ISS 2016 Statistics Paper-1 Solution: Question 21 (Expectation of Floor of Laplace Variable)
Continuing our question-by-question walk through the ISS Statistics Paper-1 archive, today we land on Question 21 of the 2016 paper — a neat little probability problem that looks intimidating the moment a "greatest integer function" shows up inside an expectation. As always in this series, the goal is not just to name the correct option but to rebuild the entire derivation from scratch, so that anyone reading along could reproduce it on their own in the exam hall.
Quick Summary
Topic:Continuous distributions — Laplace (double exponential) distribution, expectation of a floor-function transformation
Question Reference:ISS 2016, Statistics Paper-1, Question 21
Correct Answer:Option (c),E([X]) = -1/2
The Question, As Asked
Let X be a random variable with density
f(x) = (1/2) e-|x|, -∞ < x < ∞.
What is the expected value of [X], where [.] denotes the greatest integer function?
(a) 1/2
(b) 0
(c) -1/2
(d) -1
Step 1: Recognise the Distribution
The density f(x) = (1/2)e-|x| defined on the whole real line is the standard Laplace distribution (also called the double exponential distribution). You can think of it as two exponential distributions glued back to back at zero: for x ≥ 0 it behaves like an Exponential(1) density folded to have total area 1/2, and for x < 0 it is the mirror image. Because of this symmetry, the distribution is symmetric about 0, which means X and -X have exactly the same distribution. Hold on to that fact — it becomes the key trick later.
First, let's just double check it is a valid density by integrating it over the whole line:
∫-∞∞ (1/2)e-|x| dx = 2 × ∫0∞ (1/2)e-x dx = 2 × (1/2) × 1 = 1.
Good — it integrates to 1, so it is indeed a proper density, and by symmetry E(X) = 0 (though that is not what's being asked here).
Step 2: Understand What [X] Actually Means
[X] is the greatest integer function (also written as floor(X)), which rounds X down to the nearest integer. For example, [2.7] = 2, [0.3] = 0, [-0.3] = -1, and [-2.7] = -3. Notice the asymmetry: rounding down always moves you toward negative infinity, never toward zero. This is exactly why, even though X itself is symmetric around 0, [X] will NOT have an expectation of zero. This subtlety is what makes option (b) a tempting but incorrect trap answer.
Step 3: Split the Real Line into Unit Intervals
Since [X] = n whenever n ≤ X < n+1 for any integer n, we can compute E([X]) by breaking the real line into unit-length intervals [n, n+1) for every integer n, finding the probability X falls into each interval, and then summing n times that probability:
E([X]) = Σn = -∞∞ n · P(n ≤ X < n+1)
Because the density has a different formula depending on whether x is positive or negative, we must handle n ≥ 0 and n ≤ -1 separately.
Step 4: Compute the Probability for n ≥ 0
For n ≥ 0, the whole interval [n, n+1) lies on the non-negative side, where |x| = x, so f(x) = (1/2)e-x. Then:
P(n ≤ X < n+1) = ∫nn+1 (1/2)e-x dx = (1/2)[-e-x]nn+1 = (1/2)(e-n - e-(n+1)) = (1/2)e-n(1 - e-1).
Step 5: Compute the Probability for n ≤ -1
For n ≤ -1, the whole interval [n, n+1) lies on the negative side (since n+1 ≤ 0), where |x| = -x, so f(x) = (1/2)ex. Then:
P(n ≤ X < n+1) = ∫nn+1 (1/2)ex dx = (1/2)(en+1 - en) = (1/2)en(e - 1).
Step 6: Sum the Series for n ≥ 0
Let q = e-1 ≈ 0.367879 to simplify notation. The n ≥ 0 part of the sum is:
S1 = Σn=0∞ n · (1/2)qn(1-q) = (1/2)(1-q) · Σn=0∞ n qn
Here we use the standard power series result Σn=0∞ n rn = r / (1-r)2 for |r| < 1 (this comes from differentiating the geometric series Σ rn = 1/(1-r) with respect to r and multiplying by r). Applying it with r = q:
S1 = (1/2)(1-q) · q/(1-q)2 = (1/2) · q/(1-q)
Step 7: Sum the Series for n ≤ -1
For the negative side, substitute m = -n so that as n runs through -1, -2, -3, ..., m runs through 1, 2, 3, .... Also en = e-m = qm. The sum becomes:
Σn=-∞-1 n en = Σm=1∞ (-m) qm = -Σm=1∞ m qm = -q/(1-q)2
So the n ≤ -1 part of the total sum is:
S2 = (1/2)(e-1) · [ -q/(1-q)2 ]
Now notice that e - 1 = 1/q - 1 = (1-q)/q, so (e-1)·q = 1-q. Substituting this in:
S2 = (1/2) · [ -(1-q)/(1-q)2 ] = -(1/2) / (1-q)
Step 8: Add the Two Pieces Together
E([X]) = S1 + S2 = (1/2)·q/(1-q) - (1/2)/(1-q) = (1/2) · (q - 1)/(1-q)
Since (q-1) = -(1-q), this simplifies beautifully:
E([X]) = (1/2) · [-(1-q)] / (1-q) = -1/2
As a numerical check, with q = e-1 ≈ 0.367879: S1 ≈ 0.29099 and S2 ≈ -0.79099, and their sum is indeed -0.50000, confirming the algebra.
