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Random Variables and Probability
Distributions
Schaum’s Outlines of
Probability and Statistics
Chapter 2
Presented by Carol Dahl
Examples by Tyler Hodge
2-2
Outline of Topics
Topics Covered:
 Random Variables
 Discrete Probability Distributions
 Continuous Probability Distributions
 Discrete Joint Distributions
 Continuous Joint Distributions
 Joint Distributions Example
 Independent Random Variables
 Changing of Variables
 Convolutions
2-3
Random Variables
Variables
events or values
given probabilities
Examples
Drill for oil - may hit oil or a dry well
Quantity oil found
2-4
Random Variables
Notation: P (X=xk)= P(xk)
probability random variable, X, takes value xk = P(xk).
Example
Drilling oil well in Saudi Arabia
outcome is either (Dry, Wet)
5% chance of being dry
P (X = Dry well) = 0.05 or 5%
P (X = Wet well) = 0.95 or 95%
2-5
Discrete Probability Distributions
Discrete probability distribution
takes on discrete, not continuous values.
Examples:
Mineral exploration
discrete - finds deposit or not
Amount of deposit found
continuous - any amount may be found
2-6
Discrete Probability Distributions
If random variable X defined by:
P(X=xk)= P(xk) , k=1, 2, ….
Then,
0<P (xk)<1
∑k P(x) = 1
2-7
Discrete Probability Distributions
Example: Saudi oil well drilling
Drilled well
Dry
Wet
f (well)
0.05
0.95
So, ∑ f (well) = 0.05 +0.95 = 1.0
2-8
Discrete Probability Distributions
Random variable X
cumulative distribution function
probability X  x
F ( x )  P( X  x )   P(ui )
ui  x
2-9
Discrete Probability Distributions
Example: Find cumulative distribution function
wind speed at a location (miles per hour)
Wind
Speed
P(x)
F(x)
4
4/36
4/36
5
6
7
8
5/36 6/36 6/36 6/36
9/36 15/36 21/36 27/36
9
10
5/36 4/36
32/36 36/36
2-10
Continuous Probability Distributions
Definition: Continuous probability distribution
takes any value over a defined range
Similar to discrete BUT
replace summation signs with integrals
Examples:
• amount of oil from a well
• student heights & weights
2-11
Continuous Probability Distributions
Continuous probability distribution
describe probabilities in ranges
f (x) 0
 f ( x)dx  1
x
Equation for probability X lies between a and b:
b
P(a  X  b)   f ( x)dx
a
2-12
Continuous Probability Distributions
Example:
Engineer wants to know probability
gas well pressure in economical range
between 300 and 350 psi
350
P(300  X  350) 
 f (x)dx
300
2-13
Continuous Probability Distributions
Example cont: let f(x) = x/180,000 for 0<X<600
= 0 everywhere else
verify that f(x) is bonafied probability distribution
if not fix it
2-14
Continuous Probability Distributions
Example cont: let f(x) = x/180,000 for 0<X<600
= 0 everywhere else
350
P(300  X  350) 
 (x / 180,000)dx
300
x 2 350 3502
3002

