Probability density functionsAQA A-Level Further Maths: Revision notes

A discrete random variable takes separate values (for example the score on a die), and each value has a probability $P(X=x)$. A continuous random variable can take any value in an interval (for example a time or a mass), so we cannot list the values. Instead a continuous random variable is described by a probability density function (pdf) $f(x)$. Probability is the area under the curve, not the height. Because a single value has no width, $P(X=a)=0$ for every $a$, and therefore $P(X<a)=P(X\le a)$.

A function $f$ is a pdf if both conditions hold:

1. $f(x)\ge0$ for all $x$;
2. the total area is 1: $\int_{-\infty}^{\infty}f(x)\,dx=1$. A pdf is usually given in pieces and is $0$ outside the stated range, so only integrate over the range where $f$ is non-zero. If $f$ contains an unknown constant $k$, integrate over the range, set the result equal to 1 and solve. For $f(x)=kx^2$ on $[0,3]$: $\int_0^3kx^2\,dx=9k=1$, so $k=\frac19$.

The probability that $X$ lies between $a$ and $b$ is the area under $f$ between them: $P(a<X<b)=\int_a^bf(x)\,dx.$ Since $P(X=a)=0$, it makes no difference whether the inequalities are strict. For a tail, either integrate to the end of the range or use $P(X>a)=1-P(X<a)$. Example: $f(x)=\frac{x^2}{9}$ on $[0,3]$ gives $P(1<X<2)=\left[\frac{x^3}{27}\right]_1^2=\frac{8-1}{27}=\frac{7}{27}$. When $f$ is defined in pieces, split the integral at the point where the formula changes.

The median $m$ is the value with half the probability below it: $\int_{-\infty}^{m}f(x)\,dx=\frac12.$ The lower quartile $Q_1$ satisfies $\int_{-\infty}^{Q_1}f(x)\,dx=\frac14$ and the upper quartile $Q_3$ satisfies $\int_{-\infty}^{Q_3}f(x)\,dx=\frac34$. The interquartile range is $Q_3-Q_1$. Integrate from the lower end of the range (since $f=0$ below it), set the area equal to $\frac12$, $\frac14$ or $\frac34$ and solve. Reject any root outside the range of $X$.

$f(x)=\frac{x}{18}$ for $0\le x\le6$, and $0$ otherwise. Check: $\int_0^6\frac{x}{18}\,dx=\left[\frac{x^2}{36}\right]_0^6=1$. $P(X<3)=\frac{9}{36}=\frac14$, and $P(X=4)=0$. Median: $\frac{m^2}{36}=\frac12$, so $m^2=18$ and $m=3\sqrt2\approx4.24$. Lower quartile: $\frac{q^2}{36}=\frac14$, so $q=3$.