Pro Tip: There is a much faster route to the same answer, and recognising it under exam time pressure is the real skill being tested here. For any continuous random variable X (where P(X = integer) = 0), the floor function satisfies the identity [x] + [-x] = -1 for every non-integer x. Taking expectations on both sides gives E([X]) + E([-X]) = -1. Now, because the Laplace density here is symmetric about 0, -X has exactly the same distribution as X, so E([-X]) = E([X]). Substituting, 2·E([X]) = -1, which gives E([X]) = -1/2 immediately — no infinite series required. The way this shortcut is usually taught is to first spot the symmetry of the density, and only then look for an identity that converts the awkward floor function into something the symmetry can cancel out; students who drill this pattern-recognition step until it becomes automatic tend to save several minutes on exactly this kind of question.
Final Answer
E([X]) =-1/2, which is option (c). This matches the official ISS 2016 Statistics Paper-1 answer key, which also lists (c) as the correct choice for Question 21.
Why This Question Matters
This question is a favourite type in ISS and similar statistics services exams because it combines three separate skills in one short problem: recognising a standard distribution from its density formula, correctly splitting a definite integral across a piecewise-defined function, and handling an infinite series (or, better, spotting a symmetry shortcut) to avoid getting bogged down in algebra. Questions built around the Laplace distribution also test whether a candidate remembers that it is really "two exponentials stitched together," which is a useful mental model for solving several other transformation and moment-generating-function questions in the syllabus.
Frequently Asked Questions
Why isn't the answer simply 0, given that X is symmetric about 0?
It's tempting to assume that a symmetric random variable always has expectations of zero for any transformation, but that's only true for odd functions like g(x) = x or g(x) = x3. The floor function is not an odd function — [x] and -[-x] differ by exactly 1 for every non-integer x — so the symmetry of X does not carry over to [X]. That asymmetry is precisely what produces the -1/2 answer instead of 0.
What exactly is the Laplace distribution used for?
The Laplace (double exponential) distribution is often used to model data with heavier tails than the normal distribution, or situations built from the difference of two independent, identically distributed exponential variables. It shows up in signal processing, Bayesian statistics (as a prior that encourages sparsity, similar to LASSO regression), and as a natural example whenever a syllabus wants to test integration over a piecewise absolute-value density.
Is there a general formula for E([X]) for any distribution?
Not a single closed-form formula, because it depends entirely on how probability mass is distributed across each unit interval [n, n+1). However, the general method shown in Steps 3 to 8 above — summing n · P(n ≤ X < n+1) over all integers n — always works, and for symmetric distributions the shortcut E([X]) + E([-X]) = -1 (used in the Pro Tip) is a fast way to shortcut the algebra whenever the distribution is symmetric about zero.
Why does the identity [x] + [-x] = -1 hold for non-integer x?
If x is not an integer, write x = n + f where n = [x] is an integer and 0 < f < 1 is the fractional part. Then -x = -n - f, and since -f is between -1 and 0, we can write -x = (-n - 1) + (1 - f), where 0 < 1-f < 1. So [-x] = -n - 1. Adding the two floors: [x] + [-x] = n + (-n-1) = -1. The identity fails only at integer points, but since a continuous random variable takes any exact integer value with probability zero, this does not affect the expectation.
Would the answer change if the greatest integer function were replaced with rounding to the nearest integer?
Yes — rounding to the nearest integer treats values above and below the midpoint of each interval differently than flooring does, so the probabilities attached to each integer would shift and the resulting expectation would generally come out different (and for a symmetric density like this one, nearest-integer rounding would likely give an answer much closer to, or exactly, 0). It's a good exercise to redo Steps 3-8 with that alternative definition to see how the calculation changes.
How is this type of question typically weighted in the ISS Paper-1 exam?
Questions on continuous distributions, transformations, and expectation calculations make up a substantial share of the Probability section of Paper-1 year after year, so mastering the mechanics shown here — splitting integrals, handling piecewise densities, and recognising symmetry shortcuts — pays off across many similar questions, not just this one.
If any step above felt unclear, or if you spotted a different way to reach the same answer, drop a comment below — working through the disagreement is often the fastest way to actually understand it. And if you know a fellow ISS aspirant who's grinding through the same Paper-1 archive, feel free to pass this one along to them too.
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