PDF of Lecture Notes - School of Mathematical Sciences
PDF of Lecture Notes - School of Mathematical Sciences
PDF of Lecture Notes - School of Mathematical Sciences
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1. DISTRIBUTION THEORY<br />
2. Suppose Z 1 ∼ N(0, 1), Z 2 ∼ N(0, 1) independently. Let X = Z 1 + Z 2 . Find<br />
<strong>PDF</strong> <strong>of</strong> X.<br />
Solution.<br />
f X (x) =<br />
=<br />
=<br />
=<br />
∫ ∞<br />
−∞<br />
∫ ∞<br />
−∞<br />
∫ ∞<br />
−∞<br />
∫ ∞<br />
−∞<br />
φ(z)φ(x − z) dz<br />
1<br />
√<br />
2π<br />
e −z2 /2 1<br />
√<br />
2π<br />
e −(x−z)2 /2 dz<br />
1<br />
2π e− 1 2 (z2 +(x−z) 2) dz<br />
(<br />
1 1<br />
2π e− 2( z−x<br />
2 2<br />
)e )2 −x2 /4 dz<br />
= 1 ∫ ∞<br />
2 √ /4 1<br />
√<br />
π e−x2 e − (z− x 2 )2<br />
2× 1 2 dz<br />
−∞ π<br />
} {{ }<br />
= 1<br />
( x<br />
N<br />
2 , 1 <strong>PDF</strong><br />
2)<br />
i.e., X ∼ N(0, 2).<br />
= 1<br />
2 √ π e−x2 /4 ,<br />
Theorem. 1.8.3 (Ratio <strong>of</strong> continuous RVs)<br />
Suppose X 1 , X 2 are continuous with joint <strong>PDF</strong> f(x 1 , x 2 ), and let Y = X 2<br />
X 1<br />
.<br />
Then Y has <strong>PDF</strong><br />
If X 1 , X 2 independent, we obtain<br />
f Y (y) =<br />
f Y (y) =<br />
∫ ∞<br />
−∞<br />
∫ ∞<br />
−∞<br />
|x|f(x, yx) dx.<br />
|x|f X1 (x)f X2 (yx) dx.<br />
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