Continuous random variables

Continuous random variables and density functions Paper 6

A continuous random variable can take any value in an interval: a waiting time, a length, a mass. There are infinitely many possible values, so we cannot list probabilities for each one as we do for a discrete variable. Instead the distribution is described by a probability density function (pdf) $\mathrm{f}(x)$, and
$$\mathrm{P}(a<X<b)={\int }_a^b\mathrm{f}(x)\mathrm{d}x,$$
the area under the graph of $\mathrm{f}$ between $a$ and $b$.

A function can be a pdf only if

• $\mathrm{f}(x)⩾0$ for every $x$, because an area representing a probability cannot be negative;
• the total area under the graph is 1: $\int \mathrm{f}(x)\mathrm{d}x=1$ over the whole interval where $\mathrm{f}$ is defined.

Outside its stated interval, $\mathrm{f}(x)=0$. Two consequences surprise students. First, $\mathrm{P}(X=a)=0$ for any single value $a$, because a single value has no width and so no area. So $\mathrm{P}(X<2)$ and $\mathrm{P}(X⩽2)$ are the same. Second, $\mathrm{f}(x)$ itself is not a probability: it is a density, probability per unit of $x$, and it may be greater than 1 when the interval is short.

In this syllabus a density is defined by one formula over a single interval. The cumulative distribution function is not required: every probability, median and percentile is found directly from an area.

The density in Worked examples 1 and 2. The shaded area is $\mathrm{P}(X<1)$. The curve is symmetrical about $x=1.5$, which is therefore the mean.

Finding an unknown constant Paper 6

A pdf is often given with an unknown constant, such as $\mathrm{f}(x)=kx(3-x)$ for $0⩽x⩽3$. Use the total area: set the integral over the whole interval equal to 1 and solve for $k$. Then check that $\mathrm{f}(x)⩾0$ on the interval, which is why a question may restrict $k$ to be positive.

Papers 1 and 3 techniques apply: expand brackets before integrating, write $\frac{1}{x^3}$ as $x^{-3}$, and show the limits being substituted. Most pdf questions can be done exactly, with fractions; keep them exact until the final answer.

Densities on an infinite domain Paper 6

The interval may have no upper end. The syllabus example is $\mathrm{f}(x)=\frac{3}{x^4}$ for $x⩾1$. The area is an improper integral, found by letting the upper limit tend to infinity:
$${\int }_1^{\infty }3x^{-4}\mathrm{d}x=[-x^{-3}{]}_1^{\infty }=0-(-1)=1,$$
because $x^{-3}\to 0$ as $x\to \infty$. So it is a valid pdf, and, for example, $\mathrm{P}(X>2)=[-x^{-3}{]}_2^{\infty }=\frac{1}{8}$.

For a mean or variance on an infinite domain the integral must also have a finite value. With $\mathrm{f}(x)=\frac{3}{x^4}$, ${\int }_1^{\infty }x·3x^{-4}\mathrm{d}x=\frac{3}{2}$ is finite. A density that falls away too slowly, such as $\frac{1}{x^2}$ for $x⩾1$, has no finite mean, because ${\int }_1^{\infty }\frac{1}{x}\mathrm{d}x$ grows without limit.

The mean and variance Paper 6

For a discrete variable, $\mathrm{E}(X)=\sum xp$. For a continuous variable, the sum becomes an integral and the probability $p$ becomes $\mathrm{f}(x)\mathrm{d}x$:
$$\mathrm{E}(X)=\int x\mathrm{f}(x)\mathrm{d}x,\mathrm{Var}(X)=\int x^2\mathrm{f}(x)\mathrm{d}x-\{\mathrm{E}(X){\}}^2,$$
each integral taken over the interval where $\mathrm{f}$ is defined. Both formulae are in MF19.

Two checks catch most slips. The mean must lie inside the interval, and a variance must be positive. If the graph of $\mathrm{f}$ is symmetrical about $x=c$, then $\mathrm{E}(X)=c$ with no integration needed; a question may say "write down" the mean in such a case. Do not forget to subtract $\{\mathrm{E}(X){\}}^2$: $\int x^2\mathrm{f}(x)\mathrm{d}x$ alone is $\mathrm{E}(X^2)$, not the variance.

The median and other percentiles Paper 6

The median $m$ splits the area in half: it is the value with
$${\int }_a^m\mathrm{f}(x)\mathrm{d}x=\frac{1}{2},$$
where $a$ is the lower end of the interval. In the same way, the lower quartile has area $\frac{1}{4}$ to its left, the upper quartile $\frac{3}{4}$, and the 90th percentile $0.9$. Integrate from the lower end to the unknown value, set the result equal to the area you want, and solve.

The equation is often a quadratic or a cubic. It usually has more than one solution, and only one lies inside the interval: give that one, and say why the others are rejected. For a symmetrical density the median equals the mean. When the median is greater than the mean, the distribution has a longer tail to the left (negative skew); when it is less, the tail is to the right.