ISS 2016 Statistics Paper-1 Solution: Question 45 (Mean vs Variance Comparison)

Continuing our question-by-question walk through the ISS Statistics Paper-1 previous year papers, we now arrive at Question 45 of the 2016 paper. This one tests something every statistics student eventually has to memorise cold: how the mean and variance of a distribution relate to each other — and whether they are ever equal.

Quick Summary

• Topic: Comparing mean and variance across Poisson, Binomial, and Chi-square distributions

• Question Reference: ISS 2016, Statistics Paper-1, Question 45

• Correct Answer: (c) 1 and 3 only

The Question

Consider the following statements:

1. Mean and variance are equal for Poisson distribution.

2. Mean is less than variance for Binomial distribution.

3. Mean is less than variance for Chi-square distribution.

Which of the above statements are correct?

• (a) 1 and 2 only

• (b) 2 and 3 only

• (c) 1 and 3 only

• (d) 1, 2 and 3

Step 1: Check the Poisson Distribution (Statement 1)

A Poisson random variable X with parameter λ has the probability mass function

P(X = x) = e−λλx / x!,   x = 0, 1, 2, …

To find the mean, we compute E(X) directly from the definition:

E(X) = ∑x=0∞ x · e−λλx/x!

The x = 0 term vanishes (it is multiplied by x = 0), so we start the sum from x = 1. For x ≥ 1, x/x! = 1/(x−1)!, so we can rewrite each term and pull out one factor of λ:

E(X) = λ e−λ ∑x=1∞ λx−1/(x−1)! = λ e−λ · eλ = λ

(The remaining sum is just the full Poisson series re-indexed, which adds up to eλ.)

For the variance, the easiest route is through the factorial moment E[X(X−1)], which by the same index-shifting trick (now pulling out λ2) works out to λ2. Then:

E(X2) = E[X(X−1)] + E(X) = λ2 + λ

Var(X) = E(X2) − [E(X)]2 = (λ2 + λ) − λ2 = λ

So Mean = Variance = λ for a Poisson distribution.Statement 1 is correct.

Step 2: Check the Binomial Distribution (Statement 2)

Let X ~ Binomial(n, p), with q = 1 − p. The standard results (which you should know without re-deriving every time) are:

Mean = np    Variance = npq

Now compare them. Since 0 < p < 1, we also have 0 < q < 1. Multiplying np by a number strictly between 0 and 1 can only make it smaller, so:

npq < np, &nbsp; i.e. &nbsp; Variance < Mean

This is the opposite of what Statement 2 claims. The statement says mean is less than variance, but the algebra shows variance is always less than (or equal to, only in the degenerate case) the mean for a genuine binomial variable.Statement 2 is incorrect.

Pro Tip: Don't try to "remember" whether Binomial variance is bigger or smaller than its mean &mdash; derive it on the spot from Variance = Mean &times; q. Since q = 1 &minus; p is always strictly less than 1 for a non-trivial binomial variable, Variance = Mean &times; (something less than 1) must be smaller than the Mean. This one-line argument is faster and safer under exam pressure than trying to recall it as a fact, and it's exactly the kind of habit that's drilled in repeatedly when this topic is taught properly &mdash; derive the inequality from the formula instead of memorising the direction.

Step 3: Check the Chi-square Distribution (Statement 3)

A Chi-square variable with n degrees of freedom, written &chi;2n, can be built as the sum of n independent squared standard normal variables: &chi;2n = Z12 + Z22 + &hellip; + Zn2, where each Zi ~ N(0, 1).

For a single Z2: since E(Z) = 0 and Var(Z) = 1, we get E(Z2) = Var(Z) + [E(Z)]2 = 1. Its variance works out to Var(Z2) = E(Z4) &minus; [E(Z2)]2 = 3 &minus; 1 = 2 (using the standard normal's fourth moment E(Z4) = 3).

