Mathematical Methods for Physics and Engineering : A Comprehensive Guide

30.9 IMPORTANT CONTINUOUS DISTRIBUTIONS

In fact, almost all probability distributions tend towards a Gaussian when the

numbers involved become large – that this should happen is required by the

central limit theorem, which we discuss in section 30.10.

Multiple Gaussian distributions

SupposeXandYareindependentGaussian-distributed random variables, so

thatX∼N(μ 1 ,σ^21 )andY∼N(μ 2 ,σ 22 ). Let us now consider the random variable

Z=X+Y. The PDF for this random variable may be found directly using

(30.61), but it is easier to use the MGF. From (30.114), the MGFs ofXandY

are

MX(t)=exp

(

μ 1 t+^12 σ^21 t^2

)

,MY(t)=exp

(

μ 2 t+^12 σ 22 t^2

)

.

Using (30.89), sinceXandYare independent RVs, the MGF ofZ=X+Yis

simply the product ofMX(t)andMY(t). Thus, we have

MZ(t)=MX(t)MY(t)=exp

(

μ 1 t+^12 σ 12 t^2

)

exp

(

μ 2 t+^12 σ^22 t^2

)

=exp

[

(μ 1 +μ 2 )t+^12 (σ^21 +σ^22 )t^2

]

,

which we recognise as the MGF for a Gaussian with meanμ 1 +μ 2 and variance

σ^21 +σ^22. Thus,Zis also Gaussian distributed:Z∼N(μ 1 +μ 2 ,σ^21 +σ 22 ).

A similar calculation may be performed to calculate the PDF of the random

variableW=X−Y. If we introduce the variableY ̃=−YthenW=X+Y ̃,

whereY ̃ ∼N(−μ 1 ,σ^21 ). Thus, using the result above, we findW ∼N(μ 1 −

μ 2 ,σ^21 +σ 22 ).

An executive travels home from her office every evening. Her journey consists of a train

ride, followed by a bicycle ride. The timespent on the train is Gaussian distributed with

mean 52 minutes and standard deviation1.8minutes, while the time for the bicycle journey

is Gaussian distributed with mean 8 minutes and standard deviation2.6minutes. Assuming

these two factors are independent, estimate the percentage of occasions on which thewhole

journey takes more than 65 minutes.

We first define the random variables

X=timespentontrain,Y= time spent on bicycle,

so thatX∼N(52, (1.8)^2 )andY∼N(8, (2.6)^2 ). SinceXandYare independent, the total
journey timeT=X+Yis distributed as

T∼N(52 + 8,(1.8)^2 +(2.6)^2 )=N(60,(3.16)^2 ).

The standard variable is thus

Z=

T− 60

3. 16

,

and the required probability is given by

Pr(T>65) = Pr

(

Z>

65 − 60

3. 16

)

=Pr(Z> 1 .58) = 1− 0 .943 = 0. 057.

Thus the total journey time exceeds 65 minutes on 5.7% of occasions.