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S2.3 - Continuous distributions

Syllabus
2019
Topic
S2.3
Level
A2

Derive and use the continuous uniform distribution

A continuous uniform variable gives equal probability to intervals of equal length. If X∼U(a,b)X\sim U(a,b) with a<ba<b, its density is a rectangle of width b−ab-a. Unit total area fixes its height:

f(x)={1b−a,a≤x≤b,0,otherwise.f(x)=\begin{cases}\dfrac{1}{b-a},&a\le x\le b,\\0,&\text{otherwise}.\end{cases}

Probability is therefore a length ratio. Intersect the requested event with the support first: for a≤c<d≤ba\le c<d\le b,P(c<X<d)=∫cd1b−a dx=d−cb−a.P(c<X<d)=\int_c^d\frac{1}{b-a}\,dx=\frac{d-c}{b-a}.For X∼U(−5,19)X\sim U(-5,19), P(∣X∣>3.5)=[1.5+15.5]/24=17/24P(|X|>3.5)=[1.5+15.5]/24=17/24. Endpoints make no difference for a continuous variable.

Accumulating rectangular area from the left derives the cumulative distribution function:F(x)={0,x<a,x−ab−a,a≤x≤b,1,x>b.F(x)=\begin{cases}0,&x<a,\\\dfrac{x-a}{b-a},&a\le x\le b,\\1,&x>b.\end{cases}The middle piece is linear because each extra unit of xx adds the same area 1/(b−a)1/(b-a).

The mean follows fromE(X)=∫abxb−a dx=b2−a22(b−a)=a+b2.E(X)=\int_a^b\frac{x}{b-a}\,dx=\frac{b^2-a^2}{2(b-a)}=\frac{a+b}{2}.Likewise,E(X2)=b3−a33(b−a)=a2+ab+b23,E(X^2)=\frac{b^3-a^3}{3(b-a)}=\frac{a^2+ab+b^2}{3},soVar⁡(X)=E(X2)−[E(X)]2=(b−a)212.\operatorname{Var}(X)=E(X^2)-[E(X)]^2=\frac{(b-a)^2}{12}.Thus the mean is the rectangle's midpoint, while variance depends only on its width.

Uniform means constant density over a stated interval, not that every exact value has a positive equal probability. Include the zero-density outer pieces, clip transformed event intervals to the support, and use the width b-a rather than b as the denominator.

Apply a Normal approximation with continuity correction

A Normal distribution can approximate a Binomial or Poisson count when the count distribution is sufficiently spread out and not strongly skewed. Match the discrete mean and variance before converting the integer event to a continuous boundary.

Discrete model Approximating model
X∼B(n,p)X\sim B(n,p) Y∼N(np, np(1−p))Y\sim N(np,\,np(1-p))
X∼Po(λ)X\sim Po(\lambda) Y∼N(λ, λ)Y\sim N(\lambda,\,\lambda)
Discrete event Continuity-corrected Normal event
X≤kX\le k Y<k+0.5Y<k+0.5
X<kX<k Y<k−0.5Y<k-0.5
X≥kX\ge k Y>k−0.5Y>k-0.5
X>kX>k Y>k+0.5Y>k+0.5
a≤X≤ba\le X\le b a−0.5<Y<b+0.5a-0.5<Y<b+0.5

Each integer count represents a unit-width bar extending 0.5 on either side of its centre. Moving the Normal boundary to the outer edge of the last included bar preserves approximately the same area. First rewrite words such as 'fewer than 32' as the integer event X ≤ 31; then its corrected boundary is 31.5.

Suppose X∼Po(36)X\sim Po(36) and we need P(X<32)P(X<32). Use Y∼N(36,36)Y\sim N(36,36), so its standard deviation is 6. With the correction,P(X<32)≈P(Y<31.5)=P(Z<31.5−366)=P(Z<−0.75)=0.2266.P(X<32)\approx P(Y<31.5)=P\left(Z<\frac{31.5-36}{6}\right)=P(Z<-0.75)=0.2266.The value 32 is not the boundary because the count 32 is excluded.

For a Binomial example, X∼B(200,0.4)X\sim B(200,0.4) gives Y∼N(80,48)Y\sim N(80,48). ThenP(X≥90)≈P(Y>89.5)=P(Z>89.5−8048).P(X\ge90)\approx P(Y>89.5)=P\left(Z>\frac{89.5-80}{\sqrt{48}}\right).Keep the second Normal parameter as the variance, but divide by its square root when standardising.

Do not apply a continuity correction to an already continuous event, omit it for a discrete-to-Normal approximation, or use a Normal model when a small mean or extreme Binomial probability leaves the count distribution strongly skewed. This objective concerns probability approximation; hypothesis-test decisions belong to the following Topic.