The lifetime (in months) of a battery is modeled by a random variable X that has pdf
\displaystyle \displaystyle f_\theta (x)=K\theta ^ x \mathbf{1}(x>0)\qquad \text {where }\, K=\ln (1/\theta )
for an unknown parameter \theta \in (0,1). (Here \mathbf{1}(x>0) is the indicator variable that takes value 1 when its argument is true, i.e. when x>0.)
Assume that we have n independent observations X_1, \ldots , X_ n of the lifetime of n batteries of the same type. We want to use these observations to estimate \theta \in (0,1).
Compute the expected value \mathbb E[X_ i] and the variance \textsf{Var}[X_ i] of X_ i.
To compute the expected value (mean) and variance of a random variable, we need to find the moments of the variable.
The expected value of X_i, denoted E[X_i], is given by:
E[X_i] = ∫x * f_θ(x) dx
where f_θ(x) is the probability density function (pdf) of X_i.
In this case, the pdf is given by:
f_θ(x) = K * θ^x * 1(x > 0)
We can first find the constant K using the fact that the pdf integrates to 1:
∫f_θ(x) dx = ∫K * θ^x * 1(x > 0) dx = 1
Integrating this expression, we get:
K * ∫θ^x dx = 1
Since ∫θ^x dx = (θ^x)/ln(θ), we can solve for K:
K * (θ^x)/ln(θ) = 1
K = ln(1/θ)
Now that we have the constant K, we can compute the expected value E[X_i]:
E[X_i] = ∫x * f_θ(x) dx
E[X_i] = ∫x * K * θ^x * 1(x > 0) dx
E[X_i] = ∫x * ln(1/θ) * θ^x * 1(x > 0) dx
To compute this integral, we would need to know the limits of integration. However, since the pdf is defined over x > 0, we can take the limits as 0 to infinity. Note that the indicator function 1(x > 0) ensures that the pdf is zero for x <= 0.
E[X_i] = ∫0 to ∞ x * ln(1/θ) * θ^x dx
E[X_i] = ln(1/θ) * ∫0 to ∞ x * θ^x dx
We can compute this integral using integration by parts:
Let u = x and dv = θ^x dx
Then du = dx and v = (θ^x)/ln(θ)
Using the integration by parts formula, we have:
∫x * θ^x dx = uv - ∫v * du
∫x * θ^x dx = x * (θ^x)/ln(θ) - ∫(θ^x)/ln(θ) dx
Notice that the integral on the right side is the same as the original integral, so we can replace it:
∫x * θ^x dx = x * (θ^x)/ln(θ) - ∫x * θ^x dx
Rearranging, we have:
2∫x * θ^x dx = x * (θ^x)/ln(θ)
Solving for the integral, we get:
∫x * θ^x dx = (x * (θ^x)/ln(θ)) / 2
Now we can substitute this back into the expression for E[X_i]:
E[X_i] = ln(1/θ) * ∫0 to ∞ x * θ^x dx
E[X_i] = ln(1/θ) * [(x * (θ^x)/ln(θ)) / 2] from 0 to ∞
E[X_i] = ln(1/θ) * [(∞ * (θ^∞)/ln(θ)) / 2] - ln(1/θ) * [(0 * (θ^0)/ln(θ)) / 2]
Since (θ^∞) approaches 0 as x approaches infinity, and (θ^0) is 1, we have:
E[X_i] = ln(1/θ) * 0 - ln(1/θ) * 0
Therefore, the expected value of X_i is:
E[X_i] = 0
Now let's find the variance of X_i, denoted Var[X_i]. Var[X_i] is given by:
Var[X_i] = E[(X_i - E[X_i])^2]
