ISS 2016 Statistics Paper-2 Solution: Question 51 (MLE for Exponential Scale Parameter)

Continuing our question-by-question walk through the ISS Statistics Paper-2 archive, today we pick up the next item in the 2016 paper: a maximum likelihood estimation question built around two independent exponential samples. If you have been following along from Question 50, this one shifts gears from standard error back into estimation theory, so it is a good checkpoint to make sure your MLE mechanics are solid.

Quick Summary

• Topic: Maximum Likelihood Estimation (MLE) for the scale parameter of an exponential distribution, using two independent samples with related means

• Question Reference: ISS 2016, Statistics Paper-2, Question 51

• Correct Answer: Option (c), (2X̄ + Ȳ) / 4

The Question As Asked

Suppose X1, X2, X3, ..., X10, Y1, Y2, Y3, ..., Y10 are independent random variables where X's have density

fX(x) = (1/λ) e-x/λ, x > 0

and Y's have density

fY(y) = (1/2λ) e-y/2λ, y > 0.

What is the MLE of λ based on all X's and Y's?

• (a) (X̄ + Ȳ) / 2

• (b) (X̄ + Ȳ) / 3

• (c) (2X̄ + Ȳ) / 4

• (d) (X̄ + 2Ȳ) / 5

Step 1: Understand What You Are Given

Before touching any formula, read the setup slowly. There are two separate families of random variables here:

• X1, ..., X10: ten independent observations, each exponentially distributed with mean λ. Its density is fX(x) = (1/λ) e-x/λ.

• Y1, ..., Y10: ten more independent observations, each exponentially distributed, but with mean 2λ. Its density is fY(y) = (1/2λ) e-y/2λ.

So both samples are exponential, but the Y's have exactly twice the mean of the X's, and both means are tied to the same unknown parameter λ. That single shared unknown is exactly why we can pool information from both samples into one MLE instead of estimating two separate parameters.

Step 2: Write Down the Joint Likelihood Function

Since all 20 random variables are mutually independent, the joint likelihood is simply the product of all 20 individual densities.

L(λ) = [∏i=110 fX(xi)] × [∏j=110 fY(yj)]

Substituting each density:

L(λ) = [∏ (1/λ) e-xi/λ] × [∏ (1/2λ) e-yj/2λ]

Since there are 10 terms in each product, the constant factors just get raised to the 10th power, and the exponents add up across the 10 terms (because ea·eb = ea+b):

L(λ) = (1/λ)10 e-(Σxi)/λ × (1/2λ)10 e-(Σyj)/2λ

where Σxi means the sum x1 + x2 + ... + x10, and similarly for Σyj.

Step 3: Take the Log-Likelihood

Products of exponentials and powers are painful to differentiate directly, so we always take the natural log first. Logs turn products into sums and powers into multipliers, which makes differentiation much easier, and since log is a strictly increasing function, the λ that maximizes ln L(λ) is the same λ that maximizes L(λ).

ln L(λ) = -10 ln λ - (Σxi)/λ - 10 ln(2λ) - (Σyj)/2λ

Now expand ln(2λ) using the log product rule, ln(2λ) = ln 2 + ln λ, since this separates out the constant ln 2 (which has zero derivative and will not matter later):

ln L(λ) = -10 ln λ - (Σxi)/λ - 10 ln 2 - 10 ln λ - (Σyj)/2λ

Collecting the two "-10 ln λ" terms together:

ln L(λ) = -20 ln λ - 10 ln 2 - (Σxi)/λ - (Σyj)/2λ

Step 4: Differentiate With Respect to λ and Set It to Zero

We now differentiate ln L(λ) with respect to λ, term by term, using the rules d/dλ(ln λ) = 1/λ and d/dλ(1/λ) = -1/λ2.

d(ln L)/dλ = -20/λ - (Σxi)×(-1/λ2) - (Σyj)×(-1/2λ2)

d(ln L)/dλ = -20/λ + (Σxi)/λ2 + (Σyj)/2λ2

The constant term -10 ln 2 vanished because its derivative with respect to λ is zero. Setting this derivative equal to zero to find the critical point:

-20/λ + (Σxi)/λ2 + (Σyj)/2λ2 = 0

Step 5: Solve for λ

Multiply every term by λ2 to clear the denominators. This is a standard trick: multiplying through by the highest power of λ appearing in any denominator turns a fractional equation into a simple polynomial one.

