ISS 2016 Statistics Paper-1 Solution: Question 17 (Conditional Poisson Given the Sum)

We're continuing our question-by-question walk through the ISS (Indian Statistical Service) Statistics Paper-1 previous year papers, and today we land on a question that shows up in one form or another in almost every probability-heavy exam: what happens to a Poisson variable once you condition it on a sum. If you've ever felt shaky switching between joint, marginal and conditional probabilities, this is a great problem to slow down on.

Quick Summary

• Topic:Conditional distribution of independent Poisson variables given their sum

• Question Reference:ISS 2016, Statistics Paper-1, Question 17

• Correct Answer:Option (c) — Binomial

The Question, As Asked

Let X1 and X2 be the number of eggs laid by two insects which follow independent Poisson distributions with parameters λ1 and λ2, respectively. The conditional distribution of X1 given X1 + X2 will be

• (a) Poisson

• (b) Geometric

• (c) Binomial

• (d) Negative Binomial

Step 1: Write Down What We Actually Know

We're told X1 ~ Poisson(λ1) and X2 ~ Poisson(λ2), and that X1 and X2 are independent. The Poisson probability mass function, for reference, is:

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

We want the conditional distribution of X1, given that we already know the value of the sum N = X1 + X2. In plain language: if we know the two insects laid a combined total of n eggs, what's the probability distribution of how many of those n eggs came specifically from insect 1?

Step 2: Find the Distribution of the Sum First

Before we can condition on N = X1 + X2, we need to know what N itself is distributed as. This is a standard, very useful fact: the sum of two independent Poisson random variables is itself Poisson, with the parameters simply adding up.

So N = X1 + X2 ~ Poisson(λ1 + λ2), which means:

P(N = n) = e-(λ1+λ2) (λ1+λ2)n / n!

(If you want to see why this addition property holds, it comes directly from the fact that the moment generating function — or characteristic function — of a sum of independent variables is the product of their individual MGFs, and for Poisson variables that product simplifies neatly back into another Poisson MGF. It's worth memorizing as a standalone fact since it saves a lot of time in exams.)

Step 3: Set Up the Conditional Probability Definition

By the basic definition of conditional probability, for any 0 ≤ k ≤ n:

P(X1 = k | X1 + X2 = n) = P(X1 = k and X1 + X2 = n) / P(X1 + X2 = n)

Now notice something important: the event "X1 = k AND X1 + X2 = n" is exactly the same event as "X1 = k AND X2 = n − k". This substitution is the entire trick of the problem — once you make it, everything becomes a plug-and-play calculation.

So we rewrite the numerator as:

P(X1 = k, X2 = n − k) = P(X1 = k) · P(X2 = n − k)

The last step — splitting the joint probability into a simple product — is only valid because X1 and X2 are independent. If they weren't independent, we could not have made this move, and the whole derivation would need a completely different approach.

Step 4: Plug In All Three Poisson Formulas

We now substitute the individual Poisson pmfs for X1, X2, and the pmf of N that we derived in Step 2:

P(X1 = k | N = n) = [ e-λ1 λ1k / k! ] · [ e-λ2 λ2n-k / (n-k)! ] ÷ [ e-(λ1+λ2) (λ1+λ2)n / n! ]

This looks messy, but almost everything cancels. Let's clean it up piece by piece.

Step 5: Simplify — Watch the Exponentials Cancel

First, look at the exponential terms. In the numerator we have e-λ1 · e-λ2 = e-(λ1+λ2). In the denominator we also have e-(λ1+λ2) (because dividing by a fraction flips it to the numerator, but it's identical to what's already there, so it cancels completely). The exponential terms vanish entirely — this is exactly why the answer doesn't come out as another Poisson: the "memoryless-looking" exponential piece cancels out and stops playing any role.

What's left is:

P(X1 = k | N = n) = [ λ1k λ2n-k / (k! (n-k)!) ] · [ n! / (λ1+λ2)n ]

Rearranging the factorial terms together and the λ terms together:

P(X1 = k | N = n) = [ n! / (k! (n-k)!) ] · [ λ1k λ2n-k / (λ1+λ2)n ]

The first bracket, n! / (k!(n−k)!), is exactly the binomial coefficient nCk. That alone is already a strong hint about where this is heading.

