Exponential distribution model, pdf and CDFAQA A-Level Further Maths: Revision notes

The exponential distribution models the continuous time (or distance) between events that occur:

• at random and independently of each other, and
• at a constant average rate $\lambda$ per unit time. Examples: the time between calls to a helpline, or between breakdowns of a machine. If events cluster, or the rate changes (for example after servicing), the model is not appropriate.

If $X$ has parameter $\lambda>0$ the probability density function is $f(x)=\lambda e^{-\lambda x},\quad x\ge0,$ and $f(x)=0$ for $x<0$. The graph starts at height $\lambda$ and decreases: short gaps are more likely than long ones. The units of $\lambda$ are 'per unit of time', so $X$ must be in the matching unit.

Integrating the pdf from $0$: $F(x)=\int_0^x\lambda e^{-\lambda u}\,du=\left[-e^{-\lambda u}\right]_0^x=1-e^{-\lambda x},\quad x\ge0.$ Hence $\mathrm{P}(X>x)=e^{-\lambda x}$. Differentiating $F$ returns $f(x)=\lambda e^{-\lambda x}$.

Use $F(x)$ or integrate $f(x)$.

• $\mathrm{P}(X<a)=1-e^{-\lambda a}$.
• $\mathrm{P}(X>a)=e^{-\lambda a}$.
• $\mathrm{P}(a<X<b)=F(b)-F(a)=e^{-\lambda a}-e^{-\lambda b}$. Worked example: calls at $\lambda=3$ per hour. $\mathrm{P}(12<\text{gap}<30\text{ minutes})$: convert to hours, $0.2$ and $0.5$, so $e^{-0.6}-e^{-1.5}=0.326$.

If a probability is given, solve for $\lambda$ using logarithms. For $\mathrm{P}(X>2)=0.3$: $e^{-2\lambda}=0.3$, so $\lambda=\frac{-\ln0.3}{2}=0.602$. To find $t$ with $\mathrm{P}(X<t)=0.9$: $e^{-\lambda t}=0.1$, so $t=\frac{\ln10}{\lambda}$. For independent gaps multiply probabilities: three gaps each longer than $a$ has probability $\left(e^{-\lambda a}\right)^3=e^{-3\lambda a}$.