Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PROBABILITY

IfXandYare both discrete RVs then

p(z)=

∑

i,j

f(xi,yj), (30.60)

where the sum extends over all values ofiandjfor whichZ(xi,yj)=z. Similarly,

ifXandYare both continuous RVs thenp(z) is found by requiring that

p(z)dz=

∫∫

dS

f(x, y)dx dy, (30.61)

wheredS is the infinitesimal area in thexy-plane lying between the curves

Z(x, y)=zandZ(x, y)=z+dz.

SupposeXandYare independent continuous random variables in the range−∞to∞,

with PDFsg(x)andh(y)respectively. Obtain expressions for the PDFs ofZ=X+Yand

W=XY.

SinceXandYare independent RVs, their joint PDF is simplyf(x, y)=g(x)h(y). Thus,
from (30.61), the PDF of the sumZ=X+Yis given by

p(z)dz=

∫∞

−∞

dx g(x)

∫z+dz−x

z−x

dy h(y)

=

(∫∞

−∞

g(x)h(z−x)dx

)

dz.

Thusp(z)istheconvolutionof the PDFs ofgandh(i.e.p=g∗h, see subsection 13.1.7).
In a similar way, the PDF of the productW=XYis given by

q(w)dw=

∫∞

−∞

dx g(x)

∫(w+dw)/|x|

w/|x|

dy h(y)

=

(∫∞

−∞

g(x)h(w/x)

dx

|x|

)

dw

The prescription (30.61) is readily generalised to functions ofnrandom variables

Z=Z(X 1 ,X 2 ,...,Xn), in which case the infinitesimal ‘volume’ elementdSis the

region inx 1 x 2 ···xn-space between the (hyper)surfacesZ(x 1 ,x 2 ,...,xn)=zand

Z(x 1 ,x 2 ,...,xn)=z+dz. In practice, however, the integral is difficult to evaluate,

since one is faced with the complicated geometrical problem of determining the

limits of integration. Fortunately, an alternative (and powerful) technique exists

for evaluating integrals of this kind. One eliminates the geometrical problem by

integrating overallvalues of the variablesxiwithoutrestriction, while shifting

the constraint on the variables to the integrand. This is readily achieved by

multiplying the integrand by a function that equals unity in the infinitesimal

regiondSand zero elsewhere. From the discussion of the Dirac delta function in

subsection 13.1.3, we see thatδ(Z(x 1 ,x 2 ,...,xn)−z)dzsatisfies these requirements,

and so in the most general case we have

p(z)=

∫∫

···

∫

f(x 1 ,x 2 ,...,xn)δ(Z(x 1 ,x 2 ,...,xn)−z)dx 1 dx 2 ...dxn,

(30.62)