PROBABILITY
lighthouse
L beam
O
θ
coastline
y
Figure 30.8 The illumination of a coastline by the beam from a lighthouse.
A lighthouse is situated at a distanceLfrom a straight coastline, opposite a pointO, and
sends out a narrow continuous beam of light simultaneously in opposite directions. The beam
rotates with constant angular velocity. If the random variableYis the distance along the
coastline, measured fromO, of the spot that the light beam illuminates, find its probability
density function.
The situation is illustrated in figure 30.8. Since the light beam rotates at a constant angular
velocity,θis distributed uniformly between−π/2andπ/2, and sof(θ)=1/π.Now
y=Ltanθ, which possesses the single-valued inverseθ=tan−^1 (y/L), provided thatθlies
between−π/2andπ/2. Sincedy/dθ=Lsec^2 θ=L(1 + tan^2 θ)=L[1 + (y/L)^2 ], from
(30.58) we find
g(y)=
1
π
∣∣
∣
∣
dθ
dy
∣∣
∣
∣=
1
πL[1 + (y/L)^2 ]
for−∞<y<∞.
A distribution of this form is called aCauchy distributionand is discussed in subsec-
tion 30.9.5.
IfY(X) does not possess a single-valued inverse then we encounter complica-
tions, since there exist several intervals in theX-domain for whichYlies between
yandy+dy. This is illustrated in figure 30.9, which shows a functionY(X)
such thatX(Y) is a double-valued function ofY. Thus the rangeytoy+dy
corresponds toX’s being either in the rangex 1 tox 1 +dx 1 or in the rangex 2 to
x 2 +dx 2. In general, it may not be possible to obtain an expression forg(y)in
closed form, although the distribution may always be obtained numerically using
(30.57). However, a closed-form expression may be obtained in the case where
there exist single-valued functionsx 1 (y)andx 2 (y) giving the two values ofxthat
correspond to any given value ofy.Inthiscase,
g(y)dy=
∣
∣
∣
∣
∫x 1 (y+dy)
x 1 (y)
f(x)dx
∣
∣
∣
∣+
∣
∣
∣
∣
∫x 2 (y+dy)
x 2 (y)
f(x)dx
∣
∣
∣
∣,
from which we obtain
g(y)=f(x 1 (y))
∣
∣
∣
∣
dx 1
dy
∣
∣
∣
∣+f(x^2 (y))
∣
∣
∣
∣
dx 2
dy
∣
∣
∣
∣. (30.59)