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Important continuous distributions

Solution. The situation is illustrated in Fig. 6.3. Since the light beam rotates at a constant
angular velocity, θ is distributed uniformly between −π/2 and π/2, and so f(θ) = 1/π. Now
−1
y = L tan θ, which possesses the single-valued inverse θ = tan (y/L), provided that θ lies
2
2
2
between −π/2 and π/2. Since dy/dθ = L sec θ = L(1 + tan θ) = L[1 + (x/L) ], from (6.14) we
find

1 dθ 1

g(y) = = for − ∞ < y < ∞.
2
π dy πL(1 + (y/L) )

A distribution of this form is called a Cauchy distribution.
If Y (X) does not possess a single-valued inverse then we encounter complications, since there
exist several intervals in the X-domain for which Y lies between y and y + dy. Thus the range y to
y + dy corresponds to X’s being either in the range x 1 to x 1 + dx 1 or in the range x 2 to x 2 + dx 2 .
In general, it may not be possible to obtain an expression for g(y) in closed form, although the
distribution may always be obtained numerically using (6.13). However, a closed-form expression
may be obtained in the case where there exist single-valued functions x 1 (y) and x 2 (y) giving the
two values of x that correspond to any given value of y. In this case,

∫ ∫
x 1 (y+dy) x 2 (y+dy)
f(x)dx ,

g(y)dy = f(x)dx +
x 1 (y) x 2 (y)
from which we obtain

dx 1 dx 2
g(y) = f(x 1 (y)) + f(x 2 (y)) . (6.15)
dy dy

This result may be generalized straightforwardly to the case where the range y to y + dy
corresponds to more than two x-intervals.

Example 6.5. The random variable X is Gaussian distributed with mean µ and
2
2
2
variance σ . Find the PDF of the new variable Y = (X − µ) /σ . ,
Solution. It is clear that X(Y ) is a double-valued function of Y. However, in this case, it is
straightforward to obtain single-valued functions giving the two values of x that correspond to
√ √
a given value of y; these are x 1 = µ − σ y and x 2 = µ + σ y, where y is taken to mean the
positive square root. The PDF of X is given by

[ ]
1 (x − µ) 2
f(x) = √ exp − .
σ 2π 2σ 2
√ √
Since dx 1 /dy = −σ/(2 y) and dx 2 /dy = σ/(2 y), from (6.15) we obtain

1 −σ 1 σ 1 −1/2
g(y) = √ exp(−y/2) √ + √ exp(−y/2) √ = √ (y/2) exp(−y/2).
σ 2π 2 y σ 2π 2 y 2 π

Important continuous distributions

Having discussed the most commonly encountered discrete probability distributions, we now
consider some of the more important continuous probability distributions. These are
summarized for convenience in table 26.2; we refer the reader to the relevant subsection below
for an explanation of the symbols used.

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