SF3 - Mean and variance of waiting times

SF3 - Mean and variance of waiting times

The exponential rate determines both the average wait and its spread. You will prove the mean, variance and standard deviation using improper integrals, then connect those results to event rates in matching units.

Proving exponential moments by integration

If a detector's event rate doubles, do both its average wait and its standard deviation halve? The answer follows from integrating over every possible waiting time, including the unbounded tail.

Let T∼Exp⁡(λ)T\sim\operatorname{Exp}(\lambda) in the rate convention, with λ>0\lambda>0. Its density is λe−λt\lambda\mathrm{e}^{-\lambda t} for t≥0t\geq0, zero otherwise. The moment integrals are improper because there is no finite upper bound. Work to a finite upper limit RR, then take R→∞R\to\infty.

The boundary terms Re−λRR\mathrm{e}^{-\lambda R} and R2e−λRR^2\mathrm{e}^{-\lambda R} both tend to zero. This needs a reason: “RR times zero” does not evaluate a limit. For example, the exponential series gives

eλR≥(λR)36(R>0),\mathrm{e}^{\lambda R}\geq\frac{(\lambda R)^3}{6}\quad(R>0),

so

0≤R2e−λR≤6λ3R⟶0.0\leq R^2\mathrm{e}^{-\lambda R} \leq\frac6{\lambda^3R}\longrightarrow0.

For R≥1R\geq1, 0≤Re−λR≤R2e−λR0\leq R\mathrm{e}^{-\lambda R}\leq R^2\mathrm{e}^{-\lambda R}, so that term also tends to zero. The positivity of λ\lambda is essential.

Worked example 1: proving the mean

Question. Prove that E(T)=1/λE(T)=1/\lambda for T∼Exp⁡(λ)T\sim\operatorname{Exp}(\lambda), λ>0\lambda>0.

Plan. Integrate tλe−λtt\lambda\mathrm{e}^{-\lambda t} by parts up to RR, then use the boundary limits.

Choose u=tu=t and v′=λe−λtv'=\lambda\mathrm{e}^{-\lambda t}. Then u′=1u'=1 and v=−e−λtv=-\mathrm{e}^{-\lambda t}, so

I1(R)=∫0Rtλe−λt dt=[−te−λt]0R+∫0Re−λt dt=[−te−λt−1λe−λt]0R=1λ−(R+1λ)e−λR.\begin{aligned} I_1(R)&=\int_0^R t\lambda\mathrm{e}^{-\lambda t}\,\mathrm{d}t\\ &=[-t\mathrm{e}^{-\lambda t}]_0^R +\int_0^R\mathrm{e}^{-\lambda t}\,\mathrm{d}t\\ &=\left[-t\mathrm{e}^{-\lambda t} -\frac1\lambda\mathrm{e}^{-\lambda t}\right]_0^R\\ &=\frac1\lambda-\left(R+\frac1\lambda\right)\mathrm{e}^{-\lambda R}. \end{aligned}

Why choose u=tu=t? Differentiating it removes the polynomial factor and leaves a simple exponential integral.

Answer. Both boundary terms vanish, hence

E(T)=lim⁡R→∞I1(R)=1λ.E(T)=\lim_{R\to\infty}I_1(R)=\frac1\lambda.

The nonnegative first-moment integral converges, so the mean is finite.

Check. Differentiating −(t+1/λ)e−λt-(t+1/\lambda)\mathrm{e}^{-\lambda t} gives

−e−λt+λ(t+1/λ)e−λt=λte−λt,-\mathrm{e}^{-\lambda t} +\lambda(t+1/\lambda)\mathrm{e}^{-\lambda t} =\lambda t\mathrm{e}^{-\lambda t},

which checks the antiderivative and its sign.

Change only the power of the waiting time from tt to t2t^2. Integration by parts now reduces the calculation to the first moment just proved.

Worked example 2: proving variance and standard deviation

Question. Prove the variance and standard deviation of the same exponential variable.

Plan. Find E(T2)E(T^2) using parts, then subtract the squared mean and take a nonnegative square root.

Use u=t2u=t^2, v′=λe−λtv'=\lambda\mathrm{e}^{-\lambda t}, so u′=2tu'=2t, v=−e−λtv=-\mathrm{e}^{-\lambda t}:

I2(R)=∫0Rt2λe−λt dt=[−t2e−λt]0R+2∫0Rte−λt dt=−R2e−λR+2λI1(R).\begin{aligned} I_2(R)&=\int_0^R t^2\lambda\mathrm{e}^{-\lambda t}\,\mathrm{d}t\\ &=[-t^2\mathrm{e}^{-\lambda t}]_0^R +2\int_0^R t\mathrm{e}^{-\lambda t}\,\mathrm{d}t\\ &=-R^2\mathrm{e}^{-\lambda R}+\frac2\lambda I_1(R). \end{aligned}

The boundary term tends to zero and I1(R)→1/λI_1(R)\to1/\lambda, so

E(T2)=lim⁡R→∞I2(R)=2λ2.E(T^2)=\lim_{R\to\infty}I_2(R)=\frac2{\lambda^2}.

Therefore

Var⁡(T)=2λ2−(1λ)2=1λ2.\operatorname{Var}(T)=\frac2{\lambda^2} -\left(\frac1\lambda\right)^2=\frac1{\lambda^2}.

Answer. Variance 1/λ21/\lambda^2, standard deviation 1/λ1/\lambda. The square root is 1/λ1/\lambda, since λ>0\lambda>0.

Check. A full antiderivative of the second-moment integrand is

−e−λt(t2+2tλ+2λ2).-\mathrm{e}^{-\lambda t} \left(t^2+\frac{2t}{\lambda}+\frac2{\lambda^2}\right).

Its derivative is λt2e−λt\lambda t^2\mathrm{e}^{-\lambda t}; its limit at infinity is zero and its value at zero is −2/λ2-2/\lambda^2, independently confirming the second moment.

The surprising consequence is that an exponential variable's standard deviation equals its mean. Its variance equals the square of its mean, not generally the mean itself.

Exponential mean and spread

For T∼Exp⁡(λ)T\sim\operatorname{Exp}(\lambda) with rate λ>0\lambda>0,

E(T)=1λ,Var⁡(T)=1λ2,SD⁡(T)=1λ.E(T)=\frac1\lambda,\qquad \operatorname{Var}(T)=\frac1{\lambda^2},\qquad \operatorname{SD}(T)=\frac1\lambda.

These results are not printed in the AQA formula booklet; know their proofs. Integration by parts is supplied in the booklet.

A request to prove or to use integration requires the integral and its limiting argument, rather than a quotation of the final formula.

Watch out

Wrong: ∫f(t) dt\int f(t)\,\mathrm{d}t for the mean, or E(T2)=Var⁡(T)E(T^2)=\operatorname{Var}(T). Weight by tt for the mean and by t2t^2 for the second moment; then subtract the squared mean. Show actual limits at infinity.

Practise a short numerical-rate version of the proof.

Continue this lesson.

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