1. Multiple Continuous RV
Probability and Random Processes for Computer Engineering 1st semester 2009 (June – September)
Lecture # 10
Multiple Continuous RV
รศ.ดร. อนันต์ ผลเพิ่ม
Assoc.Prof.Anan Phonphoem, Ph.D.
Intelligent Wireless Network Group (IWING Lab)
Computer Engineering Department
Kasetsart University, Bangkok, Thailand

2. Joint CDF Pairs of Random Variables Discrete:
Joint PMF PX,Y(x,y) = P[X=x, Y=y]
Continuous:
PX,Y(x,y) = (PX(x) = 0, PY(y) = 0)
For 1 RV  interval on real axis
For 2 RVs  area in a plane

3. Joint CDF Definition: Joint CDF of X and Y FX,Y(x,y) = P[X  x, Y  y]

4. Interesting Properties
For Event {X  x}
FX(x) = P[X  x]
= P[X  x, Y  ] X Y
(x, )
y  
= lim FX,Y(x,y)
= FX,Y(x, )

5. Joint CDF Theorem : 0  FX,Y(x,y)  1 FX(x) = FX,Y(x, )
FY(y) = FX,Y(, y)
FX,Y(-, y) = FX,Y(x, -) = 0
If x1  x and y1  y
then FX,Y(x1,y1) > FX,Y(x,y)
(f) FX,Y(, ) = 1

6. Joint PDF  Definition: Joint PDF of X and Y is satisfied
FX,Y(x,y) = fX,Y(u,v) dv du  - x y
Theorem:
fX,Y(x,y) =
2 FX,Y(x,y)
x y

7. Joint CDF Theorem: P[x1 < X  x2, y1 < Y  y2]
= FX,Y(x2,y2) - FX,Y(x2,y1) - FX,Y(x1,y2) + FX,Y(x1,y1) X Y
(x1,y2)
(x2,y2)
(x1,y1)
(x2,y1)

8. Joint PDF   Theorem: fX,Y(x,y)  0 for all (x,y)
(b) fX,Y(x,y) dx dy = 1  - 
fX,Y(x,y)  0 for all (x,y)
Theorem:
P[A] = fX,Y(x,y) dx dy  A

9. Example I    c = 1/15 Find constant c c dx dy = 15c = 1
fX,Y(x,y) =
c  x  3, 0  y  5
Otherwise
Find constant c X Y
c dx dy = 15c = 1  3 5
P[A]
 c = 1/15
Find P[A] = P[1  x  3, 2  y  3]
P[A] = /15 dv du = 2/15  1 3 2

10. Example II A bank operates 1) Drive-up and 2) Walk-up windows
X = proportion of time Drive-up in use
Y = proportion of time Walk-up in use
The set D of possible values for (X,Y)
D = {(x,y): 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}
Joint pdf
Find probability that neither facility is busy > one-quarter of time 6 5
(x+y2) ≤ x ≤ 1, 0 ≤ y ≤ 1
Otherwise
f(x,y) =
From Devore ‘s Book

11. Example II   f(x,y) dx dy =     
To verify that this is a legitimate pdf, f(x,y) ≥ 0 and
  f(x,y) dx dy =
- -
  6 5
(x+y2) dx dy
 
0 0
1 1
= x dx dy y2 dx dy
 
0 0
1 1 6 5
= x dx y2 dy  1 6 5
= = 1 6 10 15

12. Example II
Find probability that neither facility is busy > one-quarter of time 6 5
(x+y2) dx dy
 
1/4 1/4
P(0 ≤ X ≤ , 0 ≤ Y ≤ ) = 1 4
= x dx dy y2 dx dy 6 5
 
1/4 1/4 = 6 20 x2 2 .
x = 1/4
x = 0 y3 3
y = 1/4
y = 0 +
= = 7
640

13. Marginal PDF  Theorem: fX,(x) = fX,Y(x,y) dy fY,(y) = fX,Y(x,y) dx - 
fX,(x) = fX,Y(x,y) dy

14. From Example II   Find marginal pdf of X fX(x) = fX,Y(x,y) dy 6-   1 6 5
= (x+y2) dy
x + 6 5 2 =
0 ≤ x ≤ 1
Otherwise
The probability distribution of busy time for Drive-up
without reference to Walk-up

15. From Example II   Find marginal pdf of Y fY,(y) = fX,Y(x,y) dx 6 5-  6 5
= (x+y2) dx  1
y2 + 6 5 3 =
0 ≤ x ≤ 1
Otherwise
The probability distribution of busy time for Walk-up
without reference to Drive-up

16. | Example III     c = 2 = ( cx dy) dx = cx (2x2) dx = = = 1
fX,Y(x,y) =
cx  x  1, |y| < x2
Otherwise X Y
Y = X2
Find constant c
cx dx dy  - 
= ( cx dy) dx  1
-x2 x2
= cx (2x2) dx  1 =
cx4 2 | 1 = c 2
= 1
 c = 2

17. Example III  fX,(x) = 2x dy = 4x3
Find the marginal PDF fX(x) and fY(y)
fX,(x) = 2x dy = 4x3 
-x2 x
Fixed x (X = x) then integrate all y X Y
X = x
fX(x) =
4x  x  1
Otherwise

18. Example III  fY,(y) = 2x dx = 1 - |y| 1 - |y| -1  y  1 fY(y) =
Fixed y (Y = y) then integrate all x X Y
Y = y
fY(y) =
1 - |y|  y  1
Otherwise

