ISS 2016 Statistics Paper-1 Solution: Question 1 (Joint Distribution of X and Y)

This is the first post in a new daily series solving ISS previous year Statistics questions one at a time, in order — starting with 2016 Paper-1, Question 1, and moving forward question by question, paper by paper, year by year. If you're preparing for the Indian Statistical Service (ISS) exam, this series is built to help you actually understand each question, not just memorize the answer.

Quick Summary

• Topic: Joint probability distributions and expectation

• Question: ISS 2016, Statistics Paper-1, Q1

• Skills tested: finding a marginal density, computing E(X), E(Y), and E(XY) from a joint density

• Correct answer: Option (a) — statements 1 and 2 only

The Question

Let (X, Y) be jointly distributed with density:

f(x, y) = e−y for 0 < x < y < ∞, and f(x, y) = 0 otherwise.

Consider the following statements:

1. E(X) = 1

2. E(Y) = 2

3. E(XY) = 2

Which of the above are correct?

• (a) 1 and 2 only

• (b) 2 and 3 only

• (c) 1 and 3 only

• (d) 1, 2 and 3

Step 1: Understand the Region Where the Density Is Non-Zero

Before touching any integral, always draw or picture the region first. Here, the density is non-zero only when 0 < x < y < ∞ — meaning x can range from 0 up to y, and y itself ranges from 0 to infinity. So for a fixed value of y, x runs from 0 to y.

Pro Tip: Whenever a joint density has a condition like 0 < x < y, always fix the outer variable (here, y) first and let the inner variable (x) run over its allowed range. Getting the integration limits right is where most students lose marks on this type of question — the formula itself is usually the easy part.

Step 2: Find E(Y) by First Finding the Marginal Density of Y

To find E(Y), it helps to first find the marginal density of Y, by integrating out x:

fY(y) = ∫ (from x = 0 to y) e−y dx

Since e−y does not depend on x, this integral is just e−y multiplied by the length of the interval, which is y:

fY(y) = y · e−y, for y > 0

This is a well-known form — it is the density of a Gamma distribution with shape parameter 2 and rate parameter 1. For a Gamma distribution with shape parameter k and rate parameter 1, the mean is simply k. Here k = 2, so:

E(Y) = 2

This matches statement 2 in the question, so statement 2 is true.

Step 3: Find E(X)

E(X) is found by integrating x times the joint density over the whole region:

E(X) = ∫ (y = 0 to ∞) ∫ (x = 0 to y) x · e−y dx dy

Do the inner integral (over x) first, treating y as fixed. The inner integral of x from 0 to y is a standard result: it equals y²/2. So:

E(X) = ∫ (y = 0 to ∞) e−y · (y²/2) dy = (1/2) ∫ (y = 0 to ∞) y² · e−y dy

The integral ∫ y² e−y dy from 0 to ∞ is a standard Gamma-function result equal to 2! (2 factorial), which is 2. So:

E(X) = (1/2) × 2 = 1

This matches statement 1, so statement 1 is true.

Step 4: Find E(XY)

Now the same idea, but with an extra factor of y:

E(XY) = ∫ (y = 0 to ∞) ∫ (x = 0 to y) xy · e−y dx dy

Pull y and e−y out of the inner integral (they don't depend on x), and again use the standard result that the inner integral of x from 0 to y equals y²/2:

E(XY) = ∫ (y = 0 to ∞) y · e−y · (y²/2) dy = (1/2) ∫ (y = 0 to ∞) y³ · e−y dy

The integral ∫ y³ e−y dy from 0 to ∞ equals 3! (3 factorial), which is 6. So:

E(XY) = (1/2) × 6 = 3

Statement 3 claims E(XY) = 2, but we've just calculated E(XY) = 3. So statement 3 is false.

Step 5: Put It Together

Statement 1 (E(X) = 1): true. Statement 2 (E(Y) = 2): true. Statement 3 (E(XY) = 2): false.

So the correct statements are 1 and 2 only, which is option (a)— confirmed against the official answer key.

Why This Question Matters for Your Preparation

This question is a good test of whether you can move fluently between a joint density, a marginal density, and expectation calculations — a combination that shows up repeatedly across ISS Statistics papers. At Sunrise Classes, this is exactly the kind of question we slow down on in class: not just showing the final answer, but walking through why the integration limits are set up the way they are, so the same logic transfers to the next unfamiliar problem you see — not just this one.

Frequently Asked Questions

Why do we integrate x from 0 to y instead of 0 to infinity?

Because the joint density is only defined (non-zero) for 0 < x < y — for any fixed value of y, x cannot exceed y. Integrating x beyond y would include a region where the density is actually zero, giving a wrong answer.

How do you know the integral of y² e⁻ʸ from 0 to infinity equals 2?

This is a standard result from the Gamma function: the integral of y raised to power n, times e⁻ʸ, from 0 to infinity, equals n factorial. For n = 2, that's 2! = 2. For n = 3 (used later in this question), it's 3! = 6.

Is recognizing the Gamma distribution necessary to solve this question?

It's a helpful shortcut for E(Y) specifically, since it lets you skip a full integral. But even without recognizing it, you'd reach the same answer by directly integrating y times y·e⁻ʸ from 0 to infinity, which is the same Gamma-function-based calculation.

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

Getting the integration limits wrong — either integrating x from 0 to infinity instead of 0 to y, or forgetting to integrate out one variable before working with the other. Always write out the region of integration before starting.

Have a doubt on this question, or want the next one in the series sooner? Drop it in the comments below — and if this helped, share it with another ISS aspirant.