|300 

 0.09  9%
360,000
360,000 360,000
2-15
Discrete Joint Distributions
X and Y - two discrete random variables
joint probability distribution of X and Y
P(X=x,Y=y) = f(x,y)
Where following should be satisfied:
f(x,y) 0
and
 f ( x, y)  1
x
y
2-16
Discrete Joint Distributions
joint distribution of two discrete random table
cells make up joint probabilities
column and row summations
marginal distributions.
X\Y
x1
x2
...
xm
P(Y)
y1
y2
f(x1,y1) f(x1,y2)
f(x2,y1) f(x2,y1)
...
...
f(xm,y1) f(xm,y2)
f2(y1)
f2(y2)
...
...
...
...
...
yn
f(x1,yn)
f(x1,yn)
...
f(xm,yn)
f2(yn)
P(X)
f1(x1)
f1(x2)
...
f1(xm)
1 (grand
total)
2-17
Continuous Joint Distributions
X and Y two continuous random variables
joint probability distribution of X and Y
f(x,y)
with f(x,y) 0 and
b
P(a  X  b, c  Y  d) 
d
  f (x, y) dy dx
x a y  c
2-18
Discrete Joint Distribution (Example)
Example:
Yield from two forests has distribution
f(x,y) = c(2x+y) where
x and y are integers such that:
0X2 and 0Y3 and f(x,y)=0 otherwise
2-19
Joint Distributions (Example) cont.
For previous slide, find
value of “c”
P (X=2, Y=1)
Find P(X  1, Y  2)
Find marginal probability functions of X & Y
2-20
Joint Distributions (Example) cont.
P (X=0, Y=0) = c(2X+Y) = c(2*0+0) = 0
P (X=1, Y=0) = c(2X+Y) = c(2*1+0) = 2c
Fill in all possible probabilities in table
2-21
Joint Distributions (Example) cont.
X\Y
0
1
2
3
totals
0
0
c
2c
3c
6c
1
2c
3c
4c
5c
14c
2
4c
5c
6c
7c
22c
totals
6c
9c
12c
15c
42c
Since,
  f ( x, y )  1
x
y
42c=1 which implies that c=1/42.
2-22
Joint Distributions (Example) cont.
find P( X=2 ,Y=1) from table:
X\Y
0
1
2
totals
0
0
2/42
4/42
6/42
P(X=2,Y=1)=5/42
1
1/42
3/42
5/42
9/42
2
2/42
4/42
6/42
12/42
3
3/42
5/42
7/42
15/42
totals
6/42
14/42
22/42
42/42
2-23
Joint Distributions (Example) cont.
Evaluate P(X1,1<Y2)
sum cells of shaded region below:
X\Y
0
1
2
totals
0
0
2/42
4/42
6/42
1
1/42
3/42
5/42
9/42
2
2/42
4/42
6/42
12/42
P(X1,1<Y2) = (3 + 4 + 5 + 6)/42
= 18/42=0.429
3
3/42
5/42
7/42
15/42
totals
6/42
14/42
22/42
42/42
2-24
Joint Distributions (Example) cont.
Functions: Marginal X = P(X=xi)
= j(X=xi, Y=yj)
Marginal Y = P(Y=yj)
= i(X=xi, Y=yj)
2-25
Joint Distributions (Example) cont.
Marginal probability functions
read off totals across bottom and side of table:
X\Y
0
1
2
3
P(X)
0
0
1/42
2/42
3/42
6/42
1
2/42
3/42
4/42
5/42
14/42
2
4/42
5/42
6/42
7/42
22/42
P(Y)
6/42
9/42
12/42
15/42
42/42
2-26
Joint Distributions Continuous Case
X, Y ~ f(X,Y)
fx(X) = f(X,Y)dx
fy(Y) = f(X,Y)dy
Example: ore grade for copper and zinc
f(x,y) = ce-x-y
0<x<0.4 0<Y<0.3
= 0 elsewhere
2-27
Exp-x-y
1
0.8
0.6
0.2
0
0.1
0.1
0.2
0.3
0.4
0
2-28
Joint Distributions Continuous Case
Example: ore grade for copper and zinc
f(x,y) = ce-x-y
0<X<0.4 0<Y<0.3
what is c
∫ ∫ cf(x,y)dxdy = 1
fx(x) = 0
