Cumulative distribution functionsAQA A-Level Further Maths: Revision notes

The cumulative distribution function (cdf) of a continuous random variable $X$ is $F(x)=P(X\le x)=\int_{-\infty}^{x}f(t)\,dt.$ It gives the probability that $X$ is at most $x$. Properties: $F$ is non-decreasing, $F(x)\to0$ as $x\to-\infty$ and $F(x)\to1$ as $x\to\infty$, so $0\le F(x)\le1$. For a continuous variable $F$ has no jumps.

Integrate $f$ from the bottom of its range up to $x$, using a dummy variable $t$: $F(x)=\int_{a}^{x}f(t)\,dt\quad\text{for }a\le x\le b,$ with $F(x)=0$ below the range and $F(x)=1$ above it. For a pdf in pieces, add the total area of the earlier pieces to the integral over the current piece, so $F$ is continuous at each join. Example: $f(t)=3t^{-4}$ for $t\ge1$ gives $F(x)=\left[-t^{-3}\right]_1^x=1-x^{-3}$.

Differentiate each piece: $f(x)=\frac{dF}{dx}.$ Outside the range $f(x)=0$. If $F(x)=\frac{3x^2-x^3}{4}$ on $0\le x\le2$, then $f(x)=\frac{6x-3x^2}{4}$ on that range. An unknown constant in $F$ is found from continuity or from $F=1$ at the top of the range: if $F(x)=k(x^2+2x)$ on $0\le x\le2$ then $F(2)=8k=1$, so $k=\frac18$.

Probabilities come straight from $F$ with no integration: $P(a<X<b)=F(b)-F(a),\qquad P(X>a)=1-F(a).$ The median $m$ satisfies $F(m)=\frac12$. The quartiles satisfy $F(Q_1)=\frac14$ and $F(Q_3)=\frac34$. When $F$ is in pieces, compare the target ($\frac12$, $\frac14$, $\frac34$) with the value of $F$ at the join to choose which piece to solve in. Then check the root lies in that piece.

$F(x)=\frac{x^2}{6}$ for $0\le x\le2$ and $\frac x3$ for $2<x\le3$. $P(1<X<2.5)=\frac{2.5}{3}-\frac16=\frac23$. Median: $F(2)=\frac23>\frac12$, so $\frac{m^2}{6}=\frac12$ and $m=\sqrt3$. Upper quartile: $F(2)=\frac23<\frac34$, so $\frac u3=\frac34$ and $u=\frac94$.