Page 228 - Probability and Statistical Inference

4. Functions of Random Variables and Sampling Distribution 205

Let us derive the joint pdf of Y = X + X and Y = X – X . Refer to
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2 –1/2 –1
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the Section 3.6 as needed. With c = {2πσ (1 – ρ ) } , we start
with
where –∞ < x , x < ∞. For the one-to-one transformation on hand, we have
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x = 1/2(y + y ), x = 1/2(y – y ) with 0 < y , y < ∞. One can easily verify
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that | det(J)| = 1/2, and hence the joint pdf of Y and Y would become
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for –∞ < y , y < ∞. In (4.4.23), the terms involving y , y factorize and none
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of the variables’ domains involve the other variable. Thus it is clear (Theorem
3.5.3) that the random variables Y , Y are independently distributed. Using
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such a factorization, we also observe that Y = X + X is distributed as N(0,
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2σ (1 + ρ)) whereas Y = X – X is distributed as N(0, 2σ (1 – ρ)). !
2 1 2
Suppose that the transformation from (X , ..., X ) → (Y , ..., Y ) is not
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n
one-to-one. Then, the result given in the equation (4.4.4) will need slight
adjustments. Let us briefly explain the necessary modifications. We begin by
partitioning the χ space, that is the space for (x , ..., x ), into A , ..., A in
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k
such a way that the associated transformation from A → y, that is the space
i
for (y , ..., y ), becomes separately one-to-one for each i = 1, ..., k. In other
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1
words, when we restrict the original mapping of (x , ..., x ) → (y , ..., y ) on
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A → y, given a specific (y , ..., y ) ∈ y, we can find a unique (x , ..., x ) in A i
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i
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n
which is mapped into (y , ..., y ). Given (y , ..., y ), let the associated x =
j
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b (y , ..., y ), j = 1, ..., n, i = 1, ..., k. Suppose that the corresponding
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ij
1
Jacobian matrix is denoted by J , i = 1, ..., k. Then one essentially applies
i
(4.4.4) on each piece A , ..., A and the pdf g(y , ..., y ) of Y is obtained as
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1
k
follows:
for y ∈ y. For a clear understanding, let us look at the following ex-
ample.
Example 4.4.16 Suppose that X and X are iid N(0, 1) variables.
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Let us denote Y = X /X and Y = Obviously, the transformation
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(x , x ) → (y , y ) is not one-to-one. Given (y , y ), the inverse solu-
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tion would be or We should take A =
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{(x , x ) ∈ ℜ : x > 0}, A = {(x , x ) ∈ ℜ : x < 0}. Now, we apply
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