Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
A step-by-step solution to find the joint probability density function (pdf) of two random variables z and w, given their relationship with the original random variables x and y and their joint pdf. The document also includes examples and calculations for finding the expected values of z and zy.
Typology: Study notes
1 / 7
This page cannot be seen from the preview
Don't miss anything!
Reading: G. R. Cooper & C. D. McGillem 3.6 - 3.
EE/STAT 322, #11 1
Problem Statement: Two RVs X and Y have a joint pdf fXY (x, y).
Another two RVs Z and W are related with X, Y by
Z = φ 1 (X, Y ) and W = φ 2 (X, Y );
vice verse, X = ψ 1 (Z, W ), and Y = ψ 2 (Z, W ).
Given the joint PDF fXY (x, y), try to find the joint PDF g(z, w).
If (X, Y ) have a one-to-one mapping to (Z, W ),
P (z 1 < Z < z 2 , w 1 < W < w 2 ) = P (x 1 < X < x 2 , y 1 < Y < y 2 ).
∫ (^) z 2 z 1
∫ (^) w 2 w 1
g(z, w)dzdw =
∫ (^) x 2 x 1
∫ (^) y 2 y 1
f (x, y)dxdy =
∫ (^) x 2 x 1
∫ (^) y 2 y 1
f [ψ 1 (z, w), ψ 2 (z, w)]dzdw.
X
Y
Z
W
( x 1 , y 1 ) ( z 1 , w 1 )
( x 2 (^) , y 2 )
( z 2 , w 2 )
EE/STAT 322, #11 3
Let z 1 = z, z 2 = z + ∆z; w 1 = w,
w 2 = w + ∆w,
P (z ≤ Z ≤ z + ∆z, w ≤ W ≤
w + ∆w) = g(z, w)∆z∆w =
fX,Y (x, y)∆x∆y.
X
Y
Z
W
( x 1 (^) , y 1 ) ( z 1 , w 1 )
( x 2 (^) , y 2 )
( z 2 , w 2 )
g(z, w) = fX,Y (x, y)
∆x∆y
∆z∆w
= fX,Y (x = ψ 1 (z, w), y = ψ 2 (z, w))|J|
where J is the Jacobian factor given by J =
∆x∆y ∆z∆w
∂x ∂z
∂x ∂w ∂y ∂z
∂y ∂w
Example: (Ex 3-6.2) For the previous example, show that X and Y are
independent. Find E[XY ] and E[Z], respectively.
Solution:
f (x) =
0
f (x, y)dy =
0
1 dy = 1, 0 ≤ y ≤ 1 ;
f (y) =
0 f (x, y)dx =
0 1 dx = 1, 0 ≤ x ≤ 1.
f (x, y) = f (x)f (y) for all points of 0 ≤ y ≤ 1 , ⇒X and Y are independent.
0
0
xyf (x, y)dxdy =
0
xdx
0
ydy =
x
2
1
0
y
2
1
0
EE/STAT 322, #11 7
To find E[Z], we can use two methods:
zf (x, y)dxdy =
xyf (x, y)dxdy = E[XY ] = 1/ 4.
0
zfZ (z)dz =
0
z(− ln z)dz =
0
(− ln z)d
z
2
= (− ln z)
z
2
1 0 +
0
z
2
d ln z = 0 +
0
z
2
z
dz
0
z
dz = 1/ 4.
Fourier transform: φ(u) = E[e
juX ] =
−∞ f (x)e
jux dx.
Vice versa, f (x) =
1 2 π
∞ −∞
φ(u)e
−jux du. (Inverse Fourier transform).
convenient than the PDF method, as shown below.
Given three independent RVs X 1 , X 2 , X 3 , and Z = X 1 + X 2 + X 3. Find
the PDF of Z.
PDF method: fZ (z) = fx 1 (z) ⊗ fx 2 (z) ⊗ fx 3 (z).
CHF method: φZ (z) = φX 1 (z)φX 2 (z)φX 3 (z), and
fZ (z) =
1 2 π
∞ −∞ fZ (z)e
−juz du.
EE/STAT 322, #11 9
Example: Given two independent RVs X and Y , with the PDFs
f (x) =
e
−x x ≥ 0
0 x < 0
, and f (y) =
e
−y y ≥ 0
0 y < 0
Let Z = X + Y. Find fZ (z).
Solution:
ΦX (u) = ΦY (u) =
0
e
−y e
juy dy =
e(−1+ju)y ju− 1
∞
0
1 1 −ju
⇒ΦZ (u) = [ΦX (u)]
1 (1−ju)^2
Using inverse Fourier transform, fZ (z) =
1 2 π
−∞
1 (1−ju)^2
e
−jux du =
Res[
1 (1−ju)^2
e −jux at the pole ju =1], where Res refers to the residue.
φX,Y (u, v) = E[e j(uX+vY ) ] =
−∞
−∞
f (x, y)e j(uX+vY ) dxdy.
f (x, y) =
1 (2π)^2
−∞
−∞
φX,Y (u, v)e
−j(uX+vY ) dudv.
∂ 2 φX,Y (u,v) ∂u∂v
u=v=
i Y
k ] = X i Y k =
1 ji+k
∂ i+k φX,Y (u,v)
∂ui∂vk
u=v=
For PDF approach, we use E[X i Y k ] =
x i y k f (x, y)dxdy. When i or
k is high, CHF method is more convenient.
EE/STAT 322, #11 13
Example: (Ex 3-7.2) RV X has a PDF f (x) = 2e − 2 x u(x). Using the CHF,
find the first and second moments of X.
Solution:
φX (x) =
∞ 0
2 e
− 2 x e
jux dx =
∞ 0
2 e (ju−2)x
(ju−2) dx(ju − 2)
2 e (ju−2)x
(ju−2)
∞
0
2 2 −ju
∂ 2 (2−ju) ∂ju
u=
2 ·(−1)(−1) (2−ju)^2
u=
∂ (^2 ) 2 −ju ∂(ju)^2
u=
2 ·(−2)(−1) (2−ju)^3
u=
