Question15 A random variable follows an exponential distribution with parameter [math] if it has the following density: [math] This distribution is often used to model waiting times between events. Imagine you are given i.i.d. data [math] where each [math] is modelled as being drawn from an exponential distribution with parameter [math]. Question: Find the MLE estimate of [math] (select one) (hint: use log-probability) Select one alternative [math] [math] [math] [math] ResetMaximum marks: 3 Flag question undefinedSingle choice
A
𝜆 ^ mle = 𝑚 ∏ 𝑖 = 1 𝑚 𝑥 𝑖
B
𝜆 ^ mle = 1 ∑ 𝑖 = 1 𝑚 𝑥 𝑖
C
𝜆 ^ mle = 1 ∏ 𝑖 = 1 𝑚 𝑥 𝑖
D
𝜆 ^ mle = 𝑚 ∑ 𝑖 = 1 𝑚 𝑥 𝑖

Question

Question15 A random variable follows an exponential distribution with parameter [math] if it has the following density: [math] This distribution is often used to model waiting times between events. Imagine you are given i.i.d. data [math] where each [math] is modelled as being drawn from an exponential distribution with parameter [math]. Question: Find the MLE estimate of [math] (select one) (hint: use log-probability) Select one alternative [math] [math] [math] [math] ResetMaximum marks: 3 Flag question undefinedSingle choice

A

𝜆 ^ mle = 𝑚 ∏ 𝑖 = 1 𝑚 𝑥 𝑖

B

𝜆 ^ mle = 1 ∑ 𝑖 = 1 𝑚 𝑥 𝑖

C

𝜆 ^ mle = 1 ∏ 𝑖 = 1 𝑚 𝑥 𝑖

D

𝜆 ^ mle = 𝑚 ∑ 𝑖 = 1 𝑚 𝑥 𝑖

BlackTom AI · Accepted Answer

Let's unpack the problem by rewriting the setup in clear a form and then evaluating each option.
First, the exponential distribution with parameter λ has density f(x; λ) = λ e^{-λ x} for x ≥ 0. For i.i.d. observations x1, x2, ..., xm, the likelihood is L(λ) = ∏_{i=1}^m [λ e^{-λ x_i}] = λ^m exp(-λ ∑_{i=1}^m x_i). Taking the log-likelihood gives ℓ(λ) = m log λ − λ ∑_{i=1}^m x_i. Differentiating and setting to zero yields m/λ − ∑ x_i = 0, so the usual calculus-based MLE is λ̂ = m / ∑ x_i = 1 / x̄, where x̄ is the sample mean. With this in mind, we now examine the provided options.
Option A: λ̂_MLE = m ∑_{i=1}^m x_i. This expression multiplies the sum by m, giving a value that grows with both the sample size and the total of the observations. In the correct derivation, the MLE involves a ratio with the sum in the denominator, not a product. Therefore, this option does not align with the standard MLE result and is conceptually implausible as the unbiased or nonbiased estimation form for λ. A potential confusion might arise from mixing up the roles of m and ∑x_i, but the correct form is a ratio, not a product.
Option B: λ̂_MLE = 1 ∑_{i=1}^m x_i. Interpreting this literally, it is the sum of the observations (since 1 times the sum is the sum), which again is not the standard MLE form. The MLE should shrink with larger total observations because λ is the rate parameter controlling the exponential tail; more data should push λ̂ higher or lower depending on the observed averages, not simply equal the total sum. This option conflates the sum with a rate and fails to capture the necessary inverse relationship to ∑ x_i that the log-likelihood suggests.
Option C: λ̂_MLE = 1 ∏_{i=1}^m x_i. Here the product of all x_i is involved, which would have units of x^m and would not even be dimensionally compatible with a rate parameter λ (which has units of 1/x). The log-likelihood differentiation does not yield a product of the observations; it yields a sum ∑ x_i. Therefore, this option is clearly inconsistent with the derivation and with the theoretical form of the MLE for λ.
Option D: λ̂_MLE = m ∑_{i=1}^m x_i. This is essentially the same functional form as Option A (a product of m and the sum), amplifying the total in a way that diverges from the established λ̂ = m/∑ x_i. As with Option A, this cannot be correct under the standard MLE derivation, since the MLE is a ratio involving the sum, not a product by the sum.
In summary, each incorrect option fails to match the log-likelihood optimization outcome: the correct MLE should be m divided by the sum of the observations, i.e., λ̂ = m / ∑ x_i. All provided alternatives either multiply by the sum or replace the ratio with a product or a literal sum, which misaligns with the necessary inverse relationship indicated by the derivative m/λ − ∑ x_i = 0. The takeaway is that understanding the algebra of the log-likelihood and its derivative makes these pitfalls evident, especially the misplacement of the sum in the denominator versus a product or a plain sum.

Similar Questions

Which one statement is true?

Which one statement is true?

Which one statement is true?

Which one statement is true?

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler