ISS 2016 Statistics Paper-2 Solution: Question 35 (Unbiased Estimator for Exponential Rate)

Continuing our question-by-question walk through the ISS Statistics Paper-2 archive, today we pick up right where the last post left off. This one is a classic estimation-theory question that trips up a lot of aspirants because the "obvious" answer turns out to be wrong for a subtle reason — and understanding why it is wrong is more valuable than memorising the final formula.

Quick Summary

• Topic: Point Estimation — constructing an unbiased estimator from the Exponential distribution

• Question Reference: ISS 2016, Statistics Paper-2, Question 35

• Correct Answer: Option (b), (n−1)/(n x̄)

Here is the question exactly as it appeared in the paper.

Question 35.What is an unbiased estimator of θ for the distribution, f(x, θ) = θe−θx, x ≥ 0 ?

• (a) (n−1)x̄ / n

• (b) (n−1) / (n x̄)

• (c) x̄ / (n−1)

• (d) 1 / ((n−1) x̄)

Step 1: Understand What Is Being Asked

We are told X1, X2, ..., Xn is a random sample from f(x, θ) = θe−θx, x ≥ 0. This is the Exponential distribution written in its rate form: θ is the rate parameter, and the mean of a single observation is E(X) = 1/θ, not θ itself.

The question asks us to find a statistic — built out of the sample — whose expected value equals θ exactly, for every possible value of θ. That is what "unbiased estimator of θ" means: if T is our statistic, we need E(T) = θ.

A very natural first guess is: since x̄ (the sample mean) estimates 1/θ, why not just estimate θ by 1/x̄? We will see in the next step exactly why that instinct fails, and that failure is the whole point of this question.

Step 2: Why the "Obvious" Guess 1/x̄ Does Not Work

We know E(x̄) = 1/θ. It is tempting to think that because of this, E(1/x̄) should equal θ. But expectation does not "pass through" a division like that. For any random variable Y and a non-linear function g, in general E(g(Y)) is not equal to g(E(Y)).

Here g(y) = 1/y is a convex function (its graph curves upward) for y > 0. A basic result called Jensen's Inequality tells us that for a convex function g, E(g(Y)) ≥ g(E(Y)). Applying this with Y = x̄ and g(y) = 1/y:

E(1/x̄) ≥ 1/E(x̄) = 1/(1/θ) = θ

So E(1/x̄) is always at least as large as θ, and in fact strictly larger except in degenerate cases. This means 1/x̄ systematically overestimates θ on average — it is a biased estimator. To fix this, we need to find the exact size of that bias and remove it. That is exactly what options (a) through (d) are testing: each one applies a different "correction factor" to 1/x̄ or x̄, and only one of them removes the bias completely.

Step 3: Find the Distribution of the Sample Sum

To compute E(1/x̄) exactly (not just as an inequality), we need to know the exact probability distribution of x̄, or equivalently of the sum S = X1 + X2 + ... + Xn.

A standard and important fact in statistics is:the sum of n independent Exponential(θ) random variables follows a Gamma distribution with shape parameter n and rate parameter θ. Writing S = ΣXi, the probability density function of S is:

fS(s) = [θn sn−1 e−θs] / (n−1)!,   s > 0

You can see this pattern by building it up slowly. For n = 1, this reduces to fS(s) = θe−θs, which is exactly our original Exponential density — as it must be, since S = X1 when n = 1. For general n, each time you add one more independent Exponential(θ) variable to the sum, the power of s increases by 1 and the factorial in the denominator grows to match it, which is the defining shape of the Gamma family. We will use this exact density to compute E(1/S) in the next step.

Step 4: Compute E(1/S) By Direct Integration

By the definition of expectation for a continuous random variable:

E(1/S) = ∫0∞ (1/s) · [θn sn−1 e−θs / (n−1)!] ds

Cancel one factor of s between the 1/s outside and the sn−1 inside. This lowers the power of s by exactly one:

E(1/S) = [θn / (n−1)!] · ∫0∞ sn−2 e−θs ds

Now we need the value of that remaining integral. This is a standard Gamma-function integral: for any positive integer m and any θ > 0,

∫0∞ sm−1 e−θs ds = (m−1)! / θm

Here our exponent on s is n−2, so matching sm−1 = sn−2 gives m−1 = n−2, i.e. m = n−1. Substituting m = n−1 into the standard result:

∫0∞ sn−2 e−θs ds = (n−2)! / θn−1

(This step requires n ≥ 2, since the expectation of 1/S is not finite for n = 1 — there simply is not enough data yet to average out the singularity near s = 0. In practice the ISS exam always assumes a "reasonably large" sample, so this is not a concern.)

Now substitute this back into our expression for E(1/S):

E(1/S) = [θn / (n−1)!] · [(n−2)! / θn−1]

Simplify the powers of θ: θn / θn−1 = θ1 = θ. Simplify the factorials: (n−2)! / (n−1)! = 1/(n−1), because (n−1)! = (n−1) · (n−2)!. Putting both simplifications together:

E(1/S) = θ / (n−1)

Pro Tip: This is the exact step where most students lose marks under exam pressure — they correctly set up the integral but then rush the factorial simplification and get (n−1) in the wrong place. The safe habit, the kind that is drilled question after question until it becomes automatic in a good coaching classroom, is to always write out (n−1)! = (n−1)·(n−2)! explicitly on paper before cancelling, rather than trying to do it in your head. It costs five extra seconds and eliminates the single most common arithmetic slip in this type of question.

Step 5: Build the Unbiased Estimator

We now know E(1/S) = θ/(n−1), which is close to θ but not exactly θ — it is off by a factor of 1/(n−1). To turn this into an exact unbiased estimator, simply multiply both sides by (n−1):

E[(n−1) · (1/S)] = (n−1) · [θ/(n−1)] = θ

Since expectation is linear, multiplying the random variable (n−1)/S by the constant (n−1) simply multiplies its expectation by the same constant — this is a fully legal, exact step, unlike the earlier failed shortcut in Step 2. So:

E[(n−1)/S] = θ

This tells us that T = (n−1)/S is an exact unbiased estimator of θ. The last thing to do is rewrite S in terms of the sample mean x̄, since that is how the answer options are phrased. By definition, x̄ = S/n, which rearranges to S = n x̄. Substituting this in:

T = (n−1) / S = (n−1) / (n x̄)

Step 6: Match to the Given Options and Confirm

Comparing (n−1)/(n x̄) to the four options:

• (a) (n−1)x̄/n — this is a completely different quantity (it estimates something proportional to 1/θ, not θ), so it is incorrect.

• (b) (n−1)/(n x̄) — this is exactly what we derived.Correct.

• (c) x̄/(n−1) — again estimates a multiple of 1/θ, not θ, so incorrect.

• (d) 1/((n−1) x̄) — this is our biased first guess 1/x̄ scaled by 1/(n−1), which actually overcorrects in the wrong direction. Notice from Step 5 that we needed to multiply (n−1)/S by nothing further — dividing by an extra (n−1) instead of multiplying by it flips the correction the wrong way, so this option is incorrect.

The correct answer is option (b), (n−1)/(n x̄). This matches the official ISS 2016 Statistics Paper-2 answer key, which lists the answer to Question 35 as (b). Our derivation and the official key are in full agreement.

Why This Question Matters

This question is a compact test of three ideas that show up repeatedly across the ISS Paper-2 syllabus: recognising standard sampling distributions (the Exponential-to-Gamma sum relationship), knowing that expectation does not commute with nonlinear transformations (Jensen's Inequality), and being comfortable enough with Gamma-function integrals to compute an exact moment rather than just an inequality. Aspirants who only memorise "x̄ estimates the mean" without understanding transformations like 1/x̄ tend to lose easy marks on exactly this kind of question. Building the habit of deriving these correction factors from scratch — rather than trying to recall them — is what separates a shaky guess from a confident, exam-ready answer.

Frequently Asked Questions

Why isn't 1/x̄ simply an unbiased estimator of θ?

Because expectation is only guaranteed to pass through linear functions, not general functions like 1/y. Since 1/y is convex for y > 0, Jensen's Inequality guarantees E(1/x̄) is strictly greater than 1/E(x̄) = θ, so 1/x̄ is a biased (upward-shifted) estimator of θ.

What is the difference between the MLE and an unbiased estimator here?

For this distribution, the maximum likelihood estimator of θ is 1/x̄ itself — the very estimator we showed is biased in Step 2. This is a good reminder that maximum likelihood estimators are not automatically unbiased; they only guarantee certain large-sample properties like consistency and asymptotic efficiency, not exact unbiasedness in finite samples.

Why does the sum of independent Exponential random variables follow a Gamma distribution?

The Exponential distribution is a special case of the Gamma distribution with shape parameter 1. When you add n independent Gamma random variables that share the same rate parameter, their shape parameters simply add together, which is why the sum of n independent Exponential(θ) variables becomes a Gamma(n, θ) variable.

Does the formula (n−1)/(n x̄) work when n = 1?

No. The derivation in Step 4 explicitly required n ≥ 2 for the integral defining E(1/S) to converge to a finite value. With only one observation there isn't enough information for the correction to be well defined, so this estimator is only meaningful for samples of size two or more.

Is (n−1)/(n x̄) also the minimum variance unbiased estimator of θ?

Yes. Because the sample sum S is a complete sufficient statistic for θ in this Exponential family, the Lehmann-Scheffe theorem guarantees that any unbiased function of S — including (n−1)/S — is automatically the unique minimum variance unbiased estimator (UMVUE) of θ.

How often does this type of "correct the bias" question appear in the ISS exam?

Quite often. Point estimation, and specifically constructing unbiased estimators for parameters of standard distributions (Exponential, Gamma, Uniform, Poisson), is one of the most heavily tested sub-topics across ISS Statistics Paper-2 papers, so it is well worth practising several variants of this exact derivation technique.

If any step above felt rushed or you'd like to see the derivation done differently, drop your doubt in the comments below — and if you found this walkthrough useful, please share it with a fellow ISS aspirant who is also working through the estimation theory syllabus.