Random Variables and Probability Distributions

1
Random Variables and Probability Distributions

• Schaums Outlines of
• Probability and Statistics
• Chapter 2
• Presented by Carol Dahl
• Examples by Tyler Hodge

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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

3
Random Variables

• Variables
• events or values
• given probabilities
• Examples
• Drill for oil - may hit oil or a dry well
• Quantity oil found

4
Random Variables

• Notation P (Xxk) 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

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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

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Discrete Probability Distributions

• If random variable X defined by
• P(Xxk) P(xk) , k1, 2, .
• Then,
• 0ltP (xk)lt1
• ?k P(x) 1

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Discrete Probability Distributions

• Example Saudi oil well drilling
• So, ? f (well) 0.05 0.95 1.0

Drilled well Dry Wet
f (well) 0.05 0.95
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Discrete Probability Distributions
Random variable X cumulative distribution
function probability X ? x
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Discrete Probability Distributions
Example Find cumulative distribution
function wind speed at a location (miles per
hour)
Wind Speed 4 5 6 7 8 9 10
P(x) 4/36 5/36 6/36 6/36 6/36 5/36 4/36
F(x) 4/36 9/36 15/36 21/36 27/36 32/36 36/36
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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

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Continuous Probability Distributions

• Continuous probability distribution
• describe probabilities in ranges

f (x) ?0
Equation for probability X lies between a and b
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Continuous Probability Distributions

Example Engineer wants to know probability
gas well pressure in economical
range between 300 and 350 psi
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Continuous Probability Distributions

Example cont let f(x) x/180,000 for
0ltXlt600 0 everywhere else verify that
f(x) is bonafied probability distribution if not
fix it
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Continuous Probability Distributions

Example cont let f(x) x/180,000 for
0ltXlt600 0 everywhere else
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Discrete Joint Distributions

X and Y - two discrete random variables joint
probability distribution of X and Y P(Xx,Yy)
f(x,y) Where following should be
satisfied f(x,y) ?0 and
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Discrete Joint Distributions
joint distribution of two discrete random table
cells make up joint probabilities column and
row summations marginal distributions.
X\Y y1 y2 ... yn P(X)
x1 f(x1,y1) f(x1,y2) ... f(x1,yn) f1(x1)
x2 f(x2,y1) f(x2,y1) ... f(x1,yn) f1(x2)
... ... ... ... ...
xm f(xm,y1) f(xm,y2) ... f(xm,yn) f1(xm)
P(Y) f2(y1) f2(y2) ... f2(yn) 1 (grand total)
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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
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Discrete Joint Distribution (Example)

Example Yield from two forests has
distribution f(x,y) c(2xy) where x
and y are integers such that 0?X?2 and 0?Y?3
and f(x,y)0 otherwise
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Joint Distributions (Example) cont.

For previous slide, find value of c P (X2,
Y1) Find P(X ? 1, Y ? 2) Find marginal
probability functions of X Y
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Joint Distributions (Example) cont.

P (X0, Y0) c(2XY) c(200) 0 P
(X1, Y0) c(2XY) c(210) 2c Fill in
all possible probabilities in table
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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, 42c1 which implies that c1/42.
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Joint Distributions (Example) cont.

find P( X2 ,Y1) from table
X\Y 0 1 2 3 totals
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
totals 6/42 9/42 12/42 15/42 42/42
P(X2,Y1)5/42
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Joint Distributions (Example) cont.

• Evaluate P(X?1,1ltY?2)
• sum cells of shaded region below

X\Y 0 1 2 3 totals
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
totals 6/42 9/42 12/42 15/42 42/42
P(X?1,1ltY?2) (3 4 5 6)/42
18/420.429
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Joint Distributions (Example) cont.
Functions Marginal X P(Xxi)
?j(Xxi, Yyj)
Marginal Y P(Yyj)
?i(Xxi, Yyj)
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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
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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 0ltxlt0.4 0ltYlt0.3
0 elsewhere
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Exp-x-y
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Joint Distributions Continuous Case
Example ore grade for copper and zinc
f(x,y) ce-x-y 0ltXlt0.4 0ltYlt0.3 what is
c ? ? cf(x,y)dxdy 1 fx(x) ?00.3
ce-x-y dy - ce-x-y 00.3 - ce-x-0.3
ce-x-0. ?00.4 (-ce-x-0.3 ce-x-0 )
ce-x-0.3 - ce--x 00.4 ce-0.4 -0.3 - ce
0.4 - ce-0.3 - ce -0 1 ce-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
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Independent Random Variables
Random variables X and Y are independent if
occurrence of one -gtno affect others
probability 3 tests 1. iff P(XxYy)
P(Xx) for all X and Y
P(XY) P(X) for all values X and Y
otherwise dependent
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Independent Random Variables
Random variables X and Y are independent if
occurrence of one -gtno affect others
probability 2. iff P(Xx,Yy)
P(Xx)P(Yy) P(X,Y) P(X)P(Y)
otherwise X and Y dependent
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Independent Random Variables
Random variables X and Y are independent if
occurrence of one -gtno affect others 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
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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