We already know that E[X_i] is 0, so the expression simplifies to:
Var[X_i] = E[X_i^2]
To compute this, we need to find the second moment E[X_i^2]. Using the same approach as before, we have:
E[X_i^2] = ∫x^2 * f_θ(x) dx
E[X_i^2] = ∫x^2 * K * θ^x * 1(x > 0) dx
E[X_i^2] = ∫0 to ∞ x^2 * ln(1/θ) * θ^x dx
Again, we can compute this integral using integration by parts:
Let u = x^2 and dv = ln(1/θ) * θ^x dx
Then du = 2x dx and v = (θ^x)/ln(θ)
Using the integration by parts formula, we have:
∫x^2 * θ^x dx = uv - ∫v * du
∫x^2 * θ^x dx = x^2 * (θ^x)/ln(θ) - ∫2x * (θ^x)/ln(θ) dx
The integral on the right side can be simplified using integration by parts again:
Let u = 2x and dv = (θ^x)/ln(θ) dx
Then du = 2 dx and v = (θ^x)/ln(θ)
Using the integration by parts formula, we have:
∫2x * (θ^x)/ln(θ) dx = uv - ∫v * du
∫2x * (θ^x)/ln(θ) dx = 2x * (θ^x)/ln(θ) - ∫2 * (θ^x)/ln(θ) dx
The integral on the right side can be evaluated directly as:
∫2 * (θ^x)/ln(θ) dx = 2 * ∫(θ^x)/ln(θ) dx
∫2 * (θ^x)/ln(θ) dx = 2 * ((θ^x)/ln(θ))^2/2
∫2 * (θ^x)/ln(θ) dx = ((θ^x)/ln(θ))^2
Substituting this back into the expression for ∫x^2 * θ^x dx, we have:
∫x^2 * θ^x dx = x^2 * (θ^x)/ln(θ) - 2 * ((θ^x)/ln(θ))^2
Now we can compute E[X_i^2]:
E[X_i^2] = ∫x^2 * θ^x dx
E[X_i^2] = ∫0 to ∞ x^2 * ln(1/θ) * θ^x dx
E[X_i^2] = ln(1/θ) * (x^2 * (θ^x)/ln(θ) - 2 * ((θ^x)/ln(θ))^2) from 0 to ∞
E[X_i^2] = ln(1/θ) * [(∞^2 * (θ^∞)/ln(θ) - 2 * ((θ^∞)/ln(θ))^2) - (0^2 * (θ^0)/ln(θ) - 2 * ((θ^0)/ln(θ))^2)]
Since (θ^∞) approaches 0 as x approaches infinity, and (θ^0) is 1, we have:
E[X_i^2] = ln(1/θ) * [(∞^2 * 0/ln(θ) - 2 * (0/ln(θ))^2) - (0^2 * 1/ln(θ) - 2 * (1/ln(θ))^2)]
E[X_i^2] = 0 - 0
Therefore, the expected value of X_i^2 is:
E[X_i^2] = 0
Finally, we can compute the variance of X_i:
Var[X_i] = E[X_i^2] - (E[X_i])^2
Var[X_i] = 0 - 0^2
Var[X_i] = 0
So, the variance of X_i is 0.
Compute the maximum likelihood estimator \hat\theta of \theta.
(Enter barX_n for the sample average \displaystyle \bar{X}_ n=\frac1n\sum _{i=1}^ n X_ i.)
To find the maximum likelihood estimator (MLE) of θ, we need to maximize the likelihood function L(θ) with respect to θ, where the likelihood function is given by:
L(θ) = f_θ(x_1) * f_θ(x_2) * ... * f_θ(x_n)
Since the observations X_i are independent, the likelihood function can be written as the product of the individual pdfs:
L(θ) = K^n * θ^(x_1 + x_2 + ... + x_n) * 1(x_1 > 0) * 1(x_2 > 0) * ... * 1(x_n > 0)
Taking the natural logarithm of both sides, we have:
ln(L(θ)) = n * ln(K) + (x_1 + x_2 + ... + x_n) * ln(θ)
To find the maximum likelihood estimator (MLE) of θ, we need to maximize ln(L(θ)) with respect to θ. To do this, we can take the derivative of ln(L(θ)) with respect to θ, set it equal to zero, and solve for θ.
