ISS 2016 Statistics Paper-1 Solution: Question 44 (Covariance of Linear Combinations)

This post continues our ongoing question-by-question walkthrough of the ISS Statistics Paper-1 previous year papers. Today we take up Question 44 from the 2016 paper, which tests how comfortable you are with the bilinearity property of covariance when two new variables are built as linear combinations of three original ones. As always, every algebraic step is shown in full so that even a first-time reader can follow along without skipping anything.

Quick Summary

• Topic:Covariance of linear combinations of random variables (properties of variance and covariance)

• Question Reference:ISS 2016, Statistics Paper-1, Question 44

• Correct Answer:Option (c) −82

The Question

If the random variables X, Y and Z have the means μX = 5, μY = 7 and μZ = 4; variances σ2X = 10, σ2Y = 14 and σ2Z = 20; Cov(X,Y) = 1, Cov(X,Z) = −3 and Cov(Y,Z) = 2, then what is the covariance of U = X + 4Y + 2Z and V = 3X − Y − Z?

• (a) −76

• (b) 82

• (c) −82

• (d) 76

Step 1: Notice That the Means Are a Distraction

The first thing a novice solver often does is panic at seeing μX, μY and μZ listed in the question and try to use them somewhere. But covariance, by definition, is a measure of how two variables move together around their own means — and that ‘moving together’ behavior does not change if you simply shift every value of a variable by a constant. In formula terms, for any constants a and b, Cov(X + a, Y + b) = Cov(X, Y). So the individual means μX = 5, μY = 7 and μZ = 4 play absolutely no role in computing Cov(U, V). They are given only to test whether you know which quantities actually matter. The numbers you truly need are the variances and the pairwise covariances.

Step 2: Write Down the Bilinearity Property of Covariance

Covariance behaves like a ‘multiplication’ operation that distributes over addition, in the same way ordinary multiplication distributes over a sum like (a + b)(c + d). Formally, for any constants and any random variables, covariance is bilinear: it is linear in its first argument and linear in its second argument, separately. This means:

Cov(aX + bY + cZ, dX + eY + fZ) = ad·Cov(X,X) + ae·Cov(X,Y) + af·Cov(X,Z) + bd·Cov(Y,X) + be·Cov(Y,Y) + bf·Cov(Y,Z) + cd·Cov(Z,X) + ce·Cov(Z,Y) + cf·Cov(Z,Z)

Two small facts make this manageable: first, Cov(X,X) is just Var(X), and similarly for Y and Z. Second, covariance is symmetric, so Cov(Y,X) = Cov(X,Y), Cov(Z,X) = Cov(X,Z), and Cov(Z,Y) = Cov(Y,Z). Once you accept these two facts, the whole nine-term expansion collapses into something very manageable.

Step 3: Expand Cov(U, V) Term by Term

Here U = X + 4Y + 2Z, so the coefficients of X, Y, Z in U are a = 1, b = 4, c = 2.

And V = 3X − Y − Z, so the coefficients of X, Y, Z in V are d = 3, e = −1, f = −1.

Rather than memorizing the nine-term formula, it is far safer for a beginner to expand it mechanically, exactly like multiplying out two binomial-style expressions, treating Cov as an operation that distributes like ordinary multiplication:

Cov(U, V) = Cov(X, 3X) + Cov(X, −Y) + Cov(X, −Z) + Cov(4Y, 3X) + Cov(4Y, −Y) + Cov(4Y, −Z) + Cov(2Z, 3X) + Cov(2Z, −Y) + Cov(2Z, −Z)

Now pull the constants outside each covariance (this is exactly the linearity part of bilinearity):

Cov(U, V) = 3·Cov(X,X) − Cov(X,Y) − Cov(X,Z) + 12·Cov(Y,X) − 4·Cov(Y,Y) − 4·Cov(Y,Z) + 6·Cov(Z,X) − 2·Cov(Z,Y) − 2·Cov(Z,Z)

Step 4: Substitute Var(X,X) = Var(X) and Use Symmetry

Replace Cov(X,X) with Var(X) = 10, Cov(Y,Y) with Var(Y) = 14, and Cov(Z,Z) with Var(Z) = 20. Also replace Cov(Y,X) with Cov(X,Y) = 1, Cov(Z,X) with Cov(X,Z) = −3, and Cov(Z,Y) with Cov(Y,Z) = 2. The expression becomes:

Cov(U, V) = 3(10) − (1) − (−3) + 12(1) − 4(14) − 4(2) + 6(−3) − 2(2) − 2(20)

Step 5: Group Like Terms Before Adding Numbers

It helps enormously to first collect all the terms that belong to the same quantity (Var(X), Var(Y), Var(Z), Cov(X,Y), Cov(X,Z), Cov(Y,Z)) rather than adding numbers in the order they appear. This avoids silly arithmetic slips.

