ISS 2016 Statistics Paper-1 Solution: Question 42 (Correlation Between X² and Y²)

Continuing our question-by-question walk through the ISS Statistics Paper-1 archive, today we tackle a question that looks intimidating at first glance but turns into a clean, mechanical calculation once you set it up the right way. This is Question 42 from the 2016 paper, and it tests whether you really understand what a bivariate normal distribution is telling you, not just whether you've memorized a formula.

Quick Summary

• Topic:Bivariate normal distribution – correlation between functions of correlated normal variables

• Question Reference:ISS 2016, Statistics Paper-1, Question 42

• Correct Answer:Option (b) – ρ²

The Question, As Asked

If X and Y are standard normal variates with correlation coefficient ρ between them, then what is the correlation coefficient between X² and Y²?

• (a) 2ρ − 1

• (b) ρ²

• (c) ρ

• (d) √ρ

Step 1: Understand What We're Actually Given

We're told X and Y are standard normal variates, which means E(X) = 0, E(Y) = 0, Var(X) = 1, Var(Y) = 1. They also have correlation ρ between them, and because nothing else is said, we treat (X, Y) as following a standard bivariate normal distribution with parameters (0, 0, 1, 1, ρ).

We are asked for Corr(X², Y²), not Corr(X, Y). This is a different, harder question, because squaring a variable changes its distribution completely — X² is no longer normal, it follows a chi-square distribution with 1 degree of freedom. So we cannot just reuse the formula for the correlation of X and Y; we have to build the answer from scratch using moments.

Recall the general formula for a correlation coefficient between any two random variables U and V:

Corr(U, V) = Cov(U, V) / √(Var(U)·Var(V))

Here U = X² and V = Y². So we need three things: Var(X²), Var(Y²), and Cov(X², Y²).

Step 2: Find Var(X²) Using Standard Normal Moments

Var(X²) = E(X⁴) − [E(X²)]²

For a standard normal variable, the even-order moments follow the pattern E(X⁴) = 3, E(X²) = 1 (these come from the standard normal moment formula E(X⁶ⁿ) = (2n − 1)!! for even powers; for n = 2 this gives 3!! = 3 × 1 = 3).

So: Var(X²) = 3 − (1)² = 3 − 1 = 2

By exactly the same logic, since Y is also standard normal: Var(Y²) = 2

This also matches a fact you may already know: if X is standard normal, X² follows a chi-square distribution with 1 degree of freedom, and the variance of a chi-square variable with k degrees of freedom is always 2k. Here k = 1, so variance = 2 × 1 = 2. Same answer, two routes – always a good sign.

Step 3: The Key Trick – Rewrite Y in Terms of X

This is the step that makes the whole problem manageable. When (X, Y) is standard bivariate normal with correlation ρ, we can always write Y as a linear combination of X and an independent standard normal variable Z:

Y = ρX + √(1 − ρ²)·Z, where Z ~ N(0, 1) and Z is independent of X.

Why does this work? Because conditional on X = x, Y is normally distributed with mean ρx and variance (1 − ρ²) – this is the standard conditional-distribution result for bivariate normal pairs. Writing Y this way is just another way of expressing that same conditional relationship, and it lets us compute E(X²Y²) using only moments of independent standard normal variables, which we already know how to handle.

Step 4: Compute E(X²Y²)

Square the expression for Y first:

Y² = ρ²X² + 2ρ√(1 − ρ²)·XZ + (1 − ρ²)Z²

Now multiply through by X² and take expectation term by term:

E(X²Y²) = ρ²·E(X⁴) + 2ρ√(1 − ρ²)·E(X³Z) + (1 − ρ²)·E(X²Z²)

Evaluate each piece:

• E(X⁴) = 3, as found above.

• E(X³Z) = E(X³)·E(Z), because Z is independent of X. E(X³) = 0 for a standard normal variable (all odd moments of a symmetric distribution are zero), so this entire middle term vanishes – it is 0 regardless of E(Z).

• E(X²Z²) = E(X²)·E(Z²), again by independence, which equals 1 × 1 = 1.

Substituting back:

E(X²Y²) = ρ²(3) + 2ρ√(1 − ρ²)(0) + (1 − ρ²)(1) = 3ρ² + 1 − ρ² = 1 + 2ρ²

Pro Tip: The term most students get stuck on is E(X³Z). It's tempting to start computing E(X³) and E(Z) separately and worry about signs, but the real insight is simpler – X and Z are independent, so the expectation of their product factors automatically, and since E(X³) = 0 by symmetry, the entire term disappears without any further work. Training yourself to spot "this factors, and one factor is an odd moment of a symmetric variable, so it's zero" is exactly the kind of pattern-recognition that turns a five-minute problem into a thirty-second one in the exam hall – it's the sort of shortcut that gets drilled repeatedly when this topic is taught properly, until it becomes automatic rather than something you have to rediscover under time pressure.

