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 undefined

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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.

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