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Exponential distribution

The waiting time until the next event in a Poisson process; a memoryless lifetime.

Parameters and support

$\lambda$

rate — events per unit time — $\lambda >0$

Support

$x\ge 0$

The formulas

f(x)

$$f(x)=\lambda e^{-\lambda x}$$

F(x)

$$F(x)=1-e^{-\lambda x}$$

The survival function S(x) = e^(−λx) is the form worth memorising — most questions ask for a tail.

Mean

$$E[X]=\frac{1}{\lambda }$$

Variance

$$\mathrm{Var}(X)=\frac{1}{{\lambda }^2}$$

MGF

$$M(t)=\frac{\lambda }{\lambda -t},t<\lambda$$

Where the moments come from

1. $S(x)=P(X>x)=e^{-\lambda x}$, so $F(x)=1-e^{-\lambda x}$ and $f(x)=\lambda e^{-\lambda x}$.
2. $E[X]={\int }_0^{\infty }S(x)dx={\int }_0^{\infty }{e}^{-\lambda x}dx=\frac{1}{\lambda }$.
3. $E[X^2]={\int }_0^{\infty }2xS(x)dx=\frac{2}{{\lambda }^2}$, so $\mathrm{Var}(X)=\frac{2}{{\lambda }^2}-\frac{1}{{\lambda }^2}=\frac{1}{{\lambda }^2}$.
4. Memorylessness follows straight from $e^{-\lambda (s+t)}=e^{-\lambda s}{e}^{-\lambda t}$.
5. The exponential is Gamma($\alpha =1$, $\lambda$); a sum of $n$ independent Exponential($\lambda$) is Gamma($n$, $\lambda$).

Worked example

The mean, variance, CDF and moment generating function above are re-derived numerically from this distribution’s own density on every test run — summed over the support for a discrete distribution, integrated by quadrature for a continuous one — and compared with the closed forms printed here. A typo on this page fails the build.

Traps

• Using λ where the mean θ = 1/λ belongs. Read whether the question gives a RATE or a MEAN.
• Doing the conditional probability the long way when memorylessness collapses it in one line.
• Confusing the exponential (a waiting time) with the Poisson (a count).

Related

• Poisson distribution
• Gamma distribution
• Geometric distribution
• Pareto distribution

Drill this: the Exam P question bank has original questions on this distribution, and the recall trainer builds its prompts from exactly the formulas above.