ISS 2016 Statistics Paper-1 Solution: Question 9 (PGF of a Doubled Variable)

This post continues our ongoing question-by-question walkthrough of the ISS Statistics Paper-1 previous year papers, where we take one MCQ at a time and work through it slowly enough that every algebraic step is visible. Today we are on 2016 Statistics Paper-1, Question 9, which sits in the probability generating function (PGF) portion of the paper and tests whether you actually understand what a PGF represents rather than just memorising a formula sheet.

Quick Summary

• Topic: Probability Generating Functions — transforming the PGF of a variable under a linear scale change

• Paper Reference: ISS 2016, Statistics Paper-1, Question 9

• Correct Answer: Option (d) — [P(√s) + P(−√s)] / 2

The Question As It Appears In The Paper

Let X be a random variable with probability generating function P(s) = Σk pk sk. What is the probability generating function of Y = 2X?

• (a) P(√s) + P(−√s)

• (b) [P(√s) − P(−√s)] / 2

• (c) P(√s) − P(−√s)

• (d) [P(√s) + P(−√s)] / 2

Step 1: What A PGF Actually Is

Before touching the algebra, write down what P(s) means in plain language. If X is a random variable taking non-negative integer values 0, 1, 2, 3, ... with probabilities p0, p1, p2, ..., then the probability generating function is defined as

P(s) = E(sX) = Σk pk sk = p0 + p1s + p2s2 + p3s3 + ...

Every coefficient sitting in front of sk is exactly P(X = k). This is the whole point of a PGF: it packages an entire probability distribution into a single power series, where the exponent tells you "which value" and the coefficient tells you "how likely."

Step 2: Setting Up The PGF Of The New Variable

We are told X's PGF is P(s), and we need the PGF of a second variable that is tied to X by a factor of 2. Call this second variable Y, with its own PGF Q(s) = E(sY). By definition,

Q(s) = Σj qj sj, where qj = P(Y = j).

Since the relationship between X and Y is a doubling relationship, Y only ever lands on even integers, and each even value 2j of the "doubled" variable corresponds to exactly one value j of the "halved" variable, with the same probability attached: qj = p2j. So

Q(s) = Σj p2j sj — in other words, Q(s) is built only from the even-indexed coefficients of P(s), just reindexed.

Step 3: The Trick — Pulling Out Only The Even-Indexed Terms

This is the heart of the question. We know P(s) mixes together both even-power terms (p0, p2s2, p4s4, ...) and odd-power terms (p1s, p3s3, ...). We need a way to cancel the odd-power terms completely and keep only the even ones. The standard tool for this is to evaluate P at both √s and −√s and add the results.

Write out both expansions term by term:

P(√s) = Σk pk (√s)k = Σk pk sk/2

P(−√s) = Σk pk (−√s)k = Σk pk (−1)k sk/2

Now add the two series together, term by term:

P(√s) + P(−√s) = Σk pk sk/2 [ 1 + (−1)k ]

Look closely at the bracket [1 + (−1)k]. If k is even, (−1)k = 1, so the bracket becomes 1 + 1 = 2. If k is odd, (−1)k = −1, so the bracket becomes 1 − 1 = 0. That means every single odd-indexed term is multiplied by zero and disappears entirely, while every even-indexed term survives, doubled.

So the sum collapses to:

P(√s) + P(−√s) = 2 Σk even pk sk/2 = 2 Σj p2j sj = 2 Q(s)

Dividing both sides by 2 gives exactly:

Q(s) = [P(√s) + P(−√s)] / 2

which is option (d).

Step 4: A Quick Numerical Check

To make sure this isn't just symbol-shuffling, plug in real numbers. Suppose X takes only the values 0 and 2, each with probability 1/2 (so X is already "the doubled variable"). Then P(s) = 1/2 + (1/2)s2. The corresponding Y = X/2 takes values 0 and 1, each with probability 1/2, so its true PGF is clearly Q(s) = 1/2 + (1/2)s.

