Page 209 - Probability and Statistical Inference

186 4. Functions of Random Variables and Sampling Distribution

Example 4.2.9 Let X and X be independent exponential random
1
2
variables, with their respective pdf’s given by f (x ) = e I(x ∈ ℜ )
+
–x 1
1 1 1
and f (x ) = e I(x ∈ ℜ ). What is the pdf of U = X + X ? It is obvious
+
–x 2
2
2
1
2
2
that the pdf g(u) will be zero when u ≤ 0, but it will be positive when
u > 0. Using the convolution theorem, for u > 0, we can write
which coincides with the gamma pdf in (1.7.20) with α = 2 and β = 1. In
other words, the random variable U is distributed as Gamma(2, 1). See the
related Exercise 4.2.19. !
A result analogous to the Theorem 4.2.1 for the product of two
continuous independent random variables X and X has been
1
2
set aside as the Exercise 4.2.13.
Example 4.2.10 Suppose that X and X are independent standard normal
1
2
random variables with the respective pdf’s f (x ) = ∅ (x ) and f (x ) = ∅(x )
2
1
1
2
1
2
for (x , x ) ∈ ℜ . What is the pdf of U = X + X ? It is obvious that the pdf g(u)
2
2
1
2
1
will be positive for all u ∈ ℜ. Using the convolution theorem, we can write
Now, we can simplify the expression of
g(u) as follows:
where we let But, note that for fixed u, the function
h(x) is the pdf of a N(1/2u, 1/2) variable. Thus, must be one.
Hence, from (4.2.12), we can claim that for u ∈ ℜ. But,
g(u) matches with the pdf of a N(0, 2) variable and hence U is distributed as
N(0, 2). See the related Exercise 4.2.11. !
In the next example, X and X are not independent but the
1
2
convolution or the distribution function approach still works.
Example 4.2.11 Suppose that X and X have their joint pdf given by
1 2

What is the pdf of U = X – X ? We first obtain the df G(u) of U.
1 2