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Office of Mineral Resources Management compiled
joint probability distribution
Independent Random Variables(Reclamation Example
cont.)
Size of Mine (000 st) Abandons Mine Reclaims Mine P(X)
0- 100 1/18 2/18 3/18
100 500 2/18 4/18 6/18
500 1,000 2/18 4/18 6/18
gt 1,000 1/18 2/18 3/18
P(Y) 6/18 12/18
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Independent Random Variables Check 1

• Mine Size and Reclamation Independent?
• 1. Does P(XY) P(X) for all values X and Y?
• Does P(Mine 0-100Reclaim)
• 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

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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) 3/1812/18
• 36/324 1/9
• Holds for these values
• Checking it holds for all values X and
  Y
• If not hold for any values dependent

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Independent Random Variables Check 3

• 3. Does F(X,Y) F(X)F(Y) for all values X and
  Y?
• P(Minelt500 and reclaim) 2/18 4/18 1/3
• P(Mine lt 500) P(reclaim) 6/183/18
• 9/1812/18
• 108/324 1/3
• Holds for these values
• Checking if holds for all values X and
  Y
• If not hold for any values dependent

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Are Copper and Zinc Ore Grades Independent
Continuous Case f(x,y) 0.324e-x-y

• Discrete P(XY) P(X)
• Continuous f(xy) f(x)
• f(xy) f(x,y)/f(y) 0.324e-x-y/(-0.324e-0.3
  -y0.324e-y)
• ?
  (-0.324e-0.4-x0.324e-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-x0.324e-x)(-0.324e-0.3
  -y0.324e-y)
• Discrete P(X?x,Y?y) P(X?x)P(Y?y)

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Convolutions - Example
density function of their sum UXY

• X, Y e-X-Y
• Y U-X

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Integrate Exponential Functions

• Integrate Exponential functions
• Let u x2 du/dx 2x dx du/2x

From oil well example
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Changing of Variables Example

• discrete probability distribution
• size Pennsylvania coal mine
• P(X 1) 2-1 1/2
• P(X 2) 2-2 1/4
• P(X 3) 2-3 1/8, etc.

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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 gtP(U X4 1 14 12)
  1/2
• P(X 2) 2-2 1/4 gtP(U X4 1 24 117)
  1/4
• P(X 3) 2-3 gt P(U X4 1
  )
• P(X x) 2-x gt P(U X4 1)

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Changing of Variables Discrete Example

• P(X x) 2-x 1/2x gt 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 (141,
  241, 341, . .

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Changing Variables General Case

• Discrete
• X Px(X) for X a,b,c, . .
• U g(X) gt 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 lt X lt b
• fx(X) 0 elsewhere
• U g(X) gt X g-1(U)
• U fu(U) fx(g-1(U)) for g(a) lt U lt g(b)

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Normal Practice Variable Change Problem

• X N(?, ?) ? lt X lt ?
• f(X) 1 exp(-(X- ?)2/(2 ?2)) dX
• ?(2?)0.5
• Show that (X ?) /? N (0,1) ? lt Z lt ?

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ConvolutionsExtending Variable Change to Joint
Distributions
X, Y f(X,Y) Want density function of sum U
X Y
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Convolutions
special case X and Y are independent, f (x,y)
f1(x) f2(y) and previous equation is reduced to
following
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Convolutions - Example

• Example X (Oil Production) and Y (Gas
  Production) independent random variables
• f(x,y) e-2x3e-3y
• Find density function of their sum UXY?
• g(u) ?0? 2e-2x3e-3(u-x)dx
• complete this example

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Sum Up Random Variables and Probability
Distributions (PS 2)

• Discrete Probability Distributions
• values with probabilities attached
• Cumulative Discrete Probability Functions
• Continuous Probability Distributions

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Chapter 2 Sum up

• Discrete Joint Distributions
• P(Xx,Yy) f(x,y)
• Independent Random Variables
• P(XY) 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)

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Changing Variables

• Discrete
• X Px(X) for X a,b,c, . .
• U g(X) gt 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 lt X lt b
• fx(X) 0 elsewhere
• U g(X) gt X g-1(U)
• U fu(U) fx(g-1(U)) for g(a) lt U lt g(b)

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Convolutions Extending Variable Change to Joint
Distributions
X, Y f(X,Y) Want density function of sum U
X Y
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Convolutions
Special case X and Y are independent, f (X,Y)
f1(X) f2(Y) and previous equation is reduced
to following