ISS 2016 Statistics Paper-1 Solution: Question 7 (Maximizing Variance of a Discrete Distribution)

Continuing our question-by-question walk through the ISS Statistics Paper-1 archive, we now arrive at Question 7 of the 2016 paper. This one looks like an ordinary variance calculation at first glance, but it hides a small trap around how the maximum is actually located — exactly the kind of detail that separates a rushed attempt from a fully correct one.

Quick Summary

• Topic:Mean and variance of a discrete random variable, expressed as a function of a parameter

• Question Reference:ISS 2016, Statistics Paper-1, Question 7

• Correct Answer:Option (a) — Maximum value of V(X) is 3

The Question As Asked

Let the random variable X have the distribution P(X = 0) = P(X = 3) = p, P(X = 1) = 1 − 3p, for 0 ≤ p ≤ 1/2. What is the maximum value of V(X)?

• (a) 3

• (b) 4

• (c) 5

• (d) 6

Step 1: Writing Down What We're Given

Before touching any formula, let's lay out exactly what the problem hands us. The random variable X takes only three values — 0, 1, and 3 — with the following probabilities:

• P(X = 0) = p

• P(X = 1) = 1 − 3p

• P(X = 3) = p

We are told p can range from 0 to 1/2, and we need V(X), which is standard notation for the variance of X. Our job is to find the value of p (somewhere in that range) that makes this variance as large as possible.

Step 2: Computing E(X)

The mean (expected value) of a discrete random variable is found by multiplying each value X can take by its probability, and adding everything up:

E(X) = (0 × p) + (1 × (1 − 3p)) + (3 × p)

Let's expand this term by term:

• 0 × p = 0

• 1 × (1 − 3p) = 1 − 3p

• 3 × p = 3p

Adding these three pieces together:

E(X) = 0 + (1 − 3p) + 3p = 1 − 3p + 3p

Notice that −3p and +3p cancel each other out exactly. So:

E(X) = 1

This is a neat result — no matter what value p takes (as long as it stays in a sensible range), the mean of X always works out to exactly 1. This kind of cancellation is common when a distribution is built so that the "extra" probability mass at the two endpoints (0 and 3 here) is symmetric in how it pulls the mean up and down.

Step 3: Computing E(X²)

To get the variance, we also need E(X²), the expected value of X squared. The method is identical — square each value of X, multiply by its probability, and sum:

E(X²) = (0² × p) + (1² × (1 − 3p)) + (3² × p)

Working through each term:

• 0² × p = 0 × p = 0

• 1² × (1 − 3p) = 1 × (1 − 3p) = 1 − 3p

• 3² × p = 9 × p = 9p

Adding these up:

E(X²) = 0 + (1 − 3p) + 9p = 1 − 3p + 9p

Here the p-terms don't cancel — they combine, since one is −3p and the other is +9p:

−3p + 9p = 6p

So:

E(X²) = 1 + 6p

Step 4: Assembling the Variance Formula

Variance has a standard formula that you should have memorized cold before attempting this kind of question:

V(X) = E(X²) − [E(X)]²

We already have both pieces:

• E(X²) = 1 + 6p

• E(X) = 1, so [E(X)]² = 1² = 1

Substituting these in:

V(X) = (1 + 6p) − 1 = 6p

So the variance of X, written purely as a function of the parameter p, is simply:

V(X) = 6p

Step 5: Maximizing V(X) Over the Given Range

We now need to find the value of p, somewhere between 0 and 1/2, that makes V(X) = 6p as large as possible.

Because V(X) = 6p is a straight-line (linear) function of p with a positive slope of 6, it is strictly increasing — as p goes up, V(X) goes up too, with no dips, no turning points, no maximum sitting in the middle. There is nothing to differentiate and set to zero here; a linear function on a closed interval always reaches its maximum at one of the two endpoints, and since the function is increasing, that endpoint is the right-hand one.

The given range is 0 ≤ p ≤ 1/2, so the largest allowed value of p is p = 1/2. Substituting this in:

V(X) = 6 × (1/2) = 3

So the maximum value of V(X) is 3, which is option (a) — confirmed against the official ISS 2016 answer key, which also lists (a) for this question.