19. Functions of 2 RVs Example:
Wireless based station with 2 antennas. X and Y are RVs of the signal
Find the strongest signal
W = X if |X| > |Y| or W = Y otherwise
Find the addition of 2 signals
W = X + Y
Find the addition of 2 signals with weight
W = aX + bY

20. Functions of 2 RVs
FW(w) = P[W  w] = fX,Y(x,y) dx dy 
g(x,y)  w

21. Example IV  FW(w) = P[X  w, Y  w] fX,Y(x,y) dx dy =
Otherwise
Find PDF of W = max(X,Y)
For W = max(X,Y)
 {W  w} = {X  w, Y  w}
FW(w) = P[X  w, Y  w]
fX,Y(x,y) dx dy  - w =

22. Example IV   We can divide into 2 cases FW(w) = 1/15 dx dy = w2/153
0  w  3
FW(w) = /15 dx dy = w2/15  w X Y 3
3  w  5
FW(w) = ( 1/15 dx )dy = w/5  w 3

23. Example IV 0 w < 0 1 w2/15 0  w  3 w/5 3 < w  5 1 w > 5
FW(w) =
w < 0
w2/  w  3
w/ < w  5
w > 5 1
fW(w) =
2w/  w  3
1/ < w  5
Otherwise

24. Expected Value  Theorem: E[W] = E[g(X,Y)] = g(x,y) fX,Y(x,y) dx dy -  =

25. Expected Value Theorem: For g(X,Y) = g1(X,Y) + … + gn(X,Y)
E[g(X,Y)] = E[g1(X,Y)] + … + E[gn(X,Y)]

26. Expected Value Theorem: E[ X+Y ] = E[X] + E[Y]
Find E[X]  From fX,Y(x,y)
Not necessary
 Can find from Marginal PDF fX(x)
So, we can find Var[X+Y], Cov, X,Y

27. Conditioning Joint PDF by Event
Definition:
fX,Y|B(x,y)
fX,Y(x,y)
P[B] =
(x,y)  B
Otherwise

28. Conditional PDF Definition: Theorem:
fY|X(y|x)
fX,Y(x,y)
fX(x) =
fX|Y(x|y)
fX,Y(x,y)
fY(y) =
Theorem:
fX,Y(x,y) = fY|X(y|x) fX(x) = fX|Y(x|y) fY(y)

29. Conditional Expected Value
Definition: for fY(y) > 0
E[X|Y=y] = x fX|Y(x|y) dx  - 
Definition: for fY(y) > 0
E[g(X,Y)|Y=y] = g(x,y) fX|Y(x|y) dx  - 

30. Independent RVs Definition: X and Y are independent iff
fX,Y(x,y) = fX(x) fY(y)
Example:
fX,Y(x,y) =
4xy  x  1, 0  y  1
Otherwise
Are X and Y independent ?
fX(x) =
2x 0  x  1
0 Otherwise
fY(y) =
2y 0  y  1
0 Otherwise
For all pairs are true as definition X and Y are independent

31. Independent RVs Theorem: for independent rv X and Y
E[g(X)h(Y)] = E[g(X)]E[h(Y)]
Cov [X,Y] = 0
Var [X+Y] = Var[X] + Var[Y]

32. Jointly Gaussian RV ( ) Definition: Bivariate Gaussian RV fX,Y(x,y) =
1 2
2 (x-1) (y-2) –
x-1 1
( ) 2
y-2 2 +
2(1 – 2)
exp
2 1 2  1 – 2
fX,Y(x,y) =
1 and 2  real number, 1 > 0, 2 > 0 and –1    1

33. Jointly Gaussian RV
For 1 = 2 = 0 , 1 = 2 = 1 and  = 0, 0.9, -0.9

34. Jointly Gaussian RV  =0.9  = 0 Uncorrelated  > 0
If X  (relative to mean)  Y 
If X  (relative to mean)  Y 
 < 0
If X  (relative to mean)  Y 
If X  (relative to mean)  Y 

35.  Jointly Gaussian RV e e fX,Y(x,y) = e fX(x) = e Marginal fX(x) = ?
212 1
12
–(y - 2(x))2
222
22 e ~ 1
Marginal fX(x) = ?
fX(x) = e
–(x - 1)2
212 1
22 
–(y - 2(x))2
222 e ~ dy  -

36. Jointly Gaussian RV e fX,Y(x,y) = e fX(x) = e fY(y) = e Theorem: 1
212 1
12
–(y - 2(x))2
222
22 e ~
Theorem:
fX(x) = e
–(x - 1)2
212 1
12
fY(y) = e
–(y - 2)2
222 1
22

37. Bivariate Gaussian RV e fY|X(y|x) =
Theorem: Conditional PDF of Y given X
fY|X(y|x) =
–(y - 2(x))2
222 1
22 e ~

38. Joint Gaussian PDF fY|X(y|x) = Gaussian  Bell shape cross section
For 1 = 2 = 0 , 1 = 2 = 1 and  = 0.9
fY|X(y|x) = Gaussian  Bell shape cross section

39. More Than 2 RVs 2 RVs  Bivariate Joint PDF
> 2 RVs  Multivariate Joint PDF

40. Homework HW#6 has been announced in the class website
Due date: September 1, 2009