ISS 2016 Statistics Paper-1 Solution: Question 14 (Covariance in Conditional Normal Setup)

Continuing our question-by-question walk through the ISS (Indian Statistical Service) Statistics Paper-1 previous year papers, we now arrive at Question 14 from the 2016 paper. This one looks intimidating at first glance because it mixes conditional distributions with covariance, but once you break it into small pieces, the algebra is actually very short.

Quick Summary

• Topic:Covariance using conditional normal distributions and the law of total expectation

• Question Reference:ISS 2016, Statistics Paper-1, Question 14

• Correct Answer:Option (c), ρσ2

The Question

Suppose X follows the N(0, σ2) distribution and conditional on X = x, Y has the N(ρx, σ2) distribution. What is Cov(X, Y) equal to?

• (a) ρ

• (b) ρ2

• (c) ρσ2

• (d) ρ2σ2

Step 1: Understand What "Conditional on X = x" Means

The question gives us two pieces of information, and it is important to keep them separate in your head before doing any algebra.

First, the marginal distribution of X is Normal with mean 0 and variance σ2. In symbols, X ~ N(0, σ2). This tells us E[X] = 0 and Var(X) = σ2.

Second, we are told that if we already know the value of X, say X = x, then Y behaves like a Normal random variable with mean ρx and the same variance σ2. In symbols, (Y | X = x) ~ N(ρx, σ2). This is a conditional distribution — it describes Y only after X has been fixed at a particular value x.

From this conditional distribution, two facts follow immediately, and these are the only two facts we will need:

• E[Y | X = x] = ρx (the mean of the conditional Normal distribution)

• More generally, written as a random variable rather than plugging in a fixed x, E[Y | X] = ρX

Step 2: Recall the Formula for Covariance

By definition, for any two random variables X and Y,

Cov(X, Y) = E[XY] − E[X]·E[Y]

So to find Cov(X, Y), we need three quantities: E[XY], E[X], and E[Y]. We already know E[X] = 0 from Step 1. We now need to find E[Y] and E[XY].

Step 3: Find E[Y] Using the Law of Total Expectation

Since we are not told the marginal distribution of Y directly, we find it using the law of total expectation (also called the tower property), which says:

E[Y] = E[ E[Y | X] ]

In plain words: to get the overall average of Y, first average Y assuming X is fixed at some value (this gives you a formula in terms of X), and then average that formula over all possible values of X.

We already found E[Y | X] = ρX in Step 1. So substitute that in:

E[Y] = E[ρX] = ρ·E[X]

Since E[X] = 0 (X is Normal with mean 0), we get:

E[Y] = ρ · 0 = 0

Step 4: Find E[XY] Using the Tower Property Again

This is the step where most students get stuck, so we will go slowly. We want E[XY]. Apply the law of total expectation again, but this time condition the whole product XY on X:

E[XY] = E[ E[XY | X] ]

Now here is the key trick: once we are conditioning on X, the value of X is "known" — it behaves like an ordinary constant inside the inner expectation, because there is nothing random left in X once we've fixed it. So X can be pulled straight out of the inner expectation:

E[XY | X] = X · E[Y | X]

We already know E[Y | X] = ρX from Step 1. Substituting:

E[XY | X] = X · ρX = ρX2

Now take the outer expectation over X:

E[XY] = E[ρX2] = ρ·E[X2]

To finish, we need E[X2]. Recall the basic identity relating variance and the second moment:

Var(X) = E[X2] − (E[X])2

We know Var(X) = σ2 and E[X] = 0, so (E[X])2 = 0. Rearranging the identity:

E[X2] = Var(X) + (E[X])2 = σ2 + 0 = σ2

Substituting back:

E[XY] = ρ·σ2 = ρσ2

Pro Tip: The trickiest part of this question is realizing that E[XY | X] simplifies to X·E[Y | X], not to E[X]·E[Y | X] or anything more complicated. Once you're conditioning on X, treat X exactly like a known constant — it can be pulled outside any expectation, just like a number would be. This "condition, then treat as constant, then take the outer expectation" pattern is one of the most frequently tested tricks in ISS and other statistical services exams, and the way it's usually taught is to practice pulling the conditioning variable out of at least half a dozen different expressions until it becomes automatic rather than something you have to re-derive from scratch under time pressure.

Step 5: Put It All Together and Confirm the Final Answer

Now we simply plug everything back into the covariance formula from Step 2:

Cov(X, Y) = E[XY] − E[X]·E[Y]

We found E[XY] = ρσ2, E[X] = 0, and E[Y] = 0. So:

Cov(X, Y) = ρσ2 − (0)(0) = ρσ2

This matches option (c) ρσ2, which is also exactly what the official ISS 2016 Paper-1 answer key lists for Question 14. So our independent derivation and the official answer key are in full agreement: the correct answer is (c) ρσ2.

Why This Question Matters

This question is a compact test of three ideas that show up constantly in the ISS Statistics syllabus: reading a conditional distribution correctly, applying the law of total expectation (tower property) to compute unconditional moments, and connecting variance to the second moment formula. Regression-type setups — where Y's conditional mean is a linear function of X, as it is here with E[Y|X] = ρX — are the backbone of correlation and simple linear regression theory, so getting comfortable finding covariances this way pays off far beyond this one question.

Frequently Asked Questions

Why is E[X] equal to 0 in this problem?

Because the question states that X follows the N(0, σ2) distribution, and for any Normal distribution N(μ, variance), the mean is simply μ. Here μ = 0, so E[X] = 0 directly from the definition — no calculation is needed.

What exactly is the law of total expectation, and why does it work?

The law of total expectation states that E[Y] = E[E[Y|X]]. It works because E[Y|X] is itself a random variable (it depends on the random value of X), and averaging that random variable over all values of X gives back the same answer as directly averaging Y. It is one of the most useful tools for computing moments of a variable when only its conditional distribution is given.

Could Cov(X, Y) have been negative here?

Yes, if ρ were negative. Since σ2 is always positive (it's a variance), the sign of Cov(X, Y) = ρσ2 is entirely determined by the sign of ρ. A negative ρ would mean X and Y tend to move in opposite directions, while a positive ρ means they tend to move together.

Why can't we just say Cov(X, Y) = 0 since E[Y | X = x] depends on x?

Cov(X, Y) = 0 only happens when X and Y are independent or uncorrelated. Here Y's conditional mean explicitly depends on X (through ρx), which means knowing X changes what you'd predict for Y — so X and Y are correlated, and the covariance works out to the nonzero value ρσ2, not 0.

Is there a shortcut that avoids the tower property entirely?

Not really a full shortcut, but you can speed things up by recognizing this exact structure — Y = ρX + (independent noise with variance σ2(1−ρ2) or similar) — as the standard bivariate Normal regression setup, where Cov(X,Y) = ρ·σX·σY is a known result. But for exam purposes, deriving it step by step as we did above is safer and leaves no room for misremembering a formula.

What if I get confused between Var(X) and E[X²]?

Remember they are only equal when E[X] = 0. In general, Var(X) = E[X2] − (E[X])2, so E[X2] = Var(X) + (E[X])2. Whenever E[X] = 0, as in this question, the two quantities coincide, but it's worth writing out the full identity rather than assuming they're always the same.

If any step above felt rushed or you'd solve it a different way, drop a comment below — working through the disagreement is often where the real learning happens. And if you know a fellow ISS aspirant who's grinding through the same previous year papers, feel free to share this post with them too.