1916 CHAPTER 59. BASIC PROBABILITY

Let E = X−11 (A) and

F = (X2, · · · ,Xp)−1 (B)

where A and B are Borel sets in R and Rp−1 respectively. Thus I need to verify that

P([(X1,(X2, · · · ,Xp)) ∈ (A,B)]) =

µ(X1,(X2,··· ,Xp)) (A×B) = µX1(A)µ(X2,··· ,Xp) (B) . (59.16.36)

Using 59.16.35, Fubini’s theorem, and definitions,

µ(X1,(X2,··· ,Xp)) (A×B) =

∫Rp

XA×B (x)1

(2π)p/2 det(Σ)1/2 e−12 (x−m)∗Σ−1(x−m)dx

=∫R

XA (x1)∫Rp−1

XB (X2, · · · ,Xp) ·

1

(2π)(p−1)/2√2π(σ2

1

)1/2 det(Σp−1)1/2

e−(x1−m1)

2

2σ21 ·

e−12 (x′−m′)

∗Σ−1p−1

(x′−m′

)dx′dx1

where x′ = (x2, · · · ,xp) and m′ = (m2, · · · ,mp) . Now this equals

∫R

XA (x1)1√

2πσ21

e−(x1−m1)

2

2σ21

∫B

1

(2π)(p−1)/2 det(Σp−1)1/2 · (59.16.37)

e−12 (x′−m′)

∗Σ−1p−1

(x′−m′

)dx′dx. (59.16.38)

In case B = Rp−1, the inside integral equals 1 and

µX1(A) = µ(X1,(X2,··· ,Xp))

(A×Rp−1)

=∫R

XA (x1)1√

2πσ21

e−(x1−m1)

2

2σ21 dx1

which shows X1 is normally distributed as claimed. Similarly, letting A = R,

µ(X2,··· ,Xp) (B)

= µ(X1,(X2,··· ,Xp)) (R×B)

=∫

B

1

(2π)(p−1)/2 det(Σp−1)1/2 e

−12 (x′−m′)

∗Σ−1p−1

(x′−m′

)dx′