• Var(X) terms: only 3·Var(X) = 3(10) = 30

• Var(Y) terms: only −4·Var(Y) = −4(14) = −56

• Var(Z) terms: only −2·Var(Z) = −2(20) = −40

• Cov(X,Y) terms: −1·Cov(X,Y) from the first group + 12·Cov(X,Y) from the second group = 11·Cov(X,Y) = 11(1) = 11

• Cov(X,Z) terms: +1·Cov(X,Z) (since −Cov(X,−Z) = +Cov(X,Z)) + 6·Cov(X,Z) = 5·Cov(X,Z) = 5(−3) = −15

• Cov(Y,Z) terms: −4·Cov(Y,Z) − 2·Cov(Y,Z) = −6·Cov(Y,Z) = −6(2) = −12

Pro Tip: The single most common place students lose marks on this type of question is sign-tracking across two different variables simultaneously — here X appears with coefficient +1 in U but the Cov(X,Z) term picks up a sign flip because Z appears with coefficient −1 in V. The safest habit, the one that is drilled repeatedly until it becomes automatic in a structured classroom setting, is to never combine the ‘collect coefficients’ step with the ‘multiply numbers’ step. Do them as two completely separate passes: first write out every one of the nine Cov/Var terms symbolically with its sign, and only in a second pass substitute the numerical values. Trying to do both at once under exam time pressure is exactly where a sign error creeps in.

Step 6: Add Everything Up

Now simply sum the six grouped results:

Cov(U, V) = 30 − 56 − 40 + 11 − 15 − 12

Adding step by step: 30 − 56 = −26. Then −26 − 40 = −66. Then −66 + 11 = −55. Then −55 − 15 = −70. Finally −70 − 12 = −82.

So Cov(U, V) =−82, which is option (c).

This matches the official ISS 2016 answer key, which also lists option (c) as the correct response for Question 44. Our independent derivation and the official key are in full agreement.

Why This Question Matters

Questions like this one are a staple of ISS Paper-1 because they check, in a single compact item, whether you truly understand covariance as a bilinear operation rather than just a formula you plug numbers into. The same bilinearity idea underlies far more advanced topics later in the syllabus — the variance of a sum of correlated estimators, the covariance structure in multiple regression, and the derivation of variance-covariance matrices for linear combinations of random vectors. Getting completely comfortable with expanding Cov(aX + bY, cX + dY) by hand, the way it is usually taught, pays off repeatedly across the rest of the statistics paper, not just on isolated probability questions.

Frequently Asked Questions

Why don't the individual means of X, Y and Z affect the answer?

Covariance measures joint variability around each variable's own mean, and this measure is unaffected by shifting a variable up or down by a constant. Since U and V are linear combinations with no added constant terms, the means given in the question are not needed for this particular calculation; they would only matter if the question had asked for something like E(U) or E(UV) instead.

Is there a shortcut formula instead of expanding all nine terms?

Yes, once you are confident with the expansion, you can jump straight to Cov(U,V) = (coefficient of X in U)(coefficient of X in V)Var(X) plus similar terms for Y and Z, plus cross terms for each pair using the combined coefficient on each covariance. But until that pattern feels automatic, writing out all nine terms explicitly, as done above, is the safer and more accurate route, especially under exam conditions.

What if a variable had appeared with a coefficient of zero in one of U or V?

Any term involving that variable would simply vanish, since multiplying by zero eliminates it from both the Var and Cov contributions. The remaining structure of the calculation stays exactly the same.

How is this different from finding Var(U) or Var(V) individually?

Var(U) is just the special case Cov(U, U), so the same nine-term (or in that case, six distinct-term, since cross terms repeat) expansion applies, but with U substituted in both argument slots. Finding Cov(U, V) for two different linear combinations, as here, requires keeping track of two separate sets of coefficients simultaneously, which is why sign errors are more common in this version of the question.

Does the order of X, Y, Z in the covariance matter, i.e., is Cov(X,Z) the same as Cov(Z,X)?

Yes, covariance is always symmetric: Cov(X,Z) = Cov(Z,X) for any two random variables. This symmetry is exactly what let us combine the cross terms for each pair into a single coefficient during Step 5 above.

Could this same approach be used for more than three random variables?

Absolutely. The bilinearity property generalizes to any number of variables: Cov of two linear combinations of n random variables expands into up to n2 terms, which again collapse using Cov(Xi, Xi) = Var(Xi) and the symmetry of cross-covariances. This is exactly the logic behind variance-covariance matrices used later in multivariate statistics.

If anything in this derivation felt unclear, or you solved it a different way and want a second opinion, drop a comment below — and if you found this walkthrough useful, do share it with a fellow ISS aspirant who might be stuck on the same question.