Topic 4.14 · teacher page · AA Higher Level

Zero probability, non-zero density

Why P(X = a) is exactly zero, why a density can exceed 1, and the variance shortcut.

The one thing to do with the animation

Ask what P(X = 1) is before you shrink anything.

Most will say "small" or will read off f(1) = 0.375. Shrinking the strip drives the probability to zero while the red marker at 0.375 never moves, so the two quantities visibly come apart.

Watch the third readout as it goes: probability divided by width settles on 0.375. That is what the word density means, and it arrives here as an observation rather than a definition.

A density is not a probability

f(x) = 2 on [0, 0.5] is a perfectly legitimate density, because 2 × 0.5 = 1. No probability is ever greater than 1, so a density that exceeds 1 settles the question permanently.

And P(X < a) = P(X ≤ a) for continuous variables, because the single point contributes nothing. For discrete variables they differ, and questions use both deliberately.

The answers

1. Var(X) from the table6.6 − 2.4² = 6.6 − 5.76 = 0.84.
2. P(X < 1)[x³/8] from 0 to 1 = 0.125.
3. P(X = 1.5)B, exactly 0. A point encloses no area.

Where the marks go

1 markUsing ∫ f = 1 to find an unknown constant, which is usually part (a) and everything else depends on it.

1 markLimits taken from the question, not from the whole domain.

1 markE(X²) − [E(X)]², with both quantities shown separately.

What each wrong answer tells you

They giveWhat it means
4.2 (Q1)Subtracted E(X) rather than its square.
−0.84 (Q1)Reversed the order. A negative variance is impossible, which is the self-check to teach.
0.917 (Q1)Gave the standard deviation. Fine mathematics, wrong question.
0.375 (Q2)Gave the density at x = 1 rather than the area to its left. This is the exact confusion the figure targets.
0.5 (Q2)Assumed half the range is half the probability. The density is far larger near 2, which the picture shows.
"A small positive number" (Q3)Nearly there, and worth praising before correcting: they know it is tiny, they have not yet accepted it is exactly zero.

Other things they will say

"If every value has probability zero, how does X take one?" A genuinely good question and worth sitting with. Zero probability does not mean impossible for a continuous variable; probability lives in intervals, not points.

"Can f be bigger than 1?" Yes, as long as the total area is 1. A narrow, tall density is exactly how a precise measurement looks.

"Do I integrate or sum?" Continuous integrates, discrete sums. The structure is identical otherwise, which is worth drawing side by side.

A possible order

 What is happening
1Predict P(X = 1). Shrink the strip. Let the two readouts separate.
2The conditions on a density, and the f = 2 example that exceeds 1.
3E(X) and Var(X) by integration, on this density, by hand.
4The discrete table, with the shortcut and the long way compared.
5A find-the-constant question, which is how this is nearly always examined.

Two things not to say

Do not say P(X = a) is "very small". It is zero, and "very small" leaves the misconception fully intact.

Do not introduce the density and the distribution function in the same lesson. One idea at a time; the cdf is a better second lesson than a worse first one.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

  1. Find kf(x) = kx on 0 ≤ x ≤ 3 and zero elsewhere. Find k.
    The integral must be 1: k(9/2) = 1, so k = 2/9.
  2. A probabilityFind P(X < 2).
    The integral of (2/9)x from 0 to 2 is 4/9 = 0.444.

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

  1. Mean and varianceFind E(X) and Var(X).
    E(X) = 2 and E(X²) = 4.5, so Var(X) = 4.5 − 4 = 0.5.
  2. MedianFind the median.
    Solve (2/9)(m²/2) = 0.5, so m² = 4.5 and m = 2.12.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

  1. Mean against medianThe mean is 2 and the median is 2.12. Explain the order.
    The density increases with x, so the distribution is skewed towards the left in the sense that the mass piles up at high x; the median sits above the mean here because more of the probability lies in the upper part while the lower tail stretches to 0. The point is to compute both rather than assume mean below median.
  2. Why a point has probability zeroExplain why P(X = 2) = 0 for a continuous variable, and what that means for P(X ≤ 2) against P(X < 2).
    The integral over a single point is zero, so the two are equal. The strict and non-strict inequality never differ for a continuous variable, which is the opposite of the discrete case where it matters a great deal.

Practicalities

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