ISS 2016 Statistics Paper-1 Solution: Question 4 (Variance of Poisson Combination)

Continuing the ISS Paper-1 previous-year question series, where we work through each question of the objective paper one at a time, in order, with the full reasoning shown rather than just the final option. Today we look at Question 4 of the ISS 2016 Statistics Paper-1, which combines two small ideas — finding an unknown Poisson parameter from a probability ratio, and finding the variance of a linear combination of two independent random variables — into a single question.

Quick Summary

• Topic: Poisson distribution — parameter estimation and variance of a linear combination

• Question Reference: ISS 2016 Statistics Paper-1, Question 4

• Correct Answer: Option (a) — 14

The Question

X1 and X2 are independent Poisson variables such that P(X1 = 2) = P(X1 = 1) and P(X2 = 2) = P(X2 = 3). What is the variance of (X1 − 2X2) ?

• (a) 14

• (b) 4

• (c) 3

• (d) 2

Step 1: Write Down the Poisson Probability Mass Function

A random variable X follows a Poisson distribution with parameter λ (λ > 0) if its probability mass function is

P(X = k) = e−λ λk / k!,   k = 0, 1, 2, 3, ...

Here we are told two Poisson variables, X1 and X2, each with its own (unknown) parameter — call them λ1 and λ2. We are given one equation for each variable, and from each equation we can solve for its parameter.

Step 2: Find λ1 from P(X1 = 2) = P(X1 = 1)

Using the pmf above, write both probabilities in terms of λ1:

P(X1 = 1) = e−λ1 λ11 / 1! = e−λ1 λ1

P(X1 = 2) = e−λ1 λ12 / 2!

Setting P(X1 = 2) = P(X1 = 1):

e−λ1 λ12 / 2 = e−λ1 λ1

The factor e−λ1 is common to both sides and is never zero, so it cancels out completely:

λ12 / 2 = λ1

Since λ1 > 0, we can also divide both sides by λ1 (it is never zero, so this is a safe, reversible step):

λ1 / 2 = 1  ⟹  λ1 = 2

Step 3: Find λ2 from P(X2 = 2) = P(X2 = 3)

Do exactly the same thing for X2. Write both probabilities using the pmf with parameter λ2:

P(X2 = 2) = e−λ2 λ22 / 2!

P(X2 = 3) = e−λ2 λ23 / 3!

Setting them equal:

e−λ2 λ22 / 2 = e−λ2 λ23 / 6

Cancel e−λ2 from both sides, then cross-multiply:

6 · λ22 = 2 · λ23

Divide both sides by 2λ22 (valid since λ2 > 0, so λ22 ≠ 0):

3 = λ2  ⟹  λ2 = 3

Step 4: Recall That Variance of a Poisson Variable Equals Its Parameter

One of the most useful facts about the Poisson distribution is that its mean and variance are both equal to λ:

E(X) = λ,   Var(X) = λ

So directly from what we found:

Var(X1) = λ1 = 2

Var(X2) = λ2 = 3

Step 5: Apply the Variance Formula for a Linear Combination

We need Var(X1 − 2X2). For any two independent random variables X and Y and constants a, b, the general rule is:

Var(aX + bY) = a2Var(X) + b2Var(Y)

This holds only because independence makes the covariance term 2ab·Cov(X, Y) vanish — if X1 and X2 were not independent, we would need to subtract or add a covariance term as well. Here a = 1 and b = −2, so:

Var(X1 − 2X2) = (1)2 Var(X1) + (−2)2 Var(X2) = Var(X1) + 4·Var(X2)

Step 6: Substitute and Compute the Final Answer

Now plug in the values found in Step 4:

Var(X1 − 2X2) = 2 + 4 × 3 = 2 + 12 = 14

Pro Tip: The single most common slip on this question is treating the minus sign in "X1 − 2X2" as if it should subtract the variances too. It never does —

So the correct option is (a) 14. Checking against the official answer key for ISS 2016 Statistics Paper-1, Question 4 is listed as option A — which matches our derivation exactly, confirming both the parameter values λ1 = 2, λ2 = 3 and the variance computation.

Why This Question Matters

This question is a good test of whether a candidate can chain two separate ideas correctly under exam pressure: recovering a distribution's parameter from a probability condition, and then correctly applying the algebra of variances to a combination of two such variables. Questions phrased this way — "find the parameter, then find some derived quantity" — are extremely common across ISS, IIT JAM, and other statistics entrance papers, so getting comfortable with both halves separately, and then chaining them without losing a sign, is time well spent. Students who drill this kind of two-step derivation until it becomes automatic tend to move noticeably faster through the rest of the paper, since half the battle in a 200-mark objective paper is not running out of time on questions like this one.

Frequently Asked Questions

Why is the variance of a Poisson distribution equal to its parameter λ?

For a Poisson variable, both the mean and the variance work out to λ when you compute E(X) and E(X²) − [E(X)]² from the moment generating function or directly from the pmf. This is a special and convenient property of the Poisson distribution — most other distributions have a mean and variance that are different expressions in their parameters.

Why does Var(X1 − 2X2) not equal Var(X1) − 4Var(X2)?

Because variance is a squared-deviation quantity, and squaring a coefficient always produces a non-negative number. The formula Var(aX + bY) = a²Var(X) + b²Var(Y) uses squared coefficients precisely so that a minus sign in front of a variable never makes its contribution to the total variance negative.

What would change if X1 and X2 were not independent?

If they were dependent, an extra covariance term would appear: Var(aX1 + bX2) = a²Var(X1) + b²Var(X2) + 2ab·Cov(X1, X2). Since the question explicitly states the two variables are independent, this covariance term is zero and drops out entirely.

Is there a shortcut to find λ from an equation like P(X = k) = P(X = k+1)?

Yes. Writing out the ratio P(X = k+1)/P(X = k) for a Poisson variable always simplifies to λ/(k+1). Setting this ratio equal to 1 (since the two probabilities are equal) directly gives λ = k+1, which is a faster route to the same answer once you're comfortable with the algebra shown above.

Could X1 and X2 have had non-integer parameters here?

In principle λ can be any positive real number, but in this particular question the equations happened to simplify to whole numbers (λ1 = 2 and λ2 = 3). That is a feature of how this specific question was constructed, not a general rule — other questions of this type may well give you a fractional λ.

Why is this classified under Probability Distributions rather than Statistical Inference?

The question only asks you to manipulate a known distribution's pmf and variance formula — no data is being used to estimate anything, and no hypothesis is being tested. That places it squarely in the probability distributions portion of the ISS Paper-1 syllabus rather than in inference.

If any step above wasn't clear, or you spotted a quicker way to get to the same answer, drop a comment below — and if this walkthrough helped, consider sharing it with a fellow ISS aspirant who's working through the same paper.