FS2 2.3 - Mode, median, percentiles and skewness

FS2 2.3 - Mode, median, percentiles and skewness

A peak, a halfway point and a long tail describe different features of a distribution. You will locate modes from a density, find medians and percentiles from a CDF, and justify whether a continuous distribution has positive, negative or zero skewness.

Find the highest density on the stated support

Can a density that rises all the way across its interval have negative skewness? Yes: f(x)=3x2f(x)=3x^2 on [0,1][0,1] does exactly that. Its peak is at the right-hand endpoint, while its tail stretches to the left. The direction of a slope and the direction of skewness are different ideas.

You already know

A density's height is not a point probability. A CDF gives accumulated area, and differentiating it gives a density on each interval. From calculus, stationary points and endpoints are candidates for a maximum.

A mode describes a peak of the density curve. An interior stationary point is only a candidate; it could be a minimum or lie outside the support.

Mode

A mode of a continuous distribution is a value at a peak of its density. A density can have one mode, several modes, or a flat maximum with no unique mode. An increasing density on a finite closed model interval has its boundary mode at the upper endpoint.

Use the shape of the specified density curve. Arbitrarily changing its value at one isolated point does not create a meaningful new peak of the distribution.

Finding a mode

For a differentiable density, find interior candidates from

f′(x)=0f'(x)=0

Then compare valid candidates, endpoints and any joins in the density. The mode must correspond to a greatest density height on the support, rather than merely a zero derivative.

Not in the formula booklet: learn it.

The next pair changes a density's shape while leaving its support [0,2][0,2] unchanged.

Worked example 1: an interior maximum

Question. Find the mode of

f(x)=34x(2−x),0≤x≤2f(x)=\frac34x(2-x),\qquad 0\le x\le2

with density zero elsewhere. Give a reason for your answer.

Plan. Differentiate the quadratic, check the stationary point, and compare the endpoint heights.

f′(x)=34(2−2x)differentiate the expanded quadraticf′(x)=0  ⟹  x=1interior candidatef′′(x)=−32<0strict maximumf(0)=f(2)=0,f(1)=34compare endpoints\begin{aligned} f'(x)&=\frac34(2-2x) && \text{differentiate the expanded quadratic}\\ f'(x)=0&\implies x=1 && \text{interior candidate}\\ f''(x)&=-\frac32<0 && \text{strict maximum}\\ f(0)&=f(2)=0,\qquad f(1)=\frac34 && \text{compare endpoints} \end{aligned}

Why check the interval? A maximum of the algebraic expression outside [0,2][0,2] would say nothing about this density.

Answer. The mode is 11, where the density has its unique maximum.

Check. The density rises before 11 and falls after 11; its total area is 34[x2−x33]02=1\frac34[x^2-\frac{x^3}{3}]_0^2=1.

Worked example 2 changes the factor 2−x2-x to 4−x4-x. Its normalising coefficient changes accordingly; the support stays fixed.

Worked example 2: the maximum moves to an endpoint

Question. Find the mode of

f(x)=316x(4−x),0≤x≤2f(x)=\frac3{16}x(4-x),\qquad 0\le x\le2

with density zero elsewhere. Give a reason.

Plan. Find the derivative's sign throughout the support and inspect the upper endpoint.

f′(x)=316(4−2x)differentiatef′(x)>0for 0<x<2density rises throughoutf(2)=316×2×2=34greatest height\begin{aligned} f'(x)&=\frac3{16}(4-2x) && \text{differentiate}\\ f'(x)&>0 \quad\text{for }0<x<2 && \text{density rises throughout}\\ f(2)&=\frac3{16}\times2\times2=\frac34 && \text{greatest height} \end{aligned}

Why can an endpoint be the mode? The density never turns down within the permitted interval; its greatest height occurs at the last permitted value.

Answer. The mode is 22, because the density is increasing up to that boundary.

Check. The stationary point x=2x=2 agrees with the endpoint comparison, and 316[2x2−x33]02=1\frac3{16}[2x^2-\frac{x^3}{3}]_0^2=1.

A density with no interior stationary point can still have a mode. A zero point probability at its peak does not prevent that peak describing where nearby intervals have greatest density.

Watch out

Wrong: every solution of f′(x)=0f'(x)=0 is a mode. Right: reject values outside the support and distinguish maxima from minima. Check endpoints and joins too.

Increase the same shape parameter again.

Percentiles cut off areas, not heights

The median leaves half the probability on each side. A percentile changes that fraction; the density height at the cut is irrelevant.

Median and percentile

A median mm satisfies P(X≤m)=12\mathrm P(X\le m)=\frac12 for a continuous distribution. The percentile numbered 100p100p, denoted by qpq_p, cuts off probability pp to its left, where 0<p<10<p<1. The lower and upper quartiles use p=14p=\frac14 and p=34p=\frac34.

The defining probability immediately gives the calculation method.

Quantiles from the CDF

F(m)=12,F(qp)=pF(m)=\frac12,\qquad F(q_p)=p

Choose the branch containing the required accumulated probability, solve, and reject roots outside that branch. An upper-tail probability rr corresponds to F(q)=1−rF(q)=1-r.

Not in the formula booklet: learn it.

If a CDF is flat at the target probability, more than one cut can satisfy the equation; a stated percentile convention may then select one. The strictly increasing branches used below give unique answers.

We will use the triangular CDF from the preceding density lesson:

F(x)={0x<0x220≤x≤11−(2−x)221<x≤21x>2F(x)= \begin{cases} 0 & x<0\\ \frac{x^2}{2} & 0\le x\le1\\ 1-\frac{(2-x)^2}{2} & 1<x\le2\\ 1 & x>2 \end{cases}

Its value at the join is 12\frac12. That tells you which formula to use before you solve anything.

Worked example 3: a median and an upper quartile

Question. Find the median and upper quartile of this triangular distribution.

Plan. Locate probability 12\frac12 at the join; for probability 34\frac34, use the upper branch and retain its valid root.

F(1)=12half the area is already accumulatedm=1median1−(2−q)22=34upper quartile is above 1(2−q)2=12rearrangeq=2±12solve the squareq=2−12only this root lies in (1,2)\begin{aligned} F(1)&=\frac12 && \text{half the area is already accumulated}\\ m&=1 && \text{median}\\ 1-\frac{(2-q)^2}{2}&=\frac34 && \text{upper quartile is above }1\\ (2-q)^2&=\frac12 && \text{rearrange}\\ q&=2\pm\frac1{\sqrt2} && \text{solve the square}\\ q&=2-\frac1{\sqrt2} && \text{only this root lies in }(1,2) \end{aligned}

Why use the second branch? A probability greater than 12\frac12 cannot be reached on the first branch, whose maximum is 12\frac12.

Answer. The median is 11; the upper quartile is 2−122-\frac1{\sqrt2}.

Check. 2−q=1/22-q=1/\sqrt2, so F(q)=1−12(12)=34F(q)=1-\frac12(\frac12)=\frac34. The other root exceeds the support endpoint 22.

Continue this lesson.

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