Page 263 - Probability and Statistical Inference

240 4. Functions of Random Variables and Sampling Distribution

distributed. {Hint: Let u = x + x , v = x – x , and find the joint pdf of the
2
1
1
2
transformed random variable (U, V). Integrate this joint pdf with respect to v
and show that the marginal pdf of U does not match with the pdf of a normal
random variable.}
4.7.2 (Example 3.6.3 and Exercise 4.7.1 Continued) Suppose that a pair of
random variables (X , X ) has the joint pdf given by
1
2

2 T2
with 0 < α < 1, where f(x , x ; ν, θ, σ , , ρ) stands for the bivariate normal
2
1
pdf defined in (3.6.1) with and 0 < ρ < 1.
2
2
Show that marginally X is N(µ, σ ), X is N(θ,T ), but X + X is not normally
1
2
1
2
distributed.
4.7.3 (Example 4.7.2 Continued) Provide the missing details in the Ex-
ample 4.7.2.
4.7.4 If X , ..., X were iid N(µ, σ ) and
2
1
n
2
2
was the sample variance, then we could claim that (n – 1)S /σ was distrib-
uted as a Chi-square random variable with (n–1) degrees of freedom for n ≥
2. Does such a conclusion necessarily need the assumption of the normality
or the iid nature of the X sequence? In general there can be different kinds of
scenarios on hand. The following problems indicate some of the possibilities.
Think through each case carefully and summarize the story each situation
unfolds.
(i) Consider Y , Y , ..., Y which are iid N(0, 1), and define then X i
n
0
1
= Y + Y , i = 1, ..., n. Obviously, = + Y where and
i 0 0
are respectively the sample means of X , ..., X and Y , ..., Y .
n
1
n
1
Now, , which
would be distributed as a Chi-square random variable with
(n – 1) degrees of freedom. Note that X , ..., X happen to
1
n
be normal variables, but these are dependent;
(ii) In part (i), observe that Y can be allowed to have any arbitrary
0
distribution while on the other hand Y and (Y , ..., Y ) need not
0
n
1
be independent either. Yet, the conclusion that (n – 1)S is dis
2
tributed as a Chi-square random variable with (n – 1) degrees of
freedom holds. Y can even be a discrete random variable!
0