ISS 2016 Statistics Paper-1 Solution: Question 16 (Laplace Characteristic Function Inversion)

Continuing our question-by-question walk through the ISS Statistics Paper-1 archive, we now arrive at Question 16 from the 2016 paper. This one moves away from the discrete-distribution questions we've been solving and drops us into the world of characteristic functions and Fourier inversion — a topic that trips up a lot of aspirants not because the algebra is hard, but because the logic connecting two statements is easy to misread.

Quick Summary

• Topic: Characteristic functions, the Laplace (double exponential) distribution, and the Fourier inversion theorem

• Question Reference: ISS 2016, Statistics Paper-1, Question 16

• Correct Answer: Option (a) — Both Statement I and Statement II are true, and Statement II is the correct explanation of Statement I

The Question

Consider the following statements:

Statement I:For a Laplace distribution f(x) = (1/2)e−|x|, −∞ < x < ∞,

f(x) = (1/2π) ∫−∞∞ e−itx ψ(t) dt for all x ∈ R′,

where ψ(t) is the characteristic function.

Statement II:ψ(t) is absolutely integrable.

Which one of the following is correct in respect of the above statements?

• (a) Both Statement I and Statement II are true and Statement II is the correct explanation of Statement I.

• (b) Both Statement I and Statement II are true but Statement II is not the correct explanation of Statement I.

• (c) Statement I is true but Statement II is false.

• (d) Statement I is false but Statement II is true.

Step 1: Write Down What We Actually Need to Check

Before touching any integral, let's be clear about what each statement is claiming, because that clarity is half the battle in an "assertion–reason" style question like this one.

Statement I says: if you take the characteristic function ψ(t) of the Laplace distribution and plug it into the Fourier inversion formula f(x) = (1/2π)∫e−itxψ(t)dt, you get back the original density f(x) = (1/2)e−|x|.

Statement II says: ψ(t), as a function of t, is absolutely integrable — meaning ∫−∞∞ |ψ(t)| dt is a finite number.

So we have three separate jobs: (1) find ψ(t) for the Laplace distribution, (2) check whether Statement II (absolute integrability) is true, and (3) check whether Statement I (the inversion formula actually recovering the density) is true, and if both are true, whether one logically causes the other.

Step 2: Derive the Characteristic Function of the Laplace Distribution

By definition, the characteristic function is ψ(t) = E[eitX] = ∫−∞∞ eitx f(x) dx. Substituting the Laplace density f(x) = (1/2)e−|x|, we get:

ψ(t) = (1/2) ∫−∞∞ eitx e−|x| dx

The absolute value |x| forces us to split the integral into two pieces: one for x ≥ 0 (where |x| = x) and one for x < 0 (where |x| = −x).

Piece 1 (x from 0 to ∞):Here e−|x| = e−x, so the integrand becomes eitxe−x = e(it−1)x.

∫0∞ e(it−1)x dx = [e(it−1)x / (it−1)]0∞

Since the real part of (it − 1) is −1 (negative), e(it−1)x → 0 as x → ∞. At x = 0, e0 = 1. So this evaluates to:

0 − 1/(it − 1) = −1/(it − 1) = 1/(1 − it)

Piece 2 (x from −∞ to 0):Here e−|x| = ex, so the integrand becomes eitxex = e(it+1)x.

∫−∞0 e(it+1)x dx = [e(it+1)x / (it+1)]−∞0

Now the real part of (it + 1) is +1 (positive), so as x → −∞, e(it+1)x → 0. At x = 0, e0 = 1. So this evaluates to:

1/(it + 1) − 0 = 1/(1 + it)

Adding the two pieces and putting back the (1/2) factor:

ψ(t) = (1/2) [ 1/(1 − it) + 1/(1 + it) ]

To add these fractions, use the common denominator (1 − it)(1 + it) = 1 + t2 (a standard difference-of-squares style expansion, using i2 = −1):

1/(1 − it) + 1/(1 + it) = [(1 + it) + (1 − it)] / (1 + t2) = 2 / (1 + t2)

So:

ψ(t) = (1/2) × 2/(1 + t2) =1/(1 + t2)

This is a clean, well-known result — the characteristic function of the standard Laplace distribution is 1/(1 + t²), the same functional form as a Cauchy density (just without the 1/π normalizing constant).

Step 3: Check Statement II — Is ψ(t) Absolutely Integrable?

We now need ∫−∞∞ |ψ(t)| dt = ∫−∞∞ 1/(1 + t2) dt, since 1/(1+t²) is always positive, |ψ(t)| = ψ(t) here.

This is one of the most recognizable integrals in calculus. The antiderivative of 1/(1+t²) is arctan(t). So:

∫−∞∞ 1/(1 + t2) dt = [arctan(t)]−∞∞ = arctan(∞) − arctan(−∞) = π/2 − (−π/2) = π

Since π is a finite number, the integral converges, and ψ(t)is indeed absolutely integrable. So Statement II is true.

Step 4: Check Statement I — Does the Inversion Formula Recover the Density?

This is where most candidates either panic (thinking they need to compute a nasty contour integral from scratch) or take a shortcut without understanding why the shortcut is valid. Here's the clean way to think about it.

There is a classical result in probability theory called the Fourier inversion theorem for characteristic functions. It states: if X is a random variable with characteristic function ψ(t), and if ψ(t) happens to be absolutely integrable over the whole real line, then X has a continuous density function f, and that density is exactly given by:

f(x) = (1/2π) ∫−∞∞ e−itx ψ(t) dt, for every x.

