2) = 1/3
Substituting these values into the expression for I(θ), we have:
I(θ) = 1/3 - 2(1/2)/θ + 1/θ^2
(d) As n grows large, the distribution of √nˆθ approaches a normal distribution.
(e) The asymptotic variance V(ˆθ) of ˆθ is equal to the reciprocal of the Fisher information evaluated at θ, i.e., V(ˆθ) = 1/I(θ).
Substituting the expression for I(θ) from part (c), we have:
V(ˆθ) = 1/(1/3 - 2(1/2)/θ + 1/θ^2)
(f) To find the shortest confidence interval for θ with asymptotic level 85% using the plug-in method, we need to calculate the standard error of the estimator ˆθ. The standard error is given by the square root of V(ˆθ).
Therefore, the confidence interval is given by:
[ˆθ - z(α/2) * sqrt(V(ˆθ)), ˆθ + z(α/2) * sqrt(V(ˆθ))]
where z(α/2) is the α/2 quantile of a standard normal distribution.
(a) Yes, the parameter θ is identifiable because it appears explicitly in the probability density function (pdf) of the random variable X.
(b) To compute the maximum likelihood estimator (ˆθ) of θ, we need to find the value of θ that maximizes the likelihood function. The likelihood function is the product of the pdf values of the random variables X1, X2, ..., Xn. Therefore, the likelihood function is:
L(θ) = Π[i=1 to n] fθ(Xi)
Taking the natural logarithm of both sides to simplify the calculations, we obtain the log-likelihood function:
lnL(θ) = Σ[i=1 to n] ln(θX_i^(θ-1))
To find the maximum likelihood estimator, we need to solve the equation ∂lnL(θ)/∂θ = 0. Differentiating the log-likelihood function with respect to θ, we get:
∂lnL(θ)/∂θ = Σ[i=1 to n] (1/θ - ln(X_i))
Setting this derivative equal to zero and solving for θ, we get:
1/θ - (1/n)Σ[i=1 to n] ln(X_i) = 0
Simplifying further, we have:
ˆθ = n / Σ[i=1 to n] ln(X_i)
Therefore, the maximum likelihood estimator ˆθ of θ is n divided by the sum of the natural logarithms of the random variables X1, X2, ..., Xn.
(c) To compute the Fisher information (I(θ)), we need to find the second derivative of the log-likelihood function with respect to θ. Taking the derivative of the first derivative obtained in part (b), we get:
∂²lnL(θ)/∂θ² = -n/θ² - Σ[i=1 to n] (1/θ²)
Simplifying further and rearranging, we have:
I(θ) = -E[∂²lnL(θ)/∂θ²] = n/θ²
Therefore, the Fisher information I(θ) is equal to n divided by the square of θ.
(d) The distribution of √nˆθ approaches a normal distribution as n grows large. This is a consequence of the central limit theorem, which states that the sum or average of a large number of independently and identically distributed random variables tends to follow a normal distribution.
(e) The asymptotic variance V(ˆθ) of ˆθ can be found using the Cramer-Rao lower bound, which states that the variance of any unbiased estimator is greater than or equal to the inverse of the Fisher information. Therefore, V(ˆθ) can be approximated as 1/I(θ).
Substituting the expression for I(θ) obtained in part (c), we have:
V(ˆθ) ≈ θ²/n
Therefore, the asymptotic variance V(ˆθ) of ˆθ is approximately equal to θ² divided by n.
(f) To find the shortest confidence interval for θ with asymptotic level 85% using the plug-in method, we need to determine the bounds [A, B] such that:
P(A ≤ θ ≤ B) = 0.85
Using the asymptotic distribution of ˆθ as a normal distribution, we can find the z-value corresponding to a 85% confidence level (z_α/2) using the standard normal distribution function. For an 85% confidence level, α = 0.15, so z_α/2 = z_0.15/2.