0.3
= 0
0.4
ce-x-y dy
(-ce-x-0.3
+
=-
0.3
-x-y
ce | 0 =
- ce-x-0.3 + ce-x-0.
ce-x-0 )= ce-x-0.3 - ce--x
0.4
|0
= ce-0.4 -0.3 - ce –0.4 - [ce-0.3 - ce -0] = 1
= c[e-0.7 - e –0.4 - e-0.3 + 1] = 1
c = 1/[e-0.7 - e –0.4 - e-0.3 + 1]= 1/3.084 = 0.324
2-29
Independent Random Variables
Random variables X and Y are independent
if occurrence of one ->no affect other’s probability
3 tests
1. iff
P(X=x|Y=y) = P(X=x) for all X and Y
P(X|Y) = P(X) for all values X and Y
otherwise dependent
2-30
Independent Random Variables
Random variables X and Y are independent
if occurrence of one ->no affect other’s probability
2. iff
P(X=x,Y=y) = P(X=x)*P(Y=y)
P(X,Y) = P(X)*P(Y)
otherwise X and Y dependent
2-31
Independent Random Variables
Random variables X and Y are independent
if occurrence of one ->no affect other’s probability
3. iff
P(Xx,Yy) = P(Xx)*P(Yy)
F(X,Y) = F(X)*F(Y)
otherwise X and Y dependent
2-32
Independent Random Variables
(Reclamation Example)
State of Pennsylvania
problems with abandoned coal mines
Government officials-reclamation bonding requirements
new coal mines
size of mine influence probability reclamation
2-33
Independent Random Variables
(Reclamation Example cont.)
Office of Mineral Resources Management
compiled joint probability distribution
Abandons
Size of Mine
Mine
(000 st)
0- 100
1/18
100 – 500
2/18
500 – 1,000
2/18
> 1,000
1/18
P(Y)
6/18
Reclaims
Mine
2/18
4/18
4/18
2/18
12/18
P(X)
3/18
6/18
6/18
3/18
2-34
Independent Random Variables – Check 1
Mine Size and Reclamation Independent?
1. Does P(X|Y) = P(X) for all values X and Y?
Does P(Mine 0-100|Reclaim) =
P(Mine 0-100 and reclaim)/P(Reclaim)
(2/18)/(12/18) = 2/12 = 1/6
P(mine 0 - 100 ) = 3/18 = 1/6
Holds for these values
Check if holds for all values X and Y
If not hold for any values dependent
2-35
Independent Random Variables – Check 2
2. Does P(X,Y) = P(X)*P(Y) for all values X and Y?
P(Mine 0-100 and reclaim) = 2/18 = 1/9
P(mine 0-100)* P(reclaim) = 3/18*12/18
= 36/324 = 1/9
Holds for these values
Checking it holds for all values X and Y
If not hold for any values – dependent
2-36
Independent Random Variables – Check 3
3. Does F(X,Y) = F(X)*F(Y) for all values X and Y?
P(Mine<500 and reclaim) = 2/18 + 4/18 = 1/3
P(Mine < 500)* P(reclaim) =6/18+3/18
= 9/18*12/18
=108/324 = 1/3
Holds for these values
Checking if holds for all values X and Y
If not hold for any values – dependent
2-37
Are Copper and Zinc Ore Grades Independent:
Continuous Case f(x,y) = 0.324e-x-y
Discrete: P(X|Y) = P(X)
Continuous f(x|y) = f(x)
f(x|y) = f(x,y)/f(y) = 0.324e-x-y/(-0.324e-0.3-y+0.324e-y)
?= (-0.324e-0.4-x+0.324e-x)
Discrete: P(X,Y) = P(X)*P(Y)
Continuous: f(x,y) = f(x)*f(y)
0.324e-x-y?= (-0.324e-0.4-x+0.324e-x)*(-0.324e-0.3-y+0.324e-y)
Discrete: P(Xx,Yy) = P(Xx)*P(Yy)
x
y
 