Step 6: Split the λ Ratio Into a Clean Probability p

Now handle the second bracket. We split (λ1+λ2)n in the denominator into (λ1+λ2)k · (λ1+λ2)n-k, so that we can pair one copy with λ1k and the other with λ2n-k:

λ1k λ2n-k / (λ1+λ2)n = [ λ1 / (λ1+λ2) ]k · [ λ2 / (λ1+λ2) ]n-k

If we define p = λ1 / (λ1+λ2), then automatically λ2 / (λ1+λ2) = 1 − p (check: adding the two fractions gives (λ1+λ2)/(λ1+λ2) = 1, so they must sum to 1). Substituting this in:

P(X1 = k | N = n) = nCk pk (1 − p)n-k

Step 7: Recognize the Final Form

This expression, nCk pk(1−p)n-k, is precisely the probability mass function of a Binomial(n, p)distribution, with n = X1+X2 (the total number of eggs, which we're conditioning on) and success probability p = λ1 / (λ1+λ2).

Intuitively this also makes sense: once you know the total count n, each individual egg can be thought of as independently "belonging" to insect 1 with probability λ1/(λ1+λ2), and the number of eggs out of n that fall into that category is a binomial count. That intuitive story is exactly why this result — sometimes called Poisson thinning or Poisson splitting — appears constantly in queueing theory and reliability problems.

Pro Tip: The single easiest way to fall into a trap here is to see "Poisson conditioned on something" and reflexively answer "Poisson" without doing the algebra, because Poisson variables feel self-similar. The way this result is usually taught is to always physically watch the e-λ terms cancel before guessing the family — the moment those exponential terms disappear, you know for certain the answer cannot stay Poisson, because a Poisson pmf can never exist without that exponential factor in it. Drilling that one cancellation step until it's automatic is what actually prevents this mistake on exam day.

Final Answer and Confirmation

Our full derivation shows that X1 | (X1+X2 = n) follows a Binomial(n, λ1/(λ1+λ2))distribution. That corresponds to option (c) Binomial, which matches the official ISS 2016 Statistics Paper-1 answer key exactly. No discrepancy here — the key and the independent derivation agree.

Why This Question Matters

This isn't just a one-off trick question — it's one of the most reused results in applied probability. It's the backbone of "Poisson thinning" arguments used in queueing theory (splitting one arrival stream into sub-streams), in genetics (allocating mutations between categories), and in reliability engineering. Once you've internalized the pattern — sum of independent Poissons is Poisson, and each one conditional on that sum is Binomial — you'll recognize dozens of exam questions instantly instead of re-deriving from scratch every time.

Frequently Asked Questions

Why does the sum of two independent Poisson variables also follow a Poisson distribution?

This is a special "closure" property of the Poisson family. Using moment generating functions, the MGF of a sum of independent random variables is the product of their individual MGFs. For two Poisson variables with parameters λ1 and λ2, multiplying their MGFs together gives exactly the MGF of a Poisson(λ1+λ2) variable, which proves the sum has that distribution.

What is meant by "Poisson thinning" or "Poisson splitting"?

It refers to exactly the situation in this question: when a Poisson-distributed total count is split into two (or more) independent Poisson-distributed sub-categories, then conditional on the total, each sub-category count follows a Binomial distribution. It's called "thinning" because you can think of it as randomly filtering a stream of events into separate streams.

Would the answer change if X1 and X2 were not independent?

Yes, significantly. The entire derivation relies on being able to write P(X1=k, X2=n−k) as the simple product P(X1=k)·P(X2=n−k), which is only valid under independence. Without independence, you would need the actual joint distribution of X1 and X2, and the conditional distribution would generally not simplify to a clean Binomial form at all.

How is the success probability p = λ1/(λ1+λ2) interpreted intuitively?

Think of every single egg (out of the n total) as an independent trial that can be attributed to either insect 1 or insect 2. The "rate share" of insect 1 relative to the combined rate, λ1/(λ1+λ2), turns out to be exactly the probability that any one given egg belongs to insect 1 — which is exactly what a binomial success probability represents.

Is this result specific to two variables, or does it generalize?

It generalizes cleanly. If you have k independent Poisson variables with parameters λ1, λ2, ..., λk, then conditional on their sum N = n, the joint vector (X1, X2, ..., Xk) follows a Multinomial(n, p1, p2, ..., pk) distribution, where each pi = λi / (λ1+...+λk). The two-variable binomial case in this question is just the k=2 special case of that broader multinomial result.

What's a common mistake students make when solving this type of MCQ under time pressure?

Besides guessing "Poisson" reflexively, another frequent error is forgetting to first establish the distribution of the sum N before writing the conditional probability formula — without knowing P(N=n) explicitly, you can't correctly form the denominator, and the algebra falls apart. Always write out all three pieces (the two marginal pmfs and the sum's pmf) before starting to simplify.

If any step above felt unclear, or you'd like to see the multinomial generalization worked out in full, drop a comment below — and if this helped, consider sharing it with a fellow ISS aspirant who's working through the same paper.