Part-discrete part-continuous distributionsAQA A-Level Further Maths: Revision notes

Some random variables are neither purely discrete nor purely continuous. Typical examples are a waiting time that is exactly $0$ when there is no queue, a rainfall that is $0$ on dry days, or an insurance claim that is $0$ when no claim is made. These are part discrete, part continuous. A single value (often $0$) carries a positive probability, for example $P(X=0)=0.4$, and the remaining probability is spread continuously over an interval with a pdf $f(x)$.

The point masses and the area under $f$ must add to 1: $\sum P(X=a)+\int f(x)\,dx=1.$ So the integral of $f$ is $1-P(X=0)$ and not $1$. To find an unknown constant, integrate $f$ over its range and set the result equal to $1$ minus the point mass. Example: $P(X=0)=0.4$ and $f(x)=kx$ on $0<x\le2$. Then $\int_0^2kx\,dx=2k=0.6$, so $k=0.3$.

Split every probability into a discrete part and a continuous part. $P(X\le b)=P(X=0)+\int_0^bf(x)\,dx\quad(b>0).$ For an interval that does not include the point mass, such as $P(a<X<b)$ with $a>0$, use only the integral. Because $P(X=0)>0$, strict and non-strict inequalities are no longer interchangeable at $0$: $P(X\le0)=0.4$ but $P(X<0)=0$. At any value with no point mass, nothing changes: $P(X=a)=0$. Example: $P(X\le1)=0.4+\int_0^10.3x\,dx=0.4+0.15=0.55$ and $P(X>1)=\int_1^20.3x\,dx=0.45$.

The mean is $E(X)=\sum a\,P(X=a)+\int xf(x)\,dx$. A point mass at $0$ adds $0\times P(X=0)=0$, so in these examples $E(X)=\int xf(x)\,dx$, where $f$ already carries the total probability $1-P(X=0)$. A median or quartile $q$ satisfies $P(X\le q)=\frac12$, $\frac14$ or $\frac34$. If $P(X=0)$ is bigger than that probability, the value is $0$. Otherwise solve $P(X=0)+\int_0^qf(x)\,dx=\frac12$ etc., and reject roots outside the range. Example: $P(X=0)=0.7$ means the median and lower quartile are both $0$; the upper quartile solves $0.7+\frac{x^3}{90}=0.75$.

$P(R=0)=0.6$ and $f(r)=0.05r$ on $0<r\le4$ (total $0.4$). $P(R<2)=0.6+\int_0^20.05r\,dr=0.6+0.1=0.7$. $E(R)=0\times0.6+\int_0^4r\cdot0.05r\,dr=\frac{16}{15}$. $P(R>3)=\int_3^40.05r\,dr=0.175$.