ISS 2016 Statistics Paper-1 Solution: Question 22 (Difference of Lognormal Logs)

Continuing our question-by-question walkthrough of the ISS Statistics Paper-1 previous year papers, today we pick up right where the last post left off and tackle Question 22 from the 2016 paper. This one sits in the probability distributions section and tests whether you can confidently move between a lognormal variable and its underlying normal variable.

Quick Summary

• Topic: Lognormal distribution, transformation to normal, variance of a difference of independent variables

• Question Reference: ISS 2016, Statistics Paper-1, Question 22

• Correct Answer: Option (b) — Mean = 0, Variance = 2σ²

The Question

Let X and Y be independently and identically distributed as lognormal with parameters μ and σ². What are the mean and variance of log(X/Y) respectively?

• (a) 0 and σ²

• (b) 0 and 2σ²

• (c) 2μ and σ²

• (d) 2μ and 2σ²

Step 1: Understand what "lognormal with parameters μ and σ²" actually means

This is the part that trips up a lot of students right at the start, so let's be very explicit. A random variable X is said to follow a lognormal distribution with parameters μ and σ²if, when you take its natural logarithm, the result follows a normal distribution with mean μ and variance σ². In symbols:

If X ~ Lognormal(μ, σ²), then ln(X) ~ N(μ, σ²).

So μ and σ² are NOT the mean and variance of X itself — they are the mean and variance of ln(X). This distinction is the entire key to solving this question quickly.

Step 2: Convert both X and Y into their normal counterparts

We are told X and Y are independent and identically distributed as Lognormal(μ, σ²). Define two new random variables:

U = ln(X) and V = ln(Y)

By the definition in Step 1, since X and Y are each Lognormal(μ, σ²), it follows directly that:

U ~ N(μ, σ²) and V ~ N(μ, σ²)

Also, since X and Y are independent of each other, any function of X (like ln X) is independent of any function of Y (like ln Y). So U and V are independent, identically distributed normal random variables, each with mean μ and variance σ².

Step 3: Rewrite log(X/Y) in terms of U and V

Recall the basic logarithm rule: log of a ratio is the difference of the logs.

log(X/Y) = log(X) − log(Y) = U − V

So the quantity we actually need to analyze is simply U − V, the difference of two independent normal random variables.

Step 4: Find the mean of U − V

Expectation is linear, so it distributes over subtraction regardless of whether the variables are independent or not:

E(U − V) = E(U) − E(V) = μ − μ = 0

So the mean of log(X/Y) is 0. This already rules out options (c) and (d), both of which claim the mean is 2μ.

Step 5: Find the variance of U − V (the step everyone must get right)

This is where students most often lose marks, because it feels natural to assume that if the mean subtracts, the variance should subtract too. It does not. For any two independent random variables U and V:

Var(U − V) = Var(U) + Var(−V) = Var(U) + Var(V)

The reason the minus sign disappears is that variance measures squared spread — multiplying a variable by −1 does not shrink or invert its spread, it only flips its direction around the mean. Squaring removes the sign, so Var(−V) = (−1)² Var(V) = Var(V). Since U and V are independent, there is no covariance term to worry about, and the variances simply add.

Substituting in Var(U) = σ² and Var(V) = σ²:

Var(U − V) = σ² + σ² = 2σ²

Pro Tip: Whenever a question asks for the variance of a difference of two independent random variables, resist the urge to subtract. Independence only kills the covariance term in Var(U − V) = Var(U) + Var(V) − 2·Cov(U,V); it does not change the sign in front of Var(V). The way this particular trap is usually taught to drill it out of you permanently: physically write out Var(U − V) = Var(U + (−V)) first, then apply Var(aW) = a²Var(W) to the (−1)·V term. Once you see the (−1)² become +1, you will never make this sign mistake again.

Step 6: Put it together and match with the options

We have shown that log(X/Y) = U − V is a linear combination of two independent normal random variables, and any linear combination of independent normal variables is itself normal. So:

log(X/Y) ~ N(0, 2σ²)

This means the mean of log(X/Y) is 0 and its variance is 2σ², which is exactly option (b). Cross-checking against the official ISS 2016 Statistics Paper-1 answer key, the marked correct option for Question 22 is indeed (b) — so our derivation matches the official key with no discrepancy.

Why This Question Matters

This question is a compact test of three separate skills stacked on top of each other: knowing the precise definition of the lognormal distribution (rather than confusing μ, σ² with the mean and variance of X itself), knowing that sums and differences of independent normal variables stay normal, and — most importantly — correctly computing the variance of a difference. In the actual ISS Paper-1 exam, this combination shows up repeatedly in disguised forms across probability distribution questions, so getting comfortable with this exact chain of reasoning pays off well beyond this one question. Students preparing seriously for this paper tend to build the habit of writing out every transformation explicitly — defining U and V, stating their distributions, and only then combining them — rather than trying to shortcut the algebra in their heads, and that habit is worth building early rather than under exam pressure.

Frequently Asked Questions

What is the difference between the parameters of a lognormal distribution and its actual mean and variance?

The parameters μ and σ² of a Lognormal(μ, σ²) distribution are the mean and variance of ln(X), not of X itself. The actual mean of X is e^(μ + σ²/2) and the actual variance of X is (e^(σ²) − 1)·e^(2μ + σ²), which are more complicated expressions — but this question never needed those formulas because it asked about log(X/Y), which brings us straight back to the normal world.

Why does Var(U − V) equal Var(U) + Var(V) instead of Var(U) − Var(V)?

Variance is always non-negative because it involves squaring deviations from the mean, so a negative sign in front of a random variable cannot reduce its variance. When you expand Var(U − V) using the general formula Var(U) + Var(V) − 2Cov(U,V), independence makes the covariance term zero, leaving Var(U) + Var(V) added together, never subtracted.

Is log(X/Y) itself lognormal, or is it normal?

log(X/Y) is normal, not lognormal. The ratio X/Y itself would be lognormal (since it is built from exponentials of a normal variable), but the moment you take a further logarithm of that ratio, you land back on a normal random variable, specifically N(0, 2σ²) here.

Would the answer change if X and Y had different parameters, say Lognormal(μ₁, σ₁²) and Lognormal(μ₂, σ₂²)?

Yes. In that more general case, log(X/Y) would be N(μ₁ − μ₂, σ₁² + σ₂²) using the same logic — means subtract, and variances of independent components add. This question is a special case where μ₁ = μ₂ = μ and σ₁² = σ₂² = σ² make the mean collapse to exactly 0.

How would this change if the question had asked about log(XY) instead of log(X/Y)?

log(XY) = log(X) + log(Y) = U + V. The mean would become E(U) + E(V) = 2μ, and because addition of independent variables also has no covariance term, the variance would still be Var(U) + Var(V) = 2σ². So log(XY) ~ N(2μ, 2σ²) — notice the variance answer stays the same either way, only the mean changes sign in effect.

Do I need to assume independence for the variance formula to work this way?

Yes, independence is essential here. If X and Y were correlated, you would need the full formula Var(U − V) = Var(U) + Var(V) − 2Cov(U,V), and the covariance term would generally not be zero, changing the final answer away from a clean 2σ².

If any step above felt unclear, drop a comment below and we will walk through it in more detail — and if you found this useful, do share it with a fellow ISS aspirant who might be revising the same topic this week.