Step 5: Compute Cov(X², Y²) and Finish

Cov(X², Y²) = E(X²Y²) − E(X²)·E(Y²) = (1 + 2ρ²) − (1)(1) = 2ρ²

Now plug everything into the correlation formula:

Corr(X², Y²) = Cov(X², Y²) / √(Var(X²)·Var(Y²)) = 2ρ² / √(2 × 2) = 2ρ² / 2 = ρ²

So the correlation coefficient between X² and Y² is simply ρ², which is option (b).

This matches the official answer key exactly – Series A of the 2016 answer key lists option (b) for Question 42, so our derivation and the official key agree. No discrepancy to flag here.

Sanity Check: Does This Make Sense?

A quick check is always worth doing. If ρ = 0 (X and Y independent), the formula gives Corr(X², Y²) = 0, which makes sense – independent variables should give independent (hence uncorrelated) squares too. If ρ = 1 (X and Y are literally the same variable), the formula gives Corr(X², Y²) = 1, which also makes sense, since X² and Y² would then be identical. Both edge cases behave exactly as intuition demands, which is good confirmation that 1 + 2ρ² and the final ρ² result are correct rather than an algebra slip.

Why This Question Matters

Questions like this one are a favourite in ISS and similar exams because they test three layers of understanding at once: knowing the structural representation of a bivariate normal pair, being comfortable manipulating expectations of products of independent variables, and recalling the standard normal moment sequence (0, 1, 0, 3, 0, 15, …). Students who only memorize "Corr(X,Y) = ρ" as a standalone fact get stuck the moment a question asks about a transformed version of X and Y. Practicing the regression-representation trick on a handful of similar problems – the way it's usually taught at Sunrise Classes and similar coaching setups – is the fastest way to make this kind of question routine rather than a surprise on exam day.

Frequently Asked Questions

Why can we write Y = ρX + √(1 − ρ²)Z instead of just working with the joint density directly?

This representation is mathematically equivalent to the joint density of a standard bivariate normal pair, but it converts a two-variable integration problem into simple algebra on independent normal variables. Since X and Z are independent, we can use the product rule for expectations freely, which is far easier than integrating the bivariate normal density directly. It's one of the most useful tools for this entire topic.

Is Corr(X², Y²) = ρ² true for any two correlated random variables, or only for normal ones?

This exact result, ρ², is specific to the bivariate normal case. For other joint distributions the relationship between Corr(X,Y) and Corr(X²,Y²) can look completely different, because it depends on the higher-order moments and the specific dependence structure of that distribution, not just on ρ.

Why is E(X³) = 0 for a standard normal variable?

The standard normal density is symmetric about zero, meaning f(x) = f(−x). Any odd power of X, like X³ or X&sup5;, is an odd function, and the expectation of an odd function under a symmetric density always integrates to exactly zero – the positive and negative contributions cancel perfectly.

How do we know E(X⁴) = 3 for a standard normal variable without redoing the integral every time?

Standard normal moments follow a known recursive pattern: E(X⁶ⁿ) for even n equals the double factorial (n − 1)!!, which is the product of all odd numbers down to 1. For the 4th moment, n = 4, so (4 − 1)!! = 3!! = 3 × 1 = 3. Memorizing this short sequence – 1, 3, 15, 105, … for the 2nd, 4th, 6th, 8th moments – saves significant time in the exam.

Could this question have been solved using the chi-square distribution instead?

Partly, yes. Since X² and Y² each individually follow a chi-square distribution with 1 degree of freedom, you could use chi-square variance formulas to get Var(X²) = Var(Y²) = 2 directly. However, the covariance term still requires knowing the joint behaviour of X and Y together, which the chi-square marginal distributions alone don't capture – so the regression-representation approach is still needed for that part.

What's the most common mistake students make on this type of question?

The most frequent error is assuming Corr(X², Y²) equals Corr(X, Y), i.e. picking option (c), ρ, without doing any calculation. Squaring is a nonlinear transformation, and correlation is not generally preserved under nonlinear transformations, so this shortcut is incorrect here even though it feels intuitive.

If anything in this derivation wasn't clear, or you spotted a step you'd solve differently, drop a comment below – and if you know a fellow ISS aspirant working through the same paper, consider sharing this with them too.