The plug-in interval will be:
[A, B] = [ˆθ - z_α/2 * sqrt(V(ˆθ)), ˆθ + z_α/2 * sqrt(V(ˆθ))]
Substituting the expressions for ˆθ and V(ˆθ) obtained in parts (b) and (e), we have:
[A, B] = [ˆθ - z_α/2 * sqrt(θ²/n), ˆθ + z_α/2 * sqrt(θ²/n)]
So, the shortest confidence interval for θ with asymptotic level 85% using the plug-in method is [ˆθ - z_α/2 * sqrt(θ²/n), ˆθ + z_α/2 * sqrt(θ²/n)].
(a) To determine if the parameter θ is identifiable, we need to check if different values of θ result in different probability density functions (pdf) for X. In this case, the pdf fθ(x) depends on θ as θx^(θ-1), which means that different values of θ will indeed result in different pdfs. Therefore, θ is identifiable.
(b) To compute the maximum likelihood estimator (MLE) ˆθ of θ, we need to find the value of θ that maximizes the likelihood function. The likelihood function is defined as the product of the pdf of each observed random variable. In this case, the likelihood function is L(θ) = Πfθ(Xi), where Xi are the observed random variables.
To maximize the likelihood function, we can take the logarithm of the likelihood function and differentiate it with respect to θ. This is because the logarithm is a monotonically increasing function, so maximizing the log-likelihood is equivalent to maximizing the likelihood itself.
Taking the logarithm, we have ln(L(θ)) = Σln(fθ(Xi)). Now, differentiating with respect to θ:
∂/∂θ ln(L(θ)) = ∂/∂θ Σln(fθ(Xi))
= Σ ∂/∂θ ln(fθ(Xi))
= Σ [(1/θ - 1) + ln(Xi)]
Setting the derivative equal to zero, we have Σ [(1/θ - 1) + ln(Xi)] = 0. Rearranging the equation, we get:
Σ ln(Xi) = n - θ
Solving for θ, we get:
ˆθ = n / (Σ ln(Xi)).
(c) To compute the Fisher information, we need to find the expected value of the second derivative of the log-likelihood function with respect to θ. The Fisher information is defined as I(θ) = -E(∂²/∂θ² ln(L(θ))).
Differentiating the log-likelihood function with respect to θ, we get:
∂/∂θ ln(L(θ)) = Σ [(1/θ - 1) + ln(Xi)]
Taking the second derivative, we get:
∂²/∂θ² ln(L(θ)) = -n/θ²
Taking the expected value, we have:
E(∂²/∂θ² ln(L(θ))) = -n/θ²
Therefore, the Fisher information is I(θ) = n/θ².
(d) As n grows large, the distribution of √nˆθ approaches a normal distribution. This is due to the central limit theorem, which states that the sample mean of a large number of independent and identically distributed random variables will be approximately normally distributed.
(e) The asymptotic variance V(ˆθ) of ˆθ can be computed using the inverse of the Fisher information. Therefore, V(ˆθ) = 1/I(θ).
From part (c), we know that the Fisher information is I(θ) = n/θ². Substituting this into the formula for the asymptotic variance, we have:
V(ˆθ) = 1/(n/θ²) = θ²/n.
(f) To find the shortest confidence interval for θ with asymptotic level 85% using the plug-in method, we can use the MLE ˆθ as an estimator for θ and construct a confidence interval based on the asymptotic variance.
The general formula for a confidence interval using the plug-in method is:
ˆθ ± z(α/2) * sqrt(V(ˆθ))
where ˆθ is the MLE, z(α/2) is the critical value corresponding to the desired confidence level (85% in this case), sqrt(V(ˆθ)) is the standard error of the estimator, and V(ˆθ) is the estimated asymptotic variance as computed in part (e).
Substituting the values into the formula, we have:
ˆθ ± z(0.15/2) * sqrt(V(ˆθ))
Simplifying further, we have:
ˆθ ± z(0.075) * sqrt(θ²/n)
Note: The critical value z(0.075) can be looked up from the standard normal distribution table or calculated using appropriate software.
Therefore, to find the shortest confidence interval, we need to determine the values of ˆθ, z(0.075), θ, and n.