Probability functions and cumulative distribution functionsEdexcel International A Level Maths: Revision notes

A random variable $X$ is a variable whose value depends on the outcome of a random experiment. It is discrete if it takes separate values, usually whole numbers, such as the score on a die or the number of defects in a batch. A capital letter, $X$, names the variable; a lower-case letter, $x$, is a particular value. $P(X=x)$ means 'the probability that $X$ takes the value $x$'. The possible values and their probabilities form the probability distribution of $X$.

The probability function is $p(x)=P(X=x)$. It can be given as a table, a list or a formula such as $P(X=x)=kx$ for $x=1,2,3,4$. Two rules always hold:

• $0\le p(x)\le1$ for every $x$;
• $\sum p(x)=1$ over all possible values. Use the second rule to find an unknown constant. Example: $P(X=x)=kx$ gives $k(1+2+3+4)=10k=1$, so $k=0.1$ and $P(X\ge3)=0.3+0.4=0.7$. For $P(X=x)=k(x^2+1)$, $x=0,1,2,3$: $k(1+2+5+10)=1$, $k=\frac1{18}$.

The cumulative distribution function is $F(x_0)=P(X\le x_0)=\sum_{x\le x_0}p(x).$ It adds up the probabilities from the smallest value up to $x_0$. Example: for the probabilities $0.1,0.2,0.3,0.4$ on $x=1,2,3,4$: $F(1)=0.1$, $F(2)=0.3$, $F(3)=0.6$, $F(4)=1$. $F$ never decreases, and $F$ for the largest value is 1. For a value that is not possible, use the next lower possible value: if $X$ takes whole numbers, $F(2.5)=F(2)$.

Convert each probability into differences of $F$ (for whole-number values):

• $P(X=x)=F(x)-F(x-1)$
• $P(X>a)=1-F(a)$
• $P(X\ge a)=1-F(a-1)$
• $P(a<X\le b)=F(b)-F(a)$ Example: if $F(1)=0.15$, $F(2)=0.40$, $F(3)=0.75$, $F(4)=1$ then $P(X=3)=0.75-0.40=0.35$ and $P(X\ge2)=1-F(1)=0.85$. To go the other way, add the probabilities cumulatively: $p(x)$ gives $F(x)$.

To find a distribution from a situation, list the possible values, count or calculate each probability, and check the total is 1. Example: a fair four-sided die rolled twice and $X$ the larger score. Of the 16 outcomes, $X=3$ for 5, so $P(X=3)=\frac5{16}$, and in general $P(X=x)=\frac{2x-1}{16}$ for $x=1,2,3,4$. A shortcut: $P(X\le x)=\left(\frac x4\right)^2$, so $F(x)=\frac{x^2}{16}$. For three rolls, $F(x)=\frac{x^3}{64}$ and $P(X=3)=\frac{27}{64}-\frac8{64}=\frac{19}{64}$. Working with $F$ is often quicker for 'largest' or 'smallest' questions.