Assume the lifetime (in months) of batteries is modeled as an exponential random variable with parameter
𝜆
. The probability density function of the random variable is
𝑝
𝜆
(
𝑥
)
=
𝜆
𝑒
−
𝜆
𝑥
,
with
𝑥
∈
[
0
,
∞
)
.
We have collected an i.i.d. sample of 5 batteries whose lifetimes in months are 4, 8, 6, 10, 7, respectively. The standardized log-likelihood of a sample with
𝑛
i.i.d. observations is
1
𝑛
log
𝐿
(
𝑥
,
𝜆
)
=
1
𝑛
∑
𝑖
=
1
𝑛
(
log
(
𝜆
)
−
𝜆
𝑥
𝑖
)
.
To compute the standard errors of the MLE estimate of
𝜆
we need to compute the value of
𝐵
̂
0
for this model. Given this information:

Question

Assume the lifetime (in months) of batteries is modeled as an exponential random variable with parameter 
𝜆
. The probability density function of the random variable is
𝑝
𝜆
(
𝑥
)
=
𝜆
𝑒
−
𝜆
𝑥
,
with 
𝑥
∈
[
0
,
∞
)
.
 We have collected an i.i.d. sample of 5 batteries whose lifetimes in months are 4, 8, 6, 10, 7, respectively. The standardized log-likelihood of a sample with 
𝑛
 i.i.d. observations is
1
𝑛
log
𝐿
(
𝑥
,
𝜆
)
=
1
𝑛
∑
𝑖
=
1
𝑛
(
log
(
𝜆
)
−
𝜆
𝑥
𝑖
)
.
To compute the standard errors of the MLE estimate of 
𝜆
 we need to compute the value of 
𝐵
̂
0
 for this model. Given this information:

BlackTom AI · Accepted Answer

We start by restating the setup in our own words to ensure clarity: we have iid lifetimes X1,...,X5 from an exponential distribution with rate parameter lambda, so the density is p_lambda(x) = lambda * e^{-lambda x} for x >= 0. The sample given is 4, 8, 6, 10, 7 (in months). The standardized log-likelihood for a sample with n observations is l_bar(lambda) = (1/n) * sum_i [ log(lambda) - lambda * x_i ] = log(lambda) - lambda * x_bar, where x_bar is the sample mean.

First, compute the sample mean: x_bar = (4 + 8 + 6 + 10 + 7) / 5 = 35 / 5 = 7. Thus l_bar(lambda) = log(lambda) - lambda * 7.

Next, identify the MLE for lambda in this exponential model. The standard (unstandardized) log-likelihood is l(lambda) = n log(lambda) - lambda * sum_i x_i. Its derivative is 0 at d/d lambda l = n / lambda - sum_i x_i = 0, giving lambda_hat = n / sum_i x_i = 5 / 35 = 1/7. This is the MLE for lambda; note that the MLE depends on the data through the sum or the mean, and here the result is lambda_hat = 1/7.

Now, consider the quantity B_hat0 that’s required for the standard errors. The usual Fisher information for a single observation in an exponential( lambda ) model is I(lambda) = 1 / lambda^2, so for n observations the information is n / lambda^2. When evaluating at the MLE lambda_hat, we get B_hat0 = n / (lambda_hat^2). If we plug in lambda_hat = 1/7, then lambda_hat^2 = (1/7)^2 = 1/49, and thus n / lambda_hat^2 = 5 / (1/49) = 5 * 49 = 245. None of the given answer choices matches 245, so there’s a discrepancy between the standard derivation and the options provided.

An alternative way some texts present B_hat0 is as the negative of the second derivative of the standardized log-likelihood evaluated at the MLE, i.e., B_hat0 = - d^2/d lambda^2 [ l_bar(lambda) ] |_{lambda = lambda_hat}. Since l_bar(lambda) = log(lambda) - lambda * x_bar, its second derivative is d^2/d lambda^2 l_bar(lambda) = -1 / lambda^2, which is independent of the data aside from lambda. Therefore, B_hat0 = -(-1 / lambda_hat^2) = 1 / lambda_hat^2. With lambda_hat = 1/7, this equals 1 / (1/7)^2 = 49. While this yields a clean constant 49, the options presented do not include 49; instead, they include negative values and simple fractions. The nearest match among the options, if one interprets B_hat0 with a sign convention that could yield a negative value in some formulations, is -49.

Now, evaluate each answer choice:
- There is not enough information to compute the estimate B_hat0. This claim is false because we can compute lambda_hat from the data as lambda_hat = 1/7, and with that we can, in principle, evaluate B_hat0 under standard definitions. So insufficient information is not the correct reasoning here.
- B_hat0 = -49. This aligns with the interpretation that B_hat0 equals 1 / lambda_hat^2 but with a sign convention or misalignment in the provided options that yields a negative value when evaluated in this context. Given lambda_hat = 1/7, 1 / lambda_hat^2 = 49, so a negative 49 would reflect a particular sign convention or an error in the option, but it is the closest to a computable quantity in their set.
- B_hat0 = -7. This would correspond to something like - lambda_hat^2 times a scaling, which does not match the standard Fisher information form for this model and does not use the data in a way that would give 7.
- B_hat0 = 1/7. This equals lambda_hat itself, not the information or its negative. It does not align with the standard second-derivative-based information calculation.
- B_hat0 = 1/49. This equals lambda_hat^2, which again does not match the common information expression n / lambda_hat^2 unless a particular tiny-sample scaling or different definition is used, which is not standard here.

In summary, the question asks for B_hat0 given the sample and model. The most consistent numerical value among the options, given the typical relationship involving lambda_hat = 1/7 and the common second-derivative-based information, would be the negative of 1 / lambda_hat^2 or a closely related quantity, which in the provided choices is -49. This matches the pattern where the computed 1 / lambda_hat^2 is 49 but the option presents as -49, suggesting a sign convention used in the question’s context.

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