Page 155 - Fundamentals of Probability and Statistics for Engineers

138 Fundamentals of Probability and Statistics for Engineers

determination of the distribution Y when all X j , j ˆ 1, 2, ... , n, are continuous
random variables. Consider the transformation

Y ˆ g X 1 ; ... ; X n † 5:40†

where the joint distribution of X 1 , X 2 ,. . . , and X n is assumed to be specified in
term of their joint probability density function (jpdf), f (x 1 ,..., x n ), or
X 1 ...X n
(x 1 ,..., x n ). In a
their joint probability distribution function (JPDF), F X 1 ...X n
more compact notation, they can be written as f ( x) and F X ( x), respectively,
X
where X is an n-dimensional random vector with components X 1 , X 2 ,..., X n .

The starting point of the derivation is the same as in the single-random-
variable case; that is, we consider F Y (y) ˆ P(Y y). In terms of X, this
probability is equal to P[g( X) y]. Thus:
F Y y†ˆ P Y y†ˆ P‰g X† yŠ
5:41†
ˆ F X ‰x : g x† yŠ:

The final expression in the above represents the JPDF of X for which the
argument x satisfies g( x) . In terms of f ( x), it is given by
y
X
Z Z
F X ‰x : g x† yŠˆ f x†dx 5:42†
X
R n : g x† y†

where the limits of the integrals are determined by an n-dimensional region R n
within which g( x) y is satisfied. In view of Equations (5.41) and (5.42), the
PDF of Y , F Y (y), can be determined by evaluating the n-dimensional integral in
Equation (5.42). The crucial step in this derivation is clearly the identification
n
of R , which must be carried out on a problem-to-problem basis. As n becomes
large, this can present a formidable obstacle.
The procedure outlined above can be best demonstrated through examples.
Example 5.11. Problem: let Y ˆ X 1 X 2 . Determine the pdf of Y in terms of

f (x 1 , x 2 ).
X 1 X 2
Answer: from Equations (5.41) and (5.42), we have
Z Z
F Y y†ˆ f x 1 ; x 2 †dx 1 dx 2 : 5:43†
X 1 X 2
2
R : x 1 x 2 y†
The equation x 1 x 2 ˆ y is graphed in Figure 5.16 in which the shaded area
2
represents R , or x 1 x 2 y. The limits of the double integral can thus be
determined and Equation (5.43) becomes

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