ISS 2016 Statistics Paper-1 Solution: Question 35 (Moments, Skewness and Kurtosis Inequalities)

Continuing our question-by-question walk through the ISS Statistics Paper-1 papers, today we take up Question 35 from the 2016 paper. This one tests whether you actually remember the inequalities connecting the mean, the harmonic mean, skewness, and kurtosis — not just their formulas.

Quick Summary

• Topic: Properties and inequalities of a frequency distribution (AM vs HM,β1 and β2, Karl Pearson's coefficient of skewness)

• Question Reference: ISS 2016, Statistics Paper-1, Question 35

• Correct Answer: Option (b) — Statements 2 and 3 only

The Question

Which of the following are correct in respect of a frequency distribution?

1. Arithmetic mean is less than harmonic mean.

2. β2 > 1

3. β2 −β1 − 1 ≥ 0

4. Karl Pearson's coefficient of skewness lies between −1 and +1.

Select the correct answer using the code given below:

• (a) 1 and 2 only

• (b) 2 and 3 only

• (c) 3 and 4 only

• (d) 1 and 4 only

Step 1: Setting Up the Notation

Before checking each statement, let's fix the notation, because this question is really about knowing what β1 and β2 mean and where they come from.

For any distribution, let μbe the mean, and let μ2,μ3,μ4 denote the 2nd, 3rd and 4th central moments — that is, moments taken about the mean:

μr = E[(X −μ)r]

Then the two dimensionless shape measures are defined as:

β1 =μ32 /μ23  (a measure of skewness)

β2 =μ4 /μ22  (a measure of kurtosis, or "peakedness")

Notice that β1 is built from μ32, a square, so β1 can never be negative — that single fact turns out to be the key that unlocks statement 3.

Step 2: Checking Statement 1 (AM vs HM)

Statement 1 claims the arithmetic mean (AM) is less than the harmonic mean (HM). This is one of the oldest inequalities in statistics, and it goes the other way. For any set of positive observations:

AM ≥ GM ≥ HM

with equality holding only when every observation in the data is identical. This is the AM-GM-HM chain, and you can see a two-number example instantly: for the values 4 and 16, AM = 10, GM = 8, HM = 6.4 — so AM > HM, exactly the reverse of what statement 1 claims.

So statement 1 is false.

Step 3: Proving Statement 3 First (it unlocks Statement 2)

It's easier to prove statement 3 before statement 2, because statement 2 falls out of it almost for free. We want to check whether:

β2 −β1 − 1 ≥ 0

Work with the centred variable Y = X −μ, so E[Y] = 0, E[Y2] =μ2, E[Y3] =μ3, E[Y4] =μ4.

Now apply the Cauchy–Schwarz inequality (the same inequality that forces a correlation coefficient to lie between −1 and +1) to the two random variables Y and Y2. In its covariance form it says:

[Cov(Y, Y2)]2 ≤ Var(Y) · Var(Y2)

Compute each piece using E[Y] = 0:

• Cov(Y, Y2) = E[Y3] − E[Y]E[Y2] =μ3 − 0 =μ3

• Var(Y) = E[Y2] − (E[Y])2 =μ2

• Var(Y2) = E[Y4] − (E[Y2])2 =μ4 −μ22

Substituting into the Cauchy–Schwarz inequality:

μ32 ≤μ2 (μ4 −μ22)

Expand the right side:

μ32 ≤μ2 μ4 −μ23

Rearranging:

μ2 μ4 ≥μ23 +μ32

Now divide every term by μ23 (which is always positive, since it's a variance cubed):

μ4 /μ22 ≥ 1 +μ32 /μ23

The left side is exactly β2, and the second term on the right is exactly β1. So this says:

β2 ≥ 1 +β1   i.e.  β2 −β1 − 1 ≥ 0

This holds for every distribution with finite fourth moment, with equality only for a two-point (degenerate, Bernoulli-type) distribution. So statement 3 is true.

Pro Tip: Don't try to memorize this inequality as a bare fact — recognize the pattern instead. Any time an exam question pairs "Cov(Y, Y²)" with a Cauchy–Schwarz-style bound, it is almost always building toward the β2 ≥ β1 + 1 result. The way this derivation is usually taught is to treat it as "correlation-coefficient-squared ≤ 1" applied to Y and Y² instead of two unrelated variables — once that clicks, you can reconstruct the whole proof from scratch under exam pressure instead of trying to recall it verbatim.

