Unformatted text preview: om the definition of the mean. Therefore, each crossterm in the last integral for V(Y)
is zero and V (Y ) = [∫ c ( x − μ )
2
1 1 1 2
[ f X 1 ( x1 )dx1 ... ∫ c 2 ( x p − μ p ) 2 f X p ( x p )dx p
p = c12V ( X 1 ) + ... + c 2V ( X p ).
p
a b 0 0 ∫∫ 596 a f XY ( x, y )dydx = ∫
0 b ∫ cdydx = cab . Therefore, c = 1/ab. Then, 0 b f X ( x) = ∫ cdy =
0 1
a a 1
for 0 < x < a, and fY ( y) = ∫ cdx = b for 0 < y < b. Therefore,
0 f XY (x,y)=f X (x)f Y (y) for all x and y and X and Y are independent.
597 The marginal density of X is
b b b 0 0 f X ( x) = ∫ g ( x)h(u )du = g ( x) ∫ h(u )du = kg ( x) where k = ∫ h(u )du. Also,
0 a f Y ( y ) = lh( y ) where l = ∫ g (v)dv. Because f XY (x,y) is a probability density function,
0 ⎡
⎤⎡b
⎤
g ( x)h( y )dydx = ⎢ ∫ g (v)dv ⎥ ⎢ ∫ h(u )du ⎥ = 1. Therefore, kl = 1 and
∫∫
00
⎣0
⎦⎣ 0
⎦
f XY (x,y)=f X (x)f Y (y) for all x and y.
a b a 548...
View
Full Document
 Spring '14
 Normal Distribution, Variance, Probability theory, probability density function, ½X+ ½Y

Click to edit the document details