ISS 2016 Statistics Paper-2 Solution: Question 46 (MLE for the Exponential Distribution)

Continuing our question-by-question walk through the ISS Statistics Paper-2 archive, today's post picks up right where the last one left off in the 2016 paper. This time the topic shifts from unbiasedness to one of the most frequently tested ideas in the estimation portion of the syllabus: the method of maximum likelihood.

Quick Summary

• Topic:Maximum Likelihood Estimation (MLE) for a one-parameter exponential distribution

• Question Reference:ISS 2016, Statistics Paper-2, Question 46

• Correct Answer:Option (b), 1/x̄ (the reciprocal of the sample mean)

The Question

What is the MLE of θ based on a random sample of size n drawn from a population with pdf as

f(x, θ) = θe−θx, 0 < x < ∞, and f(x, θ) = 0 otherwise?

• (a) x̄

• (b) 1/x̄

• (c) Largest observation

• (d) Smallest observation

Step 1: Identify the Distribution and Write the Likelihood Function

The pdf given, f(x, θ) = θe−θx for x > 0, is the exponential distribution with rate parameter θ. Notice this is not the version parameterised by the mean (f(x, θ) = (1/θ)e−x/θ); here θ itself multiplies the exponential term, so θ behaves as a rate, not a mean.

Suppose X1, X2, ..., Xn is a random sample from this population. Because the observations are independent and identically distributed, the joint density (the likelihood function) is simply the product of the individual densities:

L(θ) = f(x1, θ) · f(x2, θ) · ... · f(xn, θ) = θe−θx₁ · θe−θx₂ · ... · θe−θxₙ

Collecting the n copies of θ and adding the exponents (since multiplying exponential terms means adding their powers):

L(θ) = θn · e−θ(x₁ + x₂ + ... + xₙ) = θn e−θΣxᵢ

Step 2: Take the Log-Likelihood

Maximising L(θ) directly is awkward because of the product and the exponential. The standard trick, which works because the natural logarithm is a strictly increasing function (so whatever θ maximises L(θ) also maximises ln L(θ)), is to take logs first. This converts the product into a sum and brings the exponent down, which is far easier to differentiate.

ln L(θ) = ln(θn) + ln(e−θΣxᵢ) = n ln θ − θΣxi

Step 3: Differentiate with Respect to θ and Set It to Zero

To find the value of θ that maximises the log-likelihood, we differentiate with respect to θ and set the derivative equal to zero. This is just ordinary calculus:

d/dθ [ln L(θ)] = d/dθ [n ln θ] − d/dθ [θΣxi] = n/θ − Σxi

Setting this equal to zero:

n/θ − Σxi = 0

n/θ = Σxi

θ = n / Σxi

Step 4: Recognise n/Σxᵢ as 1/x̄

Remember that the sample mean is defined as x̄ = (1/n)Σxi, which means Σxi = n·x̄. Substituting this back:

θ̂ = n / Σxi = n / (n·x̄) = 1/x̄

So the maximum likelihood estimator of θ is θ̂ = 1/x̄, the reciprocal of the sample mean.

Step 5: Confirm It Is a Maximum, Not a Minimum

A careful solver always checks the second-order condition rather than assuming the stationary point is a maximum. Differentiating the log-likelihood a second time:

d²/dθ² [ln L(θ)] = −n/θ²

Since n > 0 and θ² > 0, this second derivative is always negative. A negative second derivative at the stationary point confirms it is indeed a maximum, so θ̂ = 1/x̄ is correctly the maximum likelihood estimator, not a minimum or a saddle point.

Pro Tip: The single most common slip on this type of question is misreading which parameter multiplies the exponent. If the pdf had instead been written as f(x, θ) = (1/θ)e−x/θ, the MLE would come out to x̄ itself, not 1/x̄ — the two forms are reciprocal parameterisations of the exact same exponential family, and examiners rely on this to test whether you actually derived the answer or just pattern-matched to "exponential distribution means x̄." The way this is usually taught is to always write out L(θ) and take the derivative from scratch rather than recalling a memorised formula, because that habit is what protects you the moment the question swaps the parameterisation on you.

Final Answer:Option (b), θ̂ = 1/x̄. This matches the official ISS 2016 Statistics Paper-2 answer key (Series A), which lists option B as correct for Question 46.

Why This Question Matters

MLE derivations for standard distributions — exponential, Poisson, normal, uniform — show up almost every single year across both ISS Paper-2 and related exams like IIT JAM(MS) and GATE(ST). The exponential distribution's MLE is especially popular because it is easy to set a trap in the parameterisation, exactly as this question does. Students who drill this five-step process (write the likelihood, take logs, differentiate, solve, verify via the second derivative) until it becomes automatic rarely lose marks to this kind of question, no matter how the pdf is dressed up. It's a small investment of practice that pays off across dozens of exam questions, not just this one.

Frequently Asked Questions

Why do we take the logarithm of the likelihood function instead of maximising L(θ) directly?

The logarithm converts a product of n terms into a sum, which is much easier to differentiate. Since ln(x) is a strictly increasing function, the value of θ that maximises ln L(θ) is guaranteed to be the same value that maximises L(θ), so nothing is lost by working with the log-likelihood.

Is the MLE here the same as the method of moments estimator?

Yes, in this particular case they coincide. The method of moments equates the population mean (1/θ for this parameterisation) to the sample mean x̄, which also gives θ̂ = 1/x̄. This agreement does not always hold for other distributions, so it should not be assumed as a general rule.

Is 1/x̄ an unbiased estimator of θ?

No. While θ̂ = 1/x̄ is the maximum likelihood estimator, it is actually a biased estimator of θ for finite sample size n; the bias only vanishes as n tends to infinity. MLEs are asymptotically unbiased and efficient, but they are not automatically unbiased for small samples — this is a separate property that needs to be checked independently.

How would the answer change if the pdf were f(x, θ) = (1/θ)e−x/θ instead?

In that parameterisation θ represents the mean of the exponential distribution rather than the rate, and repeating the same five steps gives θ̂ = x̄ directly. It is worth deriving both versions once by hand so that the difference between "θ as rate" and "θ as mean" becomes second nature.

What role does the domain "0 < x < ∞" play in the likelihood function?

It confirms that the support of X does not depend on θ, which is what allows us to differentiate the log-likelihood freely over the entire real line for θ. If the support had depended on θ (as it does for a uniform distribution on (0, θ), for instance), ordinary calculus would not apply and the MLE would instead be found by inspection, typically turning out to be an order statistic like the sample maximum.

Why are options (c) and (d), the largest and smallest observations, incorrect here?

Those forms of estimator (order statistics) are the correct MLEs for distributions whose support depends on the parameter, such as the uniform distribution on (0, θ). Here, the exponential distribution's support (0, ∞) does not depend on θ at all, so the MLE is obtained through ordinary differentiation of the log-likelihood rather than by picking an extreme observation, which rules out both (c) and (d).

If any step above still feels unclear, drop a comment below and we'll work through it together — and if you know a fellow ISS aspirant who's grinding through the same paper, consider sharing this with them so they can check their own working against it.