Step 4: Checking Statement 2 Using Step 3

Now statement 2 is easy. We just showed β2 ≥ 1 +β1. Since β1 =μ32/μ23 is a ratio of a square to a positive quantity,β1 ≥ 0 always. Combining the two:

β2 ≥ 1 +β1 ≥ 1 + 0 = 1

So β2 is never less than 1 for any real distribution — it only dips down toward the boundary value of 1 for a symmetric two-point distribution (like a fair coin toss coded as 0/1), and is strictly greater than 1 for essentially every distribution encountered in practice, including all the standard continuous families (normal, exponential, binomial for non-degenerate parameters, etc.). Taking this in the sense the paper intends,statement 2 is true.

Step 5: Checking Statement 4 (Karl Pearson's Coefficient of Skewness)

Karl Pearson's coefficient of skewness is defined as:

Sk = 3(Mean − Median) / Standard Deviation   (or, using the mode version, Sk = (Mean − Mode) / Standard Deviation)

The standard theoretical result — the one tested here — is that this coefficient always lies between −3 and +3, not between −1 and +1. The bound of ±3 comes directly from the empirical relation Mode = 3 Median − 2 Mean combined with the fact that (Mean − Median) can be shown to never exceed the standard deviation in magnitude by more than a factor of one, giving the factor of 3 once you multiply through. Statement 4 understates this range, so as written it is incorrect.

So statement 4 is false.

Final Answer

Putting it together: statement 1 is false, statement 2 is true, statement 3 is true, and statement 4 is false. That means only statements 2 and 3 are correct, which is option (b).

This matches the official ISS 2016 answer key, which also lists (b)for Question 35.

Why This Question Matters

This question is a good stress-test of whether you understand β1 and β2 as derived quantities rather than as formulas to be recalled by rote. Questions phrased as "which statements are correct" are common in ISS Paper-1 precisely because they can't be answered by pattern-matching a formula — you have to know which direction an inequality runs, and why. Students who drill through moment-based proofs like the Cauchy–Schwarz argument above until the steps feel automatic tend to handle these multi-statement questions much faster than those who only memorize the final inequalities, because they can rebuild whichever piece they've forgotten on the spot.

Frequently Asked Questions

What is the difference between β1 and β2 in statistics?

β1 =μ32/μ23 measures skewness, or how lopsided a distribution is; since it involves a squared term, it is always non-negative and tells you nothing about the direction of skew (for that you need the signed measure μ3/μ23/2).β2 =μ4/μ22 measures kurtosis, or how heavy the tails and how peaked the centre of a distribution are relative to a normal curve.

Why is β2 always at least 1?

Because β2 ≥ 1 + β1, and β1 itself is a square divided by a positive number, so it can never be negative. Adding a non-negative quantity to 1 can never take you below 1, so β2 is bounded below by 1 for every distribution with a finite fourth moment.

Does the arithmetic mean equal the harmonic mean for any distribution?

Yes, but only in the degenerate case where every single observation is identical. The moment any two values in the data differ, the strict inequality AM > HM holds, which is why statement 1 in this question is false in general.

What is the normal distribution's value of β2?

For the normal distribution, β2 = 3 exactly, which is why "excess kurtosis" is often defined as β2 − 3, giving zero for the normal distribution, positive for heavy-tailed (leptokurtic) distributions, and negative for light-tailed (platykurtic) ones.

Why does Karl Pearson's coefficient of skewness use the factor 3?

The factor of 3 comes from the approximate empirical relationship between the mean, median and mode: Mode ≈ 3 Median − 2 Mean. Substituting this relationship into the mode-based skewness formula and simplifying produces the median-based formula with the leading 3, and it is this same factor of 3 that sets the theoretical bound of the coefficient at ±3 rather than ±1.

Is this kind of "which statements are correct" question common in the ISS Paper-1 exam?

Yes, statement-based multiple-choice questions built around standard inequalities and identities appear frequently across the years of ISS Statistics Paper-1, and they tend to reward candidates who can re-derive a result on the spot over those who have only memorized the final formula.

If any step above felt rushed or you'd solve statement 3 differently, drop a comment below — and if you found this walkthrough useful, do share it with a fellow ISS aspirant who's working through the same paper.