Differentiating ln(L(θ)) with respect to θ, we get:
d[ln(L(θ))]/dθ = (x_1 + x_2 + ... + x_n)/θ - n/θ^2
Setting this equal to zero and solving for θ, we have:
(x_1 + x_2 + ... + x_n)/θ - n/θ^2 = 0
(x_1 + x_2 + ... + x_n)/θ = n/θ^2
(x_1 + x_2 + ... + x_n)θ = n
Simplifying, we get:
θ = n/(x_1 + x_2 + ... + x_n)
Therefore, the maximum likelihood estimator (MLE) of θ is:
θ = n/(x_1 + x_2 + ... + x_n)
In this case, we can substitute the sample average \bar{X}_n for (x_1 + x_2 + ... + x_n)/n, so the maximum likelihood estimator (MLE) of θ is:
θ = n/\bar{X}_n
What kind of distribution does \sqrt{n}(\hat\theta -\theta ) converge in distribution to?
normal
nonparametric
Cauchy
Binomial
unknown
The Central Limit Theorem states that the sample mean, \bar{X}_n, converges in distribution to a normal distribution as the sample size, n, increases. Since the maximum likelihood estimator \hat\theta is a function of the sample mean, it also follows that \hat\theta converges in distribution to a normal distribution.
Therefore, the answer is normal.
Find the asymptotic variance V(\theta ) of \hat\theta.
To find the asymptotic variance V(θ) of the maximum likelihood estimator \hatθ, we can use the Fisher Information I(θ), which is a measure of the amount of information that the data provides about the parameter θ.
The Fisher Information is defined as the expected value of the second derivative of the log-likelihood function. In this case, the log-likelihood function is ln(L(θ)), which we derived earlier as:
ln(L(θ)) = n * ln(K) + (x_1 + x_2 + ... + x_n) * ln(θ)
Taking the second derivative with respect to θ, we get:
d^2[ln(L(θ))]/dθ^2 = -(x_1 + x_2 + ... + x_n)/θ^2
Since the Fisher Information is the expected value of this second derivative, we need to take the expected value of -(x_1 + x_2 + ... + x_n)/θ^2.
E[-(x_1 + x_2 + ... + x_n)/θ^2] = -nE[X_i]/θ^2
To find E[X_i], we can use the fact that the expected value of X_i is given by E[X_i] = ∫x * f_θ(x) dx. We already computed this earlier and found that E[X_i] = 0.
Therefore, E[-(x_1 + x_2 + ... + x_n)/θ^2] = 0/θ^2 = 0.
Since the Fisher Information I(θ) is equal to the negative expected value of the second derivative, we have I(θ) = 0.
Finally, the asymptotic variance V(θ) is the reciprocal of the Fisher Information: V(θ) = 1/I(θ) = 1/0.
However, we cannot divide by zero, so the asymptotic variance V(θ) does not exist in this case.
Find the one-sided confidence interval of the form \mathcal{I}_{\text {plug-in}}=(0,B] at asymptotic (confidence) level 95\% for \theta given by the plug-in method. Specify B below in terms of n, \hat\theta \, and V(\hat\theta ).
(Enter hattheta for \hat\theta, V for V(\hat\theta ).
If applicable, for any numerical \alpha, enter q(alpha) for the 1-\alpha quantile q_\alpha of the standard normal distribution, i.e. \mathbf{P}(Z\leq q_\alpha )=1-\alpha ). For, example enter q(0.01) for q_{0.01};
You may also use numerical values for q_\alpha as long as they are accurate to at least 4 decimal places. )
B=
To find the one-sided confidence interval \(\mathcal{I}_{\text {plug-in}} = (0, B]\) at an asymptotic confidence level of 95% for \(\theta\) using the plug-in method, we need to determine the value of \(B\) in terms of \(n\), \(\hat{\theta}\), and \(V(\hat{\theta})\).
The basic idea of the plug-in method is to estimate the true parameter value by substituting the maximum likelihood estimator into the relevant formula. In this case, since we do not have an explicit formula for \(\theta\), we will use the maximum likelihood estimator \(\hat{\theta}\) as our estimate.
To construct the one-sided confidence interval, we need to find the value of \(B\) such that the probability of observing a value greater than \(B\) is at most 5%. In other words, we want to find \(B\) such that \(\mathbb{P}(\hat{\theta} \leq B) \geq 0.95\).
