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4. Functions of Random Variables and Sampling Distribution 213

each linear function has the univariate normal distribution for all
fixed, but arbitrary real numbers a , ..., a .
p
1
2
Example 4.6.1 Suppose that X , ..., X are iid N (µ, σ ), –∞ < µ < ∞, 0 <
1
n
σ < ∞. Consider the sample mean and let us look at the two-dimensional
joint distribution of (X , ). Obviously, E(X )=E( )=µ,V(X )
1
1
1
1
2
=σ and V( ) = -σ . Also, we can write Cov(X , ) =
2
1
n
Refer to the Theorem 3.4.3,
part (iii) as needed. But, the covariance between X and X is zero for 1 < j ≤
1 j
1 2
n. That is, Cov(X , ) = - Cov(X , X ) = -σ and hence Any
1
1
1
n
n
1
linear function L of X and is also a linear function of n original iid normal
1
random variables X , ..., X , and hence L itself is distributed normally. This
n
1
follows from the reproductive property of independent normal variables (Theo-
rem 4.3.2, part (i)). So, by the Definition 4.6.1 it follows that (X , ) is
1
jointly distributed as a bivariate normal variable. Thus, the conditional distri-
2
bution of X given is is N ( , σ (1 – 1/n)) which provides the
1
1
following facts: and V(X | ) = σ (1 – -). Refer
2
1 n
to the Theorem 3.6.1 as needed. Look at the Exercise 4.6.1 !
Example 4.6.2 (Example 4.6.1 Continued) One can easily find the condi-
tional expectation of aX + bX given in the following way where a
1
2
and b are fixed non-zero real numbers. Observe that E(aX + bX | )
2
1

= E(aX | ) + E(bX | ) = (a + b) . Also note, for example, that
1 2
= V(X | ) + {E(X | )} = σ (1 – 1/n) + 2 .
2
2
1
1
One can exploit similar techniques to obtain other expected values in this
context. Look at the Exercise 4.6.2. !
Example 4.6.3 (Example 4.6.1 Continued) A result such as E(X |
1

= can be alternately derived as follows. Obviously,E(X | ) = , and
1
this can be rewritten as )
E(X | ) which was the intended result in the first place. This argument
1
works here because E{X | } = E(X | ) for each i = 1, ..., n.
i
1
Observe that this particular approach merely used the fact that the X ’s are iid,
i
but it did not exploit the fact that the X ’s are iid normal to begin with. !
i
Example 4.6.4 (Example 4.6.2 Continued) A result such as

2
) = σ (1 – 1/n) + can be alternately derived as follows. Obviously,
2
2
2
σ = E(S ) = E(S | ) since S and are independent. Here, we have
2
2
indeed used the full power of the assumption of normality. Recall the Remark
4.4.4 in this context. So one can write and hence