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Validate and normalise a continuous probability density function

Start a set of 8
Step 5 of 5

Checking area without non-negativity. One of these lines is right. Which one?

Year 12, VCE Units 3 & 4, Chapter 12, 4 difficulty levels, No calculatorBuild a worksheet

Properties of a probability density

A density assigns probability through area. It must be non-negative on its support, have total area 1, and be zero outside that support

\[f(x)\ge0,\quad\int_a^b f(x)\,dx=1,\quad f(x)=0\text{ outside }[a,b]\]
  1. Check non-negativity The density must be at least zero at every point of its declared support \(\displaystyle f(x)\ge0\quad\text{for }a\le x\le b\)
  2. Check total area The area under the density across its complete support must be one \(\displaystyle \int_a^b f(x)\,dx=1\)
  3. State zero elsewhere A density is zero outside its declared support \(\displaystyle f(x)=0\quad\text{for }x<a\text{ or }x>b\)
  4. Normalise with the complete area Keep the unknown constant in the integral, set the total area equal to one and solve \(\displaystyle \int_0^2kx\,dx=2k=1\;\Rightarrow\;k=\frac{1}{2}\)

Worked example

Find the normalisation constant and check the density conditions for this proposed rule

\[f(x)=kx,\quad0\le x\le3,\quad f(x)=0\text{ elsewhere}\]
  1. \(\displaystyle \int_0^3kx\,dx=k\left[\frac{x^2}{2}\right]_0^3=\frac{9k}{2}\)Integrate the rule over the complete support and keep k as a factor
  2. \(\displaystyle \frac{9k}{2}=1\;\Rightarrow\;k=\frac{2}{9}\)Set the complete area equal to 1 and solve for the positive constant
  3. \(\displaystyle f(x)=\frac{2x}{9}\ge0\quad\text{for }0\le x\le3\)The normalised rule is non-negative throughout its support
  4. \(\displaystyle f(x)=0\quad\text{outside }[0,3]\)The zero-elsewhere statement supplies the remaining density condition

Common mistakes

Checking area without non-negativity
✕\(\displaystyle f(x)=3x-\frac{1}{2},\quad\int_0^1f(x)\,dx=1\;\Rightarrow\;\text{valid}\)✓\(\displaystyle f(0)=-\frac{1}{2}<0\;\Rightarrow\;\text{not valid}\)
A signed integral equal to 1 is not enough; the density must also be non-negative on its support
Dropping the normalisation constant
✕\(\displaystyle \int_0^2kx\,dx=\int_0^2x\,dx=2\)✓\(\displaystyle \int_0^2kx\,dx=k\int_0^2x\,dx=2k\)
The constant multiplies the whole rule, so it remains in the area calculation
Keeping a negative value for k
✕\(\displaystyle k=-\frac{1}{2}\)✓\(\displaystyle k=\frac{1}{2}\)
The chosen constant must keep the density non-negative on its support, so a negative value is rejected

Difficulty levels

introCheck a simple linear density on a bounded interval
standardNormalise a piecewise-linear density or identify one failed density condition
challengeNormalise an improper exponential or circular density and reject an invalid root
examChoose the valid or invalid density reason and calculate a normalisation consequence

Questions about this topic

Why must the total area under a density be 1?+

The total area represents the probability that the random variable lies somewhere in its support. That event has probability 1

Can a function with area 1 be a density if it is negative?+

No. Probability cannot be negative, so the density must be non-negative at every point of its support as well as having total area 1

Why is the constant kept inside the integral when normalising?+

The constant scales every density value and therefore scales the complete area. It must remain in the area equation before that equation is set equal to 1

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 12: Continuous probability distributions

In this chapter

12A Continuous random variables and density functions: Validate and normalise a continuous probability density function12A Continuous random variables and density functions: Estimate probabilities from grouped continuous frequency dataGroup numerical data into class intervalsConstruct a histogram from a frequency table12A Continuous random variables and density functions: Finite events, probability assignments and complements12A Continuous random variables and density functions: Conditional probability from exact representationsEvaluate and determine function rules12B Probability density validity and normalisation: Validate and normalise a continuous probability density function12B Probability density validity and normalisation: Estimate probabilities from grouped continuous frequency dataSolve a linear equation12B Probability density validity and normalisation: Evaluate a polynomial or recover a coefficient parameterWork with sets and interval notation12B Probability density validity and normalisation: Round to decimal places and significant figuresInterval and conditional probabilities from a densityCompound events, the addition rule and mutual exclusivity12C Interval and conditional probabilities: Finite events, probability assignments and complements12C Interval and conditional probabilities: Conditional probability from exact representationsMean, variance and standard deviation of a continuous random variableMedian, quartiles and percentiles of a continuous random variableExpand and factor polynomial expressions12D Mean, variance, standard deviation and percentiles: Evaluate a polynomial or recover a coefficient parameterMove between roots, factors, polynomial rules and equationsSolve a quadratic exactly with the quadratic formula12D Mean, variance, standard deviation and percentiles: Round to decimal places and significant figuresLinear transformations and totals of independent continuous random variables