Page 245 - Probability and Statistical Inference

222 4. Functions of Random Variables and Sampling Distribution

–1
–1
Thus, with 0 < t < min {(2(1 + ρ)) , (2(1 – ρ)) }, the mgf M (t) of Y is
Y
given by

Hence, we can express M (t) as
Y

–1
which will coincide with (1 – 2t) , the mgf of , if and only if ρ = 0. We
have shown that Y is distributed as if and only if ρ = 0, that is if and only
if X , X are independent. !
1
2
We have not said anything yet about the difference of random variables
which are individually distributed as Chi-squares.
The difference of two Chi-square random variables may or may
not be another Chi-square variable. See the Example 4.7.3.
Example 4.7.3 Suppose that we have X , X , X , X which are iid N(0, 1).
3
1
4
2

Let us denote . Since s are iid i
= 1, 2, 3, 4, by the reproductive property of Chi-square distributions we claim
that U is distributed as Obviously, U is distributed as At the same
1
2
time, we have which is distributed as Here, the dif-
ference of two Chi-square random variables is distributed as another Chi-
square variable. Next, let us we denote which is distributed as
But, and this random variable cannot have a
Chi-square distribution. We can give a compelling reason to support this con-
clusion. We note that U – U can take negative values with positive probabil-
3
1
ity. In order to validate this claim, we use the pdf of F and express
2,1
as

which simplifies to 1 – 2 –1/2 ≈ .29289. !
2
Let X , ..., X be iid N(µ, µ ) and S be the sample variance.
2
1
n
2
Then, (n – 1)S /σ is distributed as Chi-square. Is it possible
2
that (n – 1)S /σ is a Chi-square random variable without
2
2
the assumed normality of the X’s? See the Exercise 4.7.4.