x 0 y 0
y
x
f ( x, y )dxdy 

x 0
f ( x )dx

y 0
f ( y )dy
2-38
Convolutions - Example
density function of their sum U=X+Y :
X, Y ~ e-X-Y
Y = U-X
g (U )  


f ( X ,U  X )dX  
U
 0.324 Xe U |0.4

0.1296
e
0
0.4
0
0.324e X U  X dX
2-39
Integrate Exponential Functions
Integrate Exponential functions
 xe
x2
dw
Let u = x2
du/dx = 2x
du
 xe 2 x
dx = du/2x
u
1 u
1 u
e du  e  c

2
2
1 x2
 xe dx  2 e  c
x2
From oil well example :
x7
1 72
 e c
2
 1.95
2-40
Changing of Variables Example
discrete probability distribution
size Pennsylvania coal mine
2 x x  1,2,3, . . .
P( X )  
otherwise
0
P(X = 1) = 2-1 = 1/2
P(X = 2) = 2-2 = 1/4
P(X = 3) = 2-3 = 1/8, etc.
2-41
Changing of Variables
MRM wants probability distribution of
reclamation bond amounts where reclamation = U
U = g(X) = X4 + 1
What is pdf of U
P(X = 1) = 2-1 = 1/2 =>P(U =X4 + 1= 14 + 1=2)= 1/2
P(X = 2) = 2-2 = 1/4 =>P(U =X4 + 1= 24 + 1=17)= 1/4
P(X = 3) = 2-3 =
=> P(U =X4 + 1=
)=
P(X = x) = 2-x => P(U =X4 + 1)
2-42
Changing of Variables Discrete Example
P(X = x) = 2-x = 1/2x => P(U =X4 + 1) = 1/2x
Want in terms of U
U = X4 + 1
Solve for X = (U – 1)(1/4)
P(U = x) = 2-x = 2 (U – 1)^(1/4) for U = (14+1, 24+1, 34+1, . .
2-43
Changing Variables General Case
Discrete
X ~ Px(X) for X = a,b,c, . .
U = g(X) => X = g-1(U)
U ~ Pu(U) = Px(g-1(U)) for g(a), g(b), g(c), . . .
Continuous
X ~ fx(X) where fx(X) = f(X) for a < X < b
fx(X) = 0 elsewhere
U = g(X) => X = g-1(U)
U ~ fu(U) = fx(g-1(U)) for g(a) < U < g(b)
2-44
Normal Practice Variable Change Problem
X ~ N(, ) – < X < 
f(X) = 1
exp(-(X- )2/(2 2)) dX
(2)0.5
Show that (X – ) / ~ N (0,1) – < Z < 
2-45
Convolutions
Extending Variable Change to Joint
Distributions
X, Y ~ f(X,Y)
Want density function of sum
U=X+Y
2-46
Convolutions
special case X and Y are independent,
f (x,y) = f1(x) f2(y)
and previous equation is reduced to following:

g (u ) 


f 1( x) f 2(u  x)dx
2-47
Convolutions - Example
Example: X (Oil Production) and Y (Gas Production)
independent random variables
2 e
f 1(oil )  
0
2 x
x0
x0
3e 3 y
f 2( gas )  
0
f(x,y) = e-2x*3e-3y
Find density function of their sum U=X+Y?
g(u) =

0
2e-2x*3e-3(u-x)dx
= complete this example
y0
y0
2-48
Sum Up Random Variables and
Probability Distributions (PS 2)
Discrete Probability Distributions
values with probabilities attached
Cumulative Discrete Probability Functions
F ( x )  P( X  x )   P(ui )
ui  x
Continuous Probability Distributions
b
P ( a  X  b) 

a
f ( x)dx
2-49
Chapter 2 Sum up
Discrete Joint Distributions
P(X=x,Y=y) = f(x,y)
Independent Random Variables
P(X|Y) = P(X) for all values X and Y
P(X,Y) = P(X)*P(Y)
F(X,Y) = Fx(X)*Fy(Y)
f(x,y) = f1(x)*f2(y)
2-50
Changing Variables
Discrete
X ~ Px(X) for X = a,b,c, . .
U = g(X) => X = g-1(U)
U ~ Pu(U) = Px(g-1(U)) for g(a), g(b), g(c), . . .
Continuous
X ~ fx(X) where fx(X) = f(X) for a < X < b
fx(X) = 0 elsewhere
U = g(X) => X = g-1(U)
U ~ fu(U) = fx(g-1(U)) for g(a) < U < g(b)
2-51
Convolutions: Extending Variable Change to
Joint Distributions
X, Y ~ f(X,Y)
Want density function of sum
U=X+Y

g (u ) 
 f ( x, u  x)dx

2-52
Convolutions
Special case X and Y are independent,
f (X,Y) = f1(X) f2(Y)
and previous equation is reduced to
following:

g (u ) 


f 1( x) f 2(u  x)dx