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PDF of Lecture Notes - School of Mathematical Sciences

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1. DISTRIBUTION THEORY<br />

f(x 1 , x 2 ) =<br />

1<br />

(2π) (r 1+r 2 )/2<br />

∣ Σ ∣<br />

11 Σ 12∣∣∣<br />

1/2<br />

Σ 21 Σ 22<br />

=<br />

⎧<br />

⎡ ⎤<br />

⎨<br />

Σ 11 Σ 12<br />

× exp<br />

⎩ −1 2 ((x 1 − µ 1 ) T , (x 2 − µ 2 ) T ) ⎣ ⎦<br />

Σ 22<br />

1<br />

(2π) r 1/2<br />

|Σ 11 − Σ 12 Σ −1<br />

22 Σ 21 | 1/2<br />

Σ 21<br />

{<br />

× exp − 1 2 (x 1 − (µ 1 + Σ 12 Σ −1<br />

22 (x 2 − µ 2 ))) T<br />

⎞⎫<br />

x 1 − µ 1 ⎬<br />

⎝ ⎠<br />

⎭<br />

x 2 − µ 2<br />

−1 ⎛<br />

× ( ) }<br />

Σ 11 − Σ 12 Σ −1 −1(x1<br />

22 Σ 21 − (µ 1 + Σ 12 Σ −1<br />

22 (x 2 − µ 2 )))<br />

×<br />

{<br />

1<br />

(2π) ∣ ∣ r 2/2 ∣Σ exp − 1 }<br />

22 1/2<br />

2 (x 2 − µ 2 ) T Σ −1<br />

22 (x 2 − µ 2 )<br />

= h(x 1 , x 2 )g(x 2 ), where for fixed x 2 , h(x 1 , x 2 ), is the<br />

N r1<br />

(<br />

µ1 + Σ 12 Σ −1<br />

22 (x 2 − µ 2 ), Σ 11 − Σ 12 Σ −1<br />

22 Σ 21<br />

)<br />

<strong>PDF</strong>, and g(x 2 ) is the N r2 (µ 2 , Σ 22 ) <strong>PDF</strong>.<br />

Hence, the result is proved.<br />

Theorem. 1.10.5<br />

Suppose that X ∼ N r (µ, Σ) and Y = AX + b, where A p×r with linearly independent<br />

rows and b ∈ R p are fixed. Then Y ∼ N p (Aµ + b, AΣA T ).<br />

[Note: p ≤ r.]<br />

Pro<strong>of</strong>. Use method <strong>of</strong> regular transformations.<br />

Since the<br />

(<br />

rows <strong>of</strong> A are linearly independent, we can find B (r−p)×r such that the r × r<br />

B<br />

matrix is invertible.<br />

A)<br />

63

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