4.1 - Piecewise probability densities | Cambridge A-Level Further Mathematics

A probability density can change formula, jump, or even rise above one. In this lesson you will decide whether a piecewise function is a valid density, calculate probabilities across its pieces, and find expectations and variances of functions of a continuous random variable.

A valid density

Can a probability density be taller than one without giving an impossible probability? Yes: a density gives probability through area, rather than through its height at one point.

You already know from continuous distributions that an interval probability is an integral. For example, a rectangle of width $1/2$ and height $2$ has area $1$ . Its height exceeds one, but it can describe a distribution concentrated in a short interval.

Probability density function

A probability density function, or PDF, for a continuous random variable $X$ is a nonnegative function $f$ with total area one:

f(x)\ge0, \qquad \int_{-\infty}^{\infty}f(x)\,\mathrm dx=1.

Then $\mathrm P(a<X<b)=\int_a^b f(x)\,\mathrm dx$ . The support identifies where the distribution can take values; give the intervals and state that the density is zero elsewhere.

A piecewise definition tells you which expression to use on each interval. Check nonnegativity on every piece, then add their areas. The formulae need not join at the same height: a density may have a jump.

Worked example 1: normalising two pieces

Question. Find $k$ so that the following is a PDF:

f(x)= \begin{cases} kx&0\le x\le1\\ k&1<x\le3\\ 0&\text{otherwise}. \end{cases}

Plan. Integrate each nonzero piece, make their sum one, then check the sign.

\begin{aligned} 1&=\int_0^1 kx\,\mathrm dx+\int_1^3 k\,\mathrm dx\\ &=\frac{k}{2}+2k=\frac{5k}{2}\\ k&=\frac25. \end{aligned}

Both $x$ on the first interval and the constant on the second interval are nonnegative when $k=2/5$ .

Answer. $k=2/5$ , giving a valid density on $[0,3]$ .

Check. Geometry gives a triangle of area $\tfrac12(1)(2/5)=1/5$ and a rectangle of area $2(2/5)=4/5$ . Their total is one.

The next density retains a rising first piece but replaces the finite second piece by a decreasing tail. An infinite support does not require an infinite area.

Worked example 2: an unbounded support

Question. Find $k$ so that

f(x)= \begin{cases} kx&0\le x\le1\\ kx^{-5}&x>1\\ 0&x<0 \end{cases}

is a PDF.

Plan. Use an improper integral for the tail and check nonnegativity.

\begin{aligned} \int_1^\infty kx^{-5}\,\mathrm dx &=\lim_{b\to\infty}\left[-\frac{k}{4x^4}\right]_1^b =\frac{k}{4}\\ 1&=\frac{k}{2}+\frac{k}{4}=\frac{3k}{4}\\ k&=\frac43. \end{aligned}

Both nonzero expressions are nonnegative for this value of $k$ .

Answer. $k=4/3$ , with support $[0,\infty)$ .

Check. Substituting $u=1/x$ gives the tail area as $\tfrac43\int_0^1u^3\,\mathrm du=1/3$ . The first piece is a triangle of area $2/3$ , confirming total one by a different calculation. The maximum density is $4/3$ ; a height above one causes no problem.

Total signed integral one is insufficient if some of the curve lies below the axis. For example, the function $2x$ on $[0,1]$ , $2(2-x)$ on $(1,3]$ , and zero elsewhere has integral one, but is negative for $2<x\le3$ . It cannot be a PDF.

Watch out

Do not check only $\int f(x)\,\mathrm dx=1$ . A negative piece would assign a negative probability to some interval. Check both nonnegativity and total area.

Apply both conditions to a new pair of pieces.

Intervals across pieces

A probability interval may cross a point where the density changes formula, so split the integral at every such point inside the interval.

Return to the density from Worked example 1:

f(x)= \begin{cases} \frac25x&0\le x\le1\\ \frac25&1<x\le3\\ 0&\text{otherwise}. \end{cases}

The shaded area in the diagram crosses $x=1$ . Its two parts need different integrands.

[DIAGRAM: asset_name: An interval crossing two density pieces; asset_slug: cam9231-piecewise-area; caption: Split the shaded probability at x equals 1, where the density changes formula; description: Exact PDF f(x) equals two fifths x on zero to one and two fifths on one to three, with zero elsewhere. Pale blue shading from one half to one has area three twentieths; pale lavender shading from one to two has area two fifths. Dashed guides mark one half, one and two, and the total shaded probability is eleven twentieths.]
Split the shaded probability at x equals 1, where the density changes formula

For any single value $a$ , the integral from $a$ to itself is zero. Thus $\mathrm P(X=a)=0$ , and including or excluding individual endpoints does not change an interval probability. Changing a density's value at an isolated endpoint also leaves all its probabilities unchanged.

Worked example 3: staying inside one piece

Question. For this density, find $\mathrm P(1/4<X<3/4)$ .

Plan. Both limits lie below one, so use only the rising piece.

\begin{aligned} \mathrm P\left(\frac14<X<\frac34\right) &=\int_{1/4}^{3/4}\frac25x\,\mathrm dx\\ &=\left[\frac{x^2}{5}\right]_{1/4}^{3/4}\\ &=\frac15\left(\frac9{16}-\frac1{16}\right)=\frac1{10}. \end{aligned}

Answer. $\mathrm P(1/4<X<3/4)=1/10$ .

Check. The region is a trapezium of width $1/2$ , with parallel sides $1/10$ and $3/10$ . Its area is $\tfrac12(1/2)(1/10+3/10)=1/10$ .

Now change the interval so that its upper end lies in the constant piece. The probability is the sum of two areas, not one formula evaluated across the whole interval.

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4.1 - Piecewise probability densities

A valid density

Intervals across pieces

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