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

A continuous random variable can take any value in an interval, such as a time, a mass or a length, so it cannot be listed value by value. Probabilities are found as areas, not heights. A key consequence: for a continuous variable $P(X=a)=0$ for every single value $a$. So $P(a<X<b)$, $P(a\le X\le b)$ and $P(a<X\le b)$ are all equal. You can ignore whether the inequality is strict.

The probability density function (pdf) $f(x)$ describes how probability is spread out. It must satisfy

• $f(x)\ge0$ for all $x$,
• the total area under the curve is 1: $\int_{-\infty}^{\infty}f(x)\,dx=1$. Probabilities come from integration: $P(a<X\le b)=\int_a^bf(x)\,dx.$ Here $f(x)$ is restricted to simple polynomials, often defined piecewise (a different expression on each interval, and 0 outside). To find an unknown constant $k$, set the total area equal to 1.

The cumulative distribution function (cdf) gives the probability of being at or below a value: $F(x_0)=P(X\le x_0)=\int_{-\infty}^{x_0}f(x)\,dx.$ It rises from 0 to 1 and never decreases. Then $P(a<X\le b)=F(b)-F(a)$, and $P(X>a)=1-F(a)$. To find $F(x)$ from $f(x)$, integrate from the lower end of the range to $x$ (or integrate and fix the constant using $F=0$ at the start). For a piecewise pdf, the cdf for a later piece must include the total probability of the earlier pieces.

The pdf is the derivative of the cdf, on any interval where it is differentiable: $f(x)=\frac{dF(x)}{dx}.$ Differentiate $F$ to get $f$ and integrate $f$ to get $F$. Example: if $F(x)=\frac{x^3}{27}$ for $0\le x\le3$ then $f(x)=\frac{x^2}{9}$ on the same interval, and $f(x)=0$ outside it. Worked example: $f(x)=\frac{x}{2}$ for $0\le x\le2$. Then $F(x)=\frac{x^2}{4}$, and $P(1<X<1.5)=F(1.5)-F(1)=\frac{9}{16}-\frac{1}{4}=\frac{5}{16}$.