Probability density and cumulative distribution functionsEdexcel A-Level Further Maths: Revision notes

A continuous random variable can take any value in an interval. Its probabilities are found as areas under its probability density function (pdf) $\mathrm{f}(x)$. A pdf must satisfy $\mathrm{f}(x)\geq0$ and total area $\int_{-\infty}^{\infty}\mathrm{f}(x)\,\mathrm{d}x=1$. At AS, $\mathrm{f}(x)$ has the form $kx^n$ ($n$ rational, $n\neq-1$) and may be piecewise, with different expressions on different intervals. A histogram of data with narrow class widths and frequency density on the vertical axis approaches the shape of the pdf, as does a frequency polygon, and the total area is 1 in both cases. Example: $\mathrm{f}(x)=kx^2$ on $0\leq x\leq3$: $\int_0^3kx^2\,\mathrm{d}x=9k=1$, so $k=\frac19$.

$\mathrm{P}(a<X\leq b)=\int_a^b\mathrm{f}(x)\,\mathrm{d}x.$ For a continuous variable $\mathrm{P}(X=a)=0$, so $<$ and $\leq$ give the same answer. For a piecewise pdf, split the integral at each boundary and use the correct expression in each part. Example: $\mathrm{f}(x)=\frac x6$ on $0\leq x\leq2$ and $\frac13$ on $2<x\leq4$ gives $\mathrm{P}(1<X<3)=\int_1^2\frac x6\,\mathrm{d}x+\int_2^3\frac13\,\mathrm{d}x=\frac14+\frac13=\frac7{12}$.

The cumulative distribution function (cdf) is $\mathrm{F}(x_0)=\mathrm{P}(X\leq x_0)=\int_{-\infty}^{x_0}\mathrm{f}(x)\,\mathrm{d}x.$ Integrate the pdf from the lower end of the range up to $x$, using a dummy variable $t$ inside the integral. State $\mathrm{F}$ for every region: $\mathrm{F}(x)=0$ below the range and $\mathrm{F}(x)=1$ above it. For a piecewise pdf, the cdf in later pieces must include the accumulated probability from earlier pieces. Example: $\mathrm{f}(x)=\frac{x^2}{9}$ on $0\leq x\leq3$ has $\mathrm{F}(x)=\frac{x^3}{27}$ there.

Since $\mathrm{F}$ accumulates probability, $\mathrm{P}(a<X\leq b)=\mathrm{F}(b)-\mathrm{F}(a)$ and $\mathrm{P}(X>a)=1-\mathrm{F}(a)$. A cdf is non-decreasing and runs from 0 to 1, and it is continuous for a continuous variable. To solve $\mathrm{P}(X\leq q)=p$, set $\mathrm{F}(q)=p$ and solve for $q$. Example: $\mathrm{F}(x)=\frac{\sqrt x-1}{2}$ on $1\leq x\leq9$: $\mathrm{P}(X>4)=1-\frac12=\frac12$, and $\mathrm{F}(q)=0.75$ gives $\sqrt q=2.5$, $q=6.25$.

The pdf and cdf are linked by calculus: integrating $\mathrm{f}$ gives $\mathrm{F}$, and differentiating $\mathrm{F}$ gives $\mathrm{f}$: $\mathrm{f}(x)=\frac{\mathrm{d}\mathrm{F}(x)}{\mathrm{d}x}.$ Example: $\mathrm{F}(x)=\frac{x^2-1}{8}$ on $1\leq x\leq3$ gives $\mathrm{f}(x)=\frac{2x}{8}=\frac x4$, and $\mathrm{f}(x)=0$ outside. Use this when a question gives $\mathrm{F}$ and asks for $\mathrm{f}$, remembering to state the range of $x$ for which each expression holds.