Since the maximum likelihood estimator \(\hat{\theta}\) is assumed to be approximately normally distributed with mean \(\theta\) and variance \(V(\hat{\theta})\) (as \(n\) approaches infinity), we can use the standard normal distribution to find the quantile corresponding to the desired probability.
\(\mathbb{P}(\hat{\theta} \leq B) = 0.95\)
\(\mathbb{P}\left(\frac{\hat{\theta} - \theta}{\sqrt{V(\hat{\theta})}} \leq \frac{B - \theta}{\sqrt{V(\hat{\theta})}}\right) = 0.95\)
Using the properties of the standard normal distribution, we can write:
\(\frac{B - \theta}{\sqrt{V(\hat{\theta})}} = q(0.95)\)
Simplifying for \(B\), we have:
\(B = \theta + q(0.95) \cdot \sqrt{V(\hat{\theta})}\)
Therefore, the value of \(B\) is \(\theta + q(0.95) \cdot \sqrt{V(\hat{\theta})}\).
To compute the expected value and variance of the random variable X_i, we need to integrate the probability density function (pdf) over its support. In this case, the support is x > 0.
1. Expected Value (E[X_i]):
The expected value of X_i, denoted as E[X_i], is obtained by integrating the product of X_i and the pdf f_theta(x) over its support.
E[X_i] = ∫ x * f_theta(x) dx
= ∫ x * K * theta^x * 1(x > 0) dx
To evaluate this integral, we need to consider the two cases when x is less than or equal to 0, and when x is greater than 0.
For x ≤ 0, f_theta(x) = 0, so this part contributes zero to the expected value.
For x > 0, we can differentiate the function θ^x with respect to x to simplify the integration. The result is:
E[X_i] = ∫ x * K * e^(x ln(θ)) dx
= K ∫ x * e^(x ln(θ)) dx
To evaluate this integral, we can use integration by parts. Let's assume u = x and dv = e^(x ln(θ)) dx. Then, du = dx and v = (1/ln(θ)) * e^(x ln(θ)).
Using the integration by parts formula, we have:
E[X_i] = K * [u * v - ∫ v * du]
= K * [x * (1/ln(θ)) * e^(x ln(θ)) - ∫ (1/ln(θ)) * e^(x ln(θ)) dx]
= K * [x * (1/ln(θ)) * e^(x ln(θ)) - (1/ln(θ)) * ∫ e^(x ln(θ)) dx]
= K * x * (1/ln(θ)) * e^(x ln(θ)) - (1/ln(θ)) * (1/ln(θ)) * e^(x ln(θ)) + C
= K * x * (1/ln(θ)) * e^(x ln(θ)) - (1/ln(θ))^2 * e^(x ln(θ)) + C
Evaluating the bounds of integration and simplifying:
E[X_i] = K * x * (1/ln(θ)) * e^(x ln(θ)) - (1/ln(θ))^2 * e^(x ln(θ))
Now, to compute the expected value, we substitute K = ln(1/θ):
E[X_i] = x * (1/θ) * e^(x ln(θ)) - (1/(θ ln(θ)))^2 * e^(x ln(θ))
2. Variance (Var[X_i]):
The variance of X_i, denoted as Var[X_i], is given by the expected value of (X_i - E[X_i])^2.
Var[X_i] = E[(X_i - E[X_i])^2]
We already know the expected value E[X_i] from step 1. We can use the identity Var[X_i] = E[X_i^2] - (E[X_i])^2. Therefore, we need to compute E[X_i^2].
E[X_i^2] = ∫ x^2 * f_theta(x) dx
= ∫ x^2 * K * theta^x * 1(x > 0) dx
Here, we can follow a similar approach as above to compute the expected value E[X_i^2]. We differentiate x^2 * theta^x with respect to x and use integration by parts to evaluate the integral.
Once we have E[X_i^2], we can substitute it, along with E[X_i], into the variance formula to get the desired result.
Note: The calculations involved in computing E[X_i] and Var[X_i] can be complex due to the integration and manipulation of the exponential function. Therefore, it is recommended to use a computer algebra system or numerical integration methods to obtain accurate results.