Since the n terms are independent, means and variances both add up directly across the sum:

Mean of &chi;2n = n &times; 1 = n

Variance of &chi;2n = n &times; 2 = 2n

Comparing the two: for any n > 0, 2n is always greater than n, so Variance > Mean, which means Mean < Variance.Statement 3 is correct.

Final Answer

Statement 1 is correct (Poisson: mean = variance). Statement 2 is incorrect (Binomial: variance is actually less than mean, not greater). Statement 3 is correct (Chi-square: variance exceeds mean).

So the correct combination is statements 1 and 3 only, which is option (c). This matches the official ISS 2016 answer key for Series A, which also lists (c) as the correct response for Question 45.

Why This Question Matters

Questions like this one aren't really testing whether you can recall a single formula &mdash; they're testing whether you have a reliable, derivable mental map of how the mean and variance behave across the standard distributions. A candidate who has only memorised "Poisson: mean = variance" in isolation, without knowing why, often gets flustered by statement 2 and either guesses or wastes two minutes trying to recall the "rule" for binomial variance. Being able to rebuild Mean = np and Variance = npq = Mean &times; q from scratch, and then reason about the inequality algebraically, is a far more exam-safe skill, and it is exactly this kind of comparative, statement-based question that ISS Paper-1 leans on heavily.

Frequently Asked Questions

Why is Poisson the only common discrete distribution where mean equals variance?

It comes directly from how the Poisson distribution is derived as a limiting case of the Binomial, where n &rarr; &infin; and p &rarr; 0 such that np = &lambda; stays fixed. In that limit, q = 1 &minus; p &rarr; 1, so Variance = npq &rarr; np = Mean = &lambda;. The single parameter &lambda; ends up controlling both the centre and the spread simultaneously, which is a distinctive feature of the Poisson family.

Is there any value of p for which the Binomial variance equals the mean?

Only in the trivial limiting case p &rarr; 0 (or equivalently q &rarr; 1), which is exactly the Poisson limit described above. For any genuine binomial variable with 0 < p < 1, the variance npq is strictly less than the mean np, so equality never holds for an actual, non-degenerate binomial distribution.

How is the Chi-square distribution related to the Gamma distribution?

&chi;2n is a special case of the Gamma distribution with shape parameter n/2 and scale parameter 2. Using the general Gamma results Mean = (shape)(scale) and Variance = (shape)(scale)2, you instantly get Mean = (n/2)(2) = n and Variance = (n/2)(4) = 2n, which is a quicker route than summing n squared normal variables term by term if you already know the Gamma formulas.

Why does Chi-square variance grow faster than its mean?

Both quantities grow linearly in n, but the variance grows at twice the rate (2n versus n) because squaring a normal variable inflates its spread more than its average value. Each added degree of freedom contributes 1 to the mean but 2 to the variance, so the gap between them widens as n increases rather than staying fixed.

Do I need to memorise all these mean-variance formulas, or can I derive them in the exam?

You should know the standard formulas (Binomial, Poisson, Chi-square, Geometric, Exponential, Normal) by heart for speed, but you should also be comfortable re-deriving any of them from the moment generating function or first principles if your memory falters under pressure. In practice, students preparing for ISS and similar exams tend to drill a short mean-variance reference table until recalling it becomes automatic, which saves crucial time on comparison-type questions like this one.

Are "which statements are correct" questions like this common in ISS Paper-1?

Yes, this statement-based (1, 2, 3 with options combining them) format is one of the most frequently used question styles across ISS Statistics Paper-1, especially for topics involving distributional properties, inequalities, and relationships between estimators. Getting comfortable evaluating each numbered statement independently, rather than trying to intuit the answer, is a core exam skill.

If anything in this derivation felt unclear, or if you spotted a different way to approach it, drop a comment below &mdash; and if you found this walkthrough useful, consider sharing it with a fellow ISS aspirant who's working through the same paper.