ISS 2016 Statistics Paper-1 Solution: Question 43 (MGF of a Shifted Poisson Variable)

Continuing our question-by-question walk through the ISS Statistics Paper-1 archive, today we land on Question 43 from the 2016 paper. This one sits in the probability distributions section and tests whether you can combine two ideas that are usually taught separately: the moment generating function (MGF) of a Poisson variable, and what happens to an MGF when you shift a random variable by a constant.

Quick Summary

• Topic:Moment Generating Function (MGF) of a transformed Poisson variable

• Question Reference:ISS 2016, Statistics Paper-1, Question 43

• Correct Answer:Option (a)

The Question

43. If X is a random variable having Poisson distribution with parameter λ, then what is the m.g.f. of Y = X − 2?

• (a) eλet − 2t − λ

• (b) eλet − t − 1

• (c) eλ(et − t − 1)

• (d) eλ(et − 1)

Step 1: Recall What a Poisson Distribution Looks Like

A random variable X follows a Poisson distribution with parameter λ if its probability mass function is

P(X = k) = e−λ λk / k!, for k = 0, 1, 2, 3, ...

Here λ is both the mean and the variance of X. Before touching Y, we need the MGF of X itself, because every other moment generating function in this question will be built on top of it.

Step 2: Derive the MGF of X From First Principles

By definition, the moment generating function of X is MX(t) = E[etX]. Writing this out as a sum over all possible values of X:

MX(t) = Σk=0∞ etk · e−λ λk / k!

Pull the constant e−λ out of the summation, and combine etk with λk into (λet)k:

MX(t) = e−λ Σk=0∞ (λet)k / k!

Now look closely at that summation. It has exactly the form Σ xk/k!, which is the Taylor series expansion of ex, with x = λet in our case. So the sum collapses to eλet:

MX(t) = e−λ · eλet = eλet − λ = eλ(et − 1)

This last form, eλ(et − 1), is the standard, well-known MGF of a Poisson(λ) variable. You'll notice it also happens to be option (d) — that's not a coincidence, it's the examiner testing whether you remember this is the MGF of X, not of Y = X − 2.

Step 3: Apply the Shift — Find the MGF of Y = X − 2

Now we bring in the second idea the question is really testing: how an MGF behaves under a linear shift. In general, for any random variable X and constants a and b, if Y = aX + b, then

MY(t) = E[etY] = E[et(aX + b)] = etb · E[e(ta)X] = etb · MX(at)

In our problem, Y = X − 2, so a = 1 and b = −2. Substituting:

MY(t) = e−2t · MX(t)

We already derived MX(t) = eλ(et − 1) in Step 2, so we substitute that in directly:

MY(t) = e−2t · eλ(et − 1)

Since both factors are powers of e, we add the exponents:

MY(t) = eλ(et − 1) − 2t = eλet − λ − 2t

Rearranging the terms in the exponent purely for presentation (addition is commutative, so the order doesn't change the value):

MY(t) = eλet − 2t − λ

Pro Tip: The trap in this question isn't the algebra — it's stopping one step too early. Most students correctly derive the Poisson MGF eλ(et − 1), see it sitting right there as option (d), and pick it without reading the question again. The way this is usually taught is to physically underline what the question asks for — here it's the MGF of Y, not X — before you even start differentiating or summing anything, so the "obvious-looking" distractor option never gets a chance to trick you.

Step 4: Match With the Options and Confirm

Our derived result, MY(t) = eλet − 2t − λ, matches option (a)exactly, term for term. Option (d) is a near-miss trap (the MGF of X, not Y), and options (b) and (c) don't follow from the correct substitution at all.

This also matches the official ISS 2016 answer key, which lists the answer to Question 43 as (a) for Test Booklet Series A. So our independent derivation and the official key are in full agreement — no discrepancy here.

Why This Question Matters

This question is a good stress test of two fundamentals that show up again and again across the ISS and similar exams: deriving a standard distribution's MGF from the summation definition (rather than memorizing it blindly), and correctly applying the MGF transformation rule for linear combinations of a random variable. Students who drill this kind of two-step derivation until it becomes automatic tend to move much faster and more confidently through the probability section, because half the battle in MCQ-format papers is not making an algebra mistake under time pressure — it's resisting the pull of a distractor option that looks almost right.

Frequently Asked Questions

Why is the MGF of X useful if the question asks about Y = X − 2?

Because the MGF of a shifted variable is always built directly from the MGF of the original variable using the rule MaX+b(t) = etbMX(at). You essentially never derive the MGF of a transformed variable from scratch — you derive the base variable's MGF once, then apply the transformation rule.

Is the standard Poisson MGF formula worth memorizing directly?

It's worth knowing by heart for speed, but you should also be able to derive eλ(et − 1) from the summation definition in under a minute, because exam questions frequently modify the setup (as this one does) in a way that punishes rote memorization without understanding.

What's the general rule for the MGF of aX + b?

If Y = aX + b, then MY(t) = ebt · MX(at). You multiply the original MGF (evaluated at at instead of t) by ebt. This single formula covers shifting, scaling, and any combination of the two.

How can I quickly verify an MGF answer without redoing the full derivation?

Differentiate the candidate MGF once with respect to t and set t = 0; this should return E[Y]. For our answer, differentiating eλet − 2t − λ and evaluating at t = 0 gives E[Y] = λ − 2, which matches the direct calculation E[X − 2] = λ − 2 since E[X] = λ for a Poisson variable. This quick sanity check confirms option (a).

Why are options (b) and (c) wrong?

Option (b) incorrectly uses −t − 1 in place of the correctly derived −2t − λ, and option (c) incorrectly factors λ across terms that shouldn't be grouped together — both appear to come from applying the shift formula with errors in where the constant b = −2 and the parameter λ get placed in the exponent. Working the derivation out fully, as we did above, is the safest way to avoid falling for either.

Does the penalty-for-wrong-answers rule in this exam change how I should approach MCQs like this?

Yes — with a one-third negative marking penalty, it's better to spend an extra 20–30 seconds confirming your derivation (like the quick E[Y] sanity check above) than to guess between two similar-looking exponential expressions under time pressure.

If anything in this derivation felt shaky or you'd solve it differently, drop a comment below — working through the disagreement is often the fastest way to actually understand it. And if you know a fellow ISS aspirant grinding through the same probability chapter, send this one their way; it's exactly the kind of question that's worth seeing explained once, properly.