ISS 2016 Statistics Paper-1 Solution: Question 6 (Moments of a Transformed Poisson Variable)

Continuing our question-by-question walkthrough of the ISS Statistics Paper-1 archive, we now arrive at Question 6 from the 2016 paper. This one is a good test of whether you truly understand how the moments of a distribution behave when the underlying random variable is scaled, rather than just knowing the mean and variance by rote. As always, we will build the answer up from first principles rather than quoting a formula out of nowhere.

Quick Summary

• Topic:Moments of the Poisson distribution and how central moments transform under a linear change of variable

• Question Reference:ISS 2016, Statistics Paper-1, Question 6

• Correct Answer:Option (c) — Statements 1 and 3 only

The Question

A Poisson variable X has mean equal to 1/2. Consider the following for Y = 2X:

1. E(Y) = 1

2. Var(Y) = 4

3. μ3(Y) = 4

4. μ4(Y) = 28

Which of the above is/are correct?

• (a) 1 only

• (b) 2 and 4 only

• (c) 1 and 3 only

• (d) 1, 2 and 3 only

Step 1: Understand What Is Being Asked

We are told X is Poisson with mean λ = 1/2. A new variable Y is created by the transformation Y = 2X. We are given four claimed facts about Y — its mean, its variance, its third central moment μ3(Y), and its fourth central moment μ4(Y) — and we must check each one individually. To do that correctly, we first need to know the exact values of the corresponding moments of X, and then work out precisely how multiplying a variable by a constant changes each of those moments.

Step 2: Find the Raw Moments of X

For a Poisson variable with parameter λ, the probability generating function is:

G(s) = E(sX) = Σ sk e−λ λk / k! = e−λ eλs = eλ(s−1)

Differentiating this repeatedly with respect to s and setting s = 1 gives the factorial moments of X. Each differentiation brings down one extra factor of λ (because d/ds of eλ(s−1) is λeλ(s−1), and at s = 1 the exponential term becomes e0 = 1). So we get the clean pattern:

• G′(1) = λ = E(X)

• G″(1) = λ2 = E[X(X−1)]

• G′′′(1) = λ3 = E[X(X−1)(X−2)]

• G(4)(1) = λ4 = E[X(X−1)(X−2)(X−3)]

Now we convert these factorial moments into ordinary (raw) power moments, one at a time, purely by expanding the brackets algebraically.

Second raw moment:E[X(X−1)] = E[X2] − E[X] = λ2, so

E[X2] = λ2 + λ

Third raw moment:Expand X(X−1)(X−2) = X3 − 3X2 + 2X. Taking expectations,

E[X3] − 3E[X2] + 2E[X] = λ3

E[X3] = λ3 + 3(λ2 + λ) − 2λ = λ3 + 3λ2 + λ

Fourth raw moment:Expand X(X−1)(X−2)(X−3) = X4 − 6X3 + 11X2 − 6X. Taking expectations,

E[X4] − 6E[X3] + 11E[X2] − 6E[X] = λ4

E[X4] = λ4 + 6(λ3 + 3λ2 + λ) − 11(λ2 + λ) + 6λ = λ4 + 6λ3 + 7λ2 + λ

Step 3: Convert Raw Moments into Central Moments

The mean is μ = E(X) = λ. The central moments are defined as μr = E[(X − μ)r]. We expand each one using the binomial theorem and substitute the raw moments found above.

Second central moment (variance):μ2 = E[X2] − μ2 = (λ2 + λ) − λ2 = λ

Third central moment:μ3 = E[X3] − 3μE[X2] + 3μ2E[X] − μ3. Substituting μ = λ:

μ3 = (λ3+3λ2+λ) − 3λ(λ2+λ) + 3λ2(λ) − λ3

μ3 = λ3+3λ2+λ − 3λ3−3λ2 + 3λ3 − λ3 = λ

Notice how every λ3 and λ2 term cancels out, leaving just λ. This is a well-known special property of the Poisson distribution:every central moment beyond the mean is built from the same parameter λ in a very structured way, which is exactly why this question is testing it.

Fourth central moment:μ4 = E[X4] − 4μE[X3] + 6μ2E[X2] − 4μ3E[X] + μ4. Substituting μ = λ and collecting like powers of λ term by term (the λ4 terms cancel to 0, the λ3 terms cancel to 0, and what survives is):

μ4 = 3λ2 + λ

So for our question, with λ = 1/2:

• μ2(X) = Var(X) = λ = 1/2

• μ3(X) = λ = 1/2

• μ4(X) = 3λ2 + λ = 3(1/4) + 1/2 = 3/4 + 2/4 = 5/4

Step 4: How Central Moments Behave Under Y = aX + b

Here Y = 2X, so a = 2 and b = 0. First, the mean transforms simply:

E(Y) = aE(X) + b = 2(1/2) + 0 = 1

For the central moments, write Y − E(Y) in terms of X. Since E(Y) = aE(X) + b,

Y − E(Y) = [aX + b] − [aE(X) + b] = a[X − E(X)]

