Page 214 - Probability and Statistical Inference

4. Functions of Random Variables and Sampling Distribution 191

Example 4.3.1 (Example 4.2.3 Continued) Let X , ..., X be independent
1
k
and let X be Binomial(n , p), i = 1, ..., k. Write
i
i
Then invoking Theorem 4.3.1 and using the expression
of the mgf of the Binomial (n , p) random variable from (2.3.5),
i
one has The
final expression of M (t) matches with the expression for the mgf of the
U
Binomial (n, p) random variable found in (2.3.5). Thus, we can claim that the
sampling distribution of U is Binomial with parameters n and p since the
correspondence between a distribution and its finite mgf is unique. !
Example 4.3.2 Suppose that we roll a fair die n times and let X denote the
i
score, that is the number on the face of the die which lands upward on the i th
toss, i = 1, ..., n. Let the total score from the n tosses. What is
the probability that U = 15? The technique of full enumeration will be very
tedious. On the other hand, let us start with the mgf of any of the X ’s which
i
is given by M (t) = 1/6e (1 + e + ... + e ) and hence M (t) = 1/6 e (1 + e +
5t
t
t
n nt
t
U
Xi
... + e ) . In the full expansion of M (t), the coefficient of e must coincide
5t n
kt
U
with P(X = k) for k = n, n + 1, ..., 6n–1, 6n. This should be clear from the
way the mgf of any discrete distribution is constructed. Hence, if n = 4, we
have P(X = 15) = 1/6 {coefficient of e in the expansion of (1 + e + ... +
4
11t
t
e ) } = 140/6 } = But, if n = 10, we have P(X = 15) = 1/6 {coefficient of e 5t
10
4
5t 10
in the expansion of (1 + e + ... + e ) } = 2002/6 . One can find the pmf of U
t
10
t 4
as an exercise. !
Example 4.3.3 Let X , ..., X be independent random variables, X dis-
1
n
i
tributed as i = 1, ..., n. Write and
Then invoking Theorem 4.3.1 and using the expression for
the mgf of the random variable from (2.3.16), one has
The final ex-
pression of M (t) matches with the expression for the mgf of the N(µ, σ )
2
U
random variable found in (2.3.16). Thus, we can claim that the sampling
distribution of U is normal with parameters µ and σ since the correspon-
2
dence between a distribution and its finite mgf is unique. !
Example 4.3.4 Let X , ..., X be independent random variables, X having a
1
i
n
Gamma distribution with the pdf f (x) = c e –x/β αi–1 with
x
i
i
for 0 < x, α, β < ∞, i = 1, ..., n. Let us write U = Then, for 0 <
i
–1
t < β , as before one has M (t) =
U
Here we used the expression for the mgf of the Gamma(α, β) random variable
i
from (2.3.23). Thus, we can claim that the sampling distribution of U is Gamma(α,
β) since the correspondence between a distribution and its finite mgf is unique. !
Example 4.3.5 Let X , ..., X be independent random variables, X be
1 n i