-20λ + Σxi + (Σyj)/2 = 0

Rearranging:

20λ = Σxi + (Σyj)/2

λ̂ = [Σxi + (Σyj)/2] / 20

Now convert the sums into sample means. Since there are exactly 10 observations in each sample, Σxi = 10X̄ and Σyj = 10Ȳ, where X̄ and Ȳ are the respective sample means. Substitute these in:

λ̂ = [10X̄ + (10Ȳ)/2] / 20 = [10X̄ + 5Ȳ] / 20

Divide numerator and denominator by 5 to simplify:

λ̂ = (2X̄ + Ȳ) / 4

Pro Tip: The trap in this question is treating the two samples as if they share one common mean. They don't — the Y's have density parameter 2λ, not λ, so every Σyj term in the log-likelihood carries an extra factor of 1/2 relative to the X terms. The way this kind of question is usually taught is to write out the density of each sample completely separately before combining them, rather than trying to shortcut by "averaging" the two samples together. Students who drill this habit — always re-deriving the log-likelihood from the stated density instead of pattern-matching to a formula they memorized — tend to catch parameter-mismatch traps like this one on the first read instead of after they've already picked a wrong option.

Step 6: Confirm This Is a Maximum

A full derivation should technically check the second derivative is negative at this critical point to confirm it is a maximum and not a minimum. Here, d²(ln L)/dλ² = 20/λ2 - 2(Σxi)/λ3 - (Σyj)/λ3, which evaluates to a negative value at λ̂ (since substituting λ̂ back in makes the negative terms dominate), confirming we have indeed found the maximum of the likelihood function, not a minimum or saddle point. In a timed MCQ setting you would not normally write this out, but it is worth knowing it holds.

Final Answer

The maximum likelihood estimator is:

λ̂ = (2X̄ + Ȳ) / 4

This matches option (c). Checking against the official ISS 2016 Statistics Paper-2 answer key (Series A), the recorded answer for Question 51 is also C, so our independently derived solution is fully confirmed.

Why This Question Matters

This question tests three things at once: whether you can correctly write a joint likelihood for independent-but-not-identical samples, whether you remember to differentiate log-likelihoods rather than raw likelihoods, and whether you can carry a scale factor (here, the "2" inside 2λ) through an entire derivation without dropping it. This exact pattern — pooling two related samples into a single MLE — shows up repeatedly in ISS, IIT-JAM, and GATE(ST) statistics papers whenever a question involves two groups whose parameters are proportionally linked rather than identical, so it is worth being completely comfortable with the algebra here rather than memorizing this one answer.

Frequently Asked Questions

Why do we take the logarithm before differentiating the likelihood function?

The raw likelihood function is a product of many exponential and power terms, which is difficult to differentiate directly because of the product rule. Taking the natural log converts the product into a sum, which is far easier to differentiate term by term. Since the natural log function is strictly increasing, the value of λ that maximizes the log-likelihood is guaranteed to be the same value that maximizes the original likelihood.

What does it mean for X̄ and Ȳ to appear together in a single estimator?

It means the data from both samples is being pooled to estimate one shared unknown parameter. This only works because both density functions were written in terms of the same λ. If the two samples had completely unrelated parameters, you would instead get two separate MLE equations, one for each parameter, solved independently.

How do I know how many terms to include when summing Σx_i and Σy_j?

You simply count how many independent observations are stated in the problem. Here the question explicitly lists X1 through X10 and Y1 through Y10, so each sum runs over exactly 10 terms. If the sample sizes for the two groups are different, the two product terms in the likelihood will have different exponents, and that difference will carry through to the final answer.

What is the most common mistake students make on this type of question?

The most common mistake is forgetting that ln(2λ) splits into ln 2 + ln λ, and either dropping the ln λ part or carrying the ln 2 term incorrectly into the derivative. Since ln 2 is a constant, it disappears entirely once you differentiate, but students who do not expand it explicitly sometimes differentiate ln(2λ) as if it behaves differently from ln λ, leading to an incorrect coefficient in the final answer.

Is there a quicker way to solve this without writing the full log-likelihood?

Not reliably for an exam like this. Because the two samples have different scale parameters (λ versus 2λ), there is no shortcut that avoids writing out the log-likelihood and differentiating carefully — any attempt to average or combine the samples informally is likely to miss the factor of 2 and lead to one of the incorrect options. The safest and fastest real-world approach is still the systematic one shown above.

Does this method generalize to more than two samples with linked parameters?

Yes. If you had a third sample Z with density depending on, say, 3λ, you would simply add a third term to the log-likelihood following the same pattern, differentiate once more, and solve the resulting linear equation in λ. The number of samples does not change the underlying method, only the number of terms you are tracking.

If any step here felt unclear, drop a comment below and we will walk through it in more detail — and if this helped, consider sharing it with a fellow ISS aspirant who might be stuck on the same question.