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
- Continuous rather than discrete A continuous model can take any real value in an interval, such as an unrounded waiting time. A count of arrivals takes separate whole-number values \(\displaystyle X\in[a,b]\)
- Probability is area, not height The density height is not the probability of an exact value. A single value has zero area, so including or excluding an interval endpoint does not change its probability \(\displaystyle \begin{gathered}\Pr(X=c)=0\\\Pr(a<X<b)=\Pr(a\le X\le b)\end{gathered}\)
- Check non-negativity The density must be at least zero at every point of its declared support \(\displaystyle f(x)\ge0\quad\text{for }a\le x\le b\)
- Check total area The area under the density across its complete support must be one \(\displaystyle \int_a^b f(x)\,dx=1\)
- State zero elsewhere A density is zero outside its declared support \(\displaystyle f(x)=0\quad\text{for }x<a\text{ or }x>b\)
- Normalise with the complete area Keep the unknown constant in the integral, set the total area equal to one and solve \(\displaystyle \int_0^2kx\,dx=2k=1\;\Rightarrow\;k=\frac{1}{2}\)
- Connect data to a density model For a relative-frequency histogram, divide each class proportion by its width. Each rectangle area is then its class proportion. A density can model these class areas without matching the raw bar heights \(\displaystyle \text{class area}=\text{width}\times\text{density height}\)
Worked example
The proposed density rule and complete support are shown. Find the normalisation constant and check both density conditions
- \(\displaystyle \int_0^3kx\,dx=k\left[\frac{x^2}{2}\right]_0^3\)Keep the constant as a factor while integrating over the full support
- \(\displaystyle \int_0^3kx\,dx=k\left(\frac{3^2}{2}-\frac{0^2}{2}\right)=\frac{9k}{2}\)Evaluate at the upper endpoint and subtract the lower-endpoint value
- \(\displaystyle \frac{9k}{2}=1\quad\Rightarrow\quad k=\frac29\)The complete area must equal one
- \(\displaystyle f(x)=\frac{2x}{9}\ge0\quad\text{for }0\le x\le3\)The coefficient is positive and every input on the support is non-negative
- \(\displaystyle \int_0^3\frac{2x}{9}\,dx=1\)Substitute the constant to verify the complete area
- \(\displaystyle f(x)=0\quad\text{outside }[0,3]\)The supplied zero-elsewhere rule completes the model
Common mistakes
Difficulty levels
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