Now test the formula: P(√s) = 1/2 + (1/2)(√s)2 = 1/2 + (1/2)s. And P(−√s) = 1/2 + (1/2)(−√s)2 = 1/2 + (1/2)s as well, because squaring erases the minus sign. Adding gives 1 + s, and dividing by 2 gives 1/2 + (1/2)s — exactly Q(s), confirming the formula works.

Pro Tip: The trickiest part of this question isn't the algebra itself — it's noticing that squaring a negative root always kills the sign, which is exactly why the odd-power terms are the only ones that vanish when you add P(√s) and P(−√s), while the even-power terms double up instead of cancelling. The way this parity-extraction trick is usually taught is to first practice it on a plain, probability-free power series like (1 + s)n before ever seeing it dressed up in PGF notation — once you've drilled the pattern in its bare algebraic form, spotting it inside a probability question like this one becomes almost automatic.

Final Answer

Q(s) = [P(√s) + P(−√s)] / 2, which is option (d). This matches the official ISS 2016 answer key, which also lists (d) as the correct response for Question 9.

One honest note for careful readers: taken completely literally, "Y = 2X" (Y is twice X) would give Q(s) = E(s2X) = P(s2) directly — a one-line result that does not match any of the four listed options at all. The four options only become algebraically consistent under the reading used above, where the doubling relationship connects X and Y so that X sits at twice the value of Y (equivalently, Y = X/2 whenever X is even) — that is the interpretation that produces option (d) cleanly and lines up exactly with the official key, and it is the one used throughout this solution.

Why This Question Matters

Questions like this rarely test whether you can recite the definition of a PGF — they test whether you can manipulate one. ISS Paper-1 repeatedly rewards candidates who can take an abstract generating function and extract a piece of information from it (an even-indexed sum, a specific probability, a moment) using substitution tricks rather than brute-force expansion. The "plug in √s and −√s" method shown here is not a one-off gimmick — it is a general filter for separating even-power and odd-power terms in any power series, and it reappears in moment generating function problems, characteristic function problems, and combinatorial identity questions across the syllabus. Students who drill this kind of substitution until it becomes second nature, the way it's typically practised in a structured Sunrise Classes problem set, tend to recognise it in under a minute instead of getting stuck rederiving it from scratch under exam pressure.

Frequently Asked Questions

What exactly does a probability generating function tell you that a normal probability distribution doesn't?

A PGF packages the entire distribution into one algebraic object, which makes operations like finding moments, combining independent variables, or extracting specific probabilities a matter of differentiating or substituting into a single function rather than working with an infinite list of separate probabilities.

Why can't you just substitute s → 2s to get the PGF of Y = 2X?

Substituting s → 2s would give P(2s) = Σk pk 2k sk, which scales every coefficient by a power of 2 — that is not a valid probability generating function at all, since its coefficients no longer sum to 1. The correct scaling substitution for E(s2X) is s → s2, not s → 2s; the two are easy to confuse under time pressure.

Is the "add P(√s) and P(−√s)" trick something I need to memorise separately for every exam?

No — it is worth understanding as one instance of a general parity-extraction identity: adding P(s) and P(−s) isolates even-indexed terms, and subtracting them isolates odd-indexed terms, of any power series. Once you understand why the signs cancel, you can rebuild the exact formula in seconds instead of memorising it.

How common are PGF-manipulation questions like this one in ISS Paper-1?

Fairly common. Generating functions (probability, moment, and cumulant) show up almost every year in some form, often testing transformations of a variable, sums of independent variables, or recovering moments by differentiating the PGF at s = 1.

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

Treating "doubling a variable" and "scaling its generating function by the same factor" as the same operation. They are not — a linear transformation of the random variable itself always affects the argument of the PGF (through substitution), never the output value directly.

Do I need to memorise this exact result, or understand the derivation?

Understand the derivation. ISS examiners frequently reframe the same underlying identity with different variables, different scale factors, or a shift added on top, and a memorised final formula breaks the moment the question changes shape even slightly.

If any step above felt unclear, or if you worked through it differently, drop a comment below — and if you know another ISS aspirant grinding through the same PGF chapter, consider sharing this one with them, it tends to be a sticking point for a lot of first-time readers.