Notice something important: this is precisely the equation written in Statement I. We already showed in Step 3 that ψ(t) = 1/(1+t²) is absolutely integrable. Because the Laplace density is a genuine, continuous probability density function, and its characteristic function satisfies the absolute-integrability hypothesis of the theorem, the theorem's conclusion applies directly: plugging ψ(t) back into that inversion integral must return f(x) = (1/2)e−|x|, exactly as Statement I claims.

(For anyone who wants to double check this by direct computation rather than by invoking the theorem: ∫−∞∞ e−itx/(1+t²) dt is a standard tabulated Fourier transform pair, and it evaluates to π e−|x|. Multiplying by the 1/2π in front of the integral gives (1/2π)(π e−|x|) = (1/2)e−|x| — the original Laplace density, confirming the theorem's prediction.)

So Statement I is also true.

Step 5: Decide the Relationship Between the Two Statements

Both statements are individually true. The remaining question is whether Statement II is the reason Statement I holds, or whether they're just two unrelated true facts sitting next to each other.

Look again at the inversion theorem stated in Step 4: absolute integrability of ψ(t) is exactly the hypothesis that guarantees the inversion formula holds and reproduces the density. In other words, Statement II is not a coincidental fact about this particular ψ(t) — it is the very condition that makes Statement I true in the first place. If ψ(t) were not absolutely integrable, the inversion formula in Statement I could fail to hold in this simple pointwise sense (you'd need more delicate arguments, like inversion in terms of distribution functions rather than densities). So Statement II correctly explains Statement I.

Pro Tip: In assertion–reason MCQs built around theorems, don't just verify that both statements are numerically or algebraically true and stop there — that only rules out options (c) and (d). The real discriminator between (a) and (b) is whether Statement II is stated as a *hypothesis* of a theorem whose *conclusion* is Statement I. The way this distinction is usually taught is to ask: "if I strip away Statement II, does the standard proof of Statement I still go through unchanged?" If the answer is no — the proof needs that exact condition — then it's the explanation, and the answer is (a). Students who drill this kind of assertion-reason reasoning until it becomes automatic, rather than treating each statement as an isolated fact to verify, tend to stop losing marks on these theorem-based questions.

Final Answer: Option (a)— Both Statement I and Statement II are true, and Statement II is the correct explanation of Statement I. This matches the official answer key, which also lists (a) as correct for Question 16.

Why This Question Matters

This question quietly tests three layers of understanding at once: your ability to actually compute a characteristic function from a density using basic integration (Step 2), your familiarity with a standard convergence check like absolute integrability (Step 3), and — most importantly — your conceptual grasp of *why* the Fourier inversion theorem needs that convergence condition in the first place (Step 5). It's easy to memorize that "the characteristic function of Laplace is 1/(1+t²)" without ever internalizing why the inversion formula is guaranteed to work for it. Questions like this are the exam's way of separating candidates who've memorized formulas from those who understand the theorems behind them — which matters a great deal once you reach the descriptive papers, where you're asked to state and apply such theorems explicitly rather than just pick an option.

Frequently Asked Questions

What exactly is the Laplace (double exponential) distribution?

It's a continuous probability distribution with density f(x) = (1/2)e−|x| for all real x, symmetric about zero. It looks like two mirrored exponential distributions glued back-to-back at zero, which is why it's often called the "double exponential" distribution.

Why do we split the integral at x = 0 when computing the characteristic function?

Because the density involves |x|, which is defined piecewise: |x| = x when x ≥ 0, and |x| = −x when x < 0. Since the exponent −|x| behaves differently on each side of zero, you cannot integrate eitxe−|x| as a single smooth expression over the whole real line — you must handle the two halves separately and add the results.

What does "absolutely integrable" mean, and why does it matter here?

A function g(t) is absolutely integrable over the real line if ∫|g(t)|dt is finite (not infinite). For characteristic functions, this property is the key hypothesis in the Fourier inversion theorem: when it holds, you're guaranteed that the underlying random variable has an honest, continuous density function that can be recovered exactly from the characteristic function via the inversion integral.

Is the characteristic function of the Laplace distribution the same as that of the Cauchy distribution?

They share the same functional form, 1/(1+t²), but they are not literally the same function pair. What's actually happening is a duality: the Cauchy density has the functional form 1/(π(1+x²)), and its characteristic function is e−|t|; the Laplace density has the form (1/2)e−|x|, and its characteristic function is 1/(1+t²). Each is essentially the "mirror image" of the other under the Fourier transform.

How would the answer change if Statement II were false instead?

If ψ(t) were not absolutely integrable, you could not immediately conclude the pointwise inversion formula in Statement I holds using this particular theorem — you'd need a different or more careful argument (for instance, inversion formulas stated in terms of the distribution function rather than the density). In that scenario, Statement I might still happen to be true by some other route, but Statement II could no longer serve as its explanation, which is exactly the kind of trap option (c) or (d) would represent.

Do ISS Paper-1 questions usually go this deep into characteristic function theory?

Yes — characteristic functions, moment generating functions, and their inversion/uniqueness properties are a recurring theme across ISS Paper-1 papers, since they underpin large parts of distribution theory and limit theorems. It's worth being comfortable deriving characteristic functions for standard distributions (binomial, Poisson, exponential, normal, Laplace, Cauchy) directly from their definitions, rather than only memorizing the final formulas.

If any step above wasn't clear, or you spotted something you think should be explained differently, drop a comment below — working through the confusion together is often the fastest way to actually lock in a topic like this. And if you know a fellow ISS aspirant grinding through the same papers, feel free to pass this one along; a second explanation of a tricky theorem rarely hurts.