Raising both sides to the power r and taking expectation:

μr(Y) = E{(Y − E(Y))r} = E{ar(X − E(X))r} = ar · μr(X)

Pro Tip: This is exactly where students lose marks on this kind of question. The shift constant b disappears completely from every central moment (it only ever affects the mean), but the scale constant a does not appear as a itself — it appears as ar, with r matching the order of the moment you're computing. So for variance you multiply by a2, for the third central moment by a3, and for the fourth by a4. Mixing these powers up is the single most common slip on transformation questions like this one, and the way it is usually taught is to write out the r = 1, 2, 3, 4 cases side by side until the ar pattern becomes automatic rather than something you have to re-derive under exam pressure.

Step 5: Check Each of the Four Statements

With a = 2, we now compute a2 = 4, a3 = 8, and a4 = 16, and apply μr(Y) = arμr(X):

Statement 1:E(Y) = 2 × (1/2) = 1. The statement claims E(Y) = 1.This is correct.

Statement 2:Var(Y) = μ2(Y) = a2 · μ2(X) = 4 × (1/2) = 2. The statement claims Var(Y) = 4. Since 2 ≠ 4,this is incorrect.

Statement 3:μ3(Y) = a3 · μ3(X) = 8 × (1/2) = 4. The statement claims μ3(Y) = 4.This is correct.

Statement 4:μ4(Y) = a4 · μ4(X) = 16 × (5/4) = 20. The statement claims μ4(Y) = 28. Since 20 ≠ 28,this is incorrect.

Final Answer

Only Statements 1 and 3 hold true. This matches option (c) 1 and 3 only, which is also exactly what the official ISS 2016 Statistics Paper-1 answer key records for Question 6 (key value: C). Our independent derivation and the official key are in full agreement.

Why This Question Matters

This question is not really about Poisson distributions specifically — it is about whether you understand moment transformation rules well enough to apply them correctly under time pressure, which is a recurring theme across ISS Paper-1. The Poisson distribution is simply a convenient, well-structured example because its central moments have such a clean closed form in terms of λ. In the actual exam hall, questions built this way are designed to catch candidates who have memorised "variance of Y = 2X is 4 times variance of X" without internalising that the same ar scaling rule applies, with the matching exponent, to every higher-order central moment as well. Being able to rebuild the general rule μr(Y) = arμr(X) from scratch, as we did here starting from the generating function, is a far more durable skill than memorising the variance case alone.

Frequently Asked Questions

Why does the constant b disappear from every central moment?

Central moments measure spread around the mean, not the mean itself. When you shift a variable by adding a constant b, its mean shifts by exactly the same amount, so the deviation (Y − mean of Y) is unaffected by b entirely — the b terms cancel exactly, as shown in Step 4. This is why central moments are sometimes described as location-invariant measures.

Why is it ar and not just a for higher central moments?

Because μr(Y) involves raising the deviation (Y − mean) to the power r, and that deviation itself contains a factor of a. Raising a factor of a to the power r naturally produces ar, as derived algebraically in Step 4.

What makes the Poisson distribution special in terms of its moments?

For a Poisson variable with parameter λ, every cumulant equals λ. Since the second and third cumulants coincide exactly with the second and third central moments, both the variance and the third central moment of a Poisson variable equal λ. The fourth central moment is slightly different because the fourth cumulant does not equal the fourth central moment directly — there is an extra 3λ2 correction term, which is why μ4 = 3λ2 + λ rather than simply λ.

Is there a shortcut to avoid deriving E[X3] and E[X4] from scratch every time?

Yes — once you have derived and memorised μ2 = λ, μ3 = λ, and μ4 = 3λ2 + λ for the Poisson distribution a single time, you can reuse these standard results directly in any future question. What still needs to be worked out fresh each time is how a linear transformation like Y = aX + b changes those moments, since the value of a differs from question to question.

How would the answer change if the question had asked about Y = 2X + 5 instead?

The additive constant 5 would change E(Y) to 2(1/2) + 5 = 6, but it would have absolutely no effect on Var(Y), μ3(Y), or μ4(Y), since central moments ignore shifts entirely. Those three values would come out exactly the same as computed in Step 5: 2, 4, and 20 respectively.

What is the quickest way to double check an answer like this before moving on?

A fast sanity check is dimensional consistency: since Var(X) = 1/2 is a fairly small number, Var(Y) for Y = 2X should scale by exactly 4, giving 2 — not jump all the way to a value like 4, which would imply an inconsistent scaling factor. Running this kind of rough magnitude check against each option can catch an arithmetic slip before you commit to a final answer.

If any step above felt unclear, drop a comment below and we'll work through it together — and if you know a fellow ISS aspirant who is currently grinding through Paper-1 statistics, sharing this post with them could save them the trouble of re-deriving these moment formulas on their own.