Pro Tip: The trickiest part of this question isn't the algebra — it's resisting the urge to differentiate. Many students, on autopilot, take dV/dp, set it to zero, and go looking for an interior critical point, only to find there isn't one because V(X) = 6p has no turning point at all. The way this kind of question is usually taught is to first check whether the expression you've derived is linear in the parameter — if it is, the extremum has to be at an endpoint of the allowed range, and you simply plug in both ends and compare. It's also worth noticing, if you look closely, that P(X = 1) = 1 − 3p turns negative once p exceeds 1/3, which technically breaks the distribution for p beyond that point — the question doesn't dwell on this, and neither does the official key, but recognizing it is a sign you've understood the setup rather than just mechanically pushed through the formula.

Why This Question Matters

This question is really testing two separate skills stitched together: whether you can correctly compute E(X) and E(X²) for a discrete distribution defined in terms of a parameter, and whether you understand that optimizing a function over a closed interval doesn't always mean setting a derivative to zero — sometimes the function is monotonic, and the answer sits at a boundary. Questions built this way appear repeatedly across ISS and similar statistics papers, precisely because they catch out candidates who solve optimization problems by reflex rather than by first checking the shape of the function. Students who make a habit of writing out E(X) and E(X²) explicitly before jumping to variance, and who train themselves to glance at whether an expression is linear before reaching for calculus, tend to move through this style of question much faster in the exam hall.

Frequently Asked Questions

Why does E(X) come out to exactly 1, independent of p?

It happens because of how the probabilities were constructed. The value 0 contributes nothing to the sum, and the +3p term from X = 3 exactly cancels the −3p term hidden inside 1 − 3p at X = 1, leaving a constant. This isn't a coincidence you should expect in every problem — it's specific to how this particular distribution was set up, and it's worth verifying by direct substitution rather than assuming it will always happen.

Do I always need both E(X) and E(X²) to find variance?

Yes, whenever you're using the standard shortcut formula V(X) = E(X²) − [E(X)]², which is almost always the fastest route for a discrete distribution with a small number of outcomes. The alternative, working directly from V(X) = E[(X − E(X))²], involves computing E(X) first anyway and then squaring deviations, which is slower for hand calculation in an exam setting.

How do I know a function has no interior maximum without doing a full derivative test?

If, after simplification, the expression you're maximizing turns out to be a straight line in the parameter (degree 1, like 6p here), it has no interior critical point at all — a linear function's slope never changes sign, so it just keeps increasing or keeps decreasing across the whole interval. Only when the expression is quadratic or higher degree do you need to check for an interior turning point using calculus.

What if the maximum had occurred at p = 0 instead of p = 1/2?

You'd simply substitute p = 0 into V(X) = 6p and get V(X) = 0. Always check both endpoints when a function is monotonic on a closed interval — for an increasing function, the right endpoint gives the maximum, but it's good practice to verify the direction of monotonicity rather than assuming it.

Is it a problem that P(X = 1) becomes negative for p greater than 1/3?

Strictly speaking, yes — a valid probability distribution can never assign a negative probability to any outcome, so this particular "distribution" stops being valid once p exceeds 1/3. The question and its official answer key don't account for this, and simply maximize the algebraic expression over the full stated range of p. In an exam setting, follow the method the key expects; in a purely theoretical sense, it's a detail worth noticing.

Would the answer change if the range of p were 0 ≤ p ≤ 1/3 instead?

Yes — since V(X) = 6p is still increasing, the maximum would then occur at p = 1/3, giving V(X) = 6 × (1/3) = 2, which isn't even one of the listed options. This is a good way to double check your own understanding: plug in the boundary the question actually gives you, not the boundary you might expect from probability theory.

If any step here felt rushed or you'd solve it differently, drop your working in the comments — questions like this are much easier to internalize when you talk through them, so feel free to share this with a fellow ISS aspirant who's also working through the Paper-1 archive.