2. discrete random variables part iv: joint pmfs
TRANSCRIPT
2.DiscreteRandomVariablesPartIV:JointPMFs
ECE302Fall2009TR3‐4:15pmPurdueUniversity,SchoolofECE
Prof.IlyaPollak
JointPMFpX,YofXandY• IfXandYarediscreterandomvariables,theirjointprobabilitymassfuncQonisdefinedaspX,Y(x,y)=P(X=xandY=y).
JointPMFpX,YofXandY• IfXandYarediscreterandomvariables,theirjointprobabilitymassfuncQonisdefinedaspX,Y(x,y)=P(X=xandY=y).
• TheindividualPMFsofXandYarecalledthemarginalPMFsandcanbeobtainedfromthejointPMF:
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pX (x) = pX ,Y (x,y)y∑
pY (y) = pX ,Y (x,y)x∑
Example2.9(Fig2.10)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
1
1
2
2
3
3 4
4
Example2.9(Fig2.10)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
Column sums: marginal PMF pX(x)
1
1
2
2
3
3 4
4
Example2.9(Fig2.10)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
Column sums: marginal PMF pX(x)
E.g., pX(1) = P(X=1) = P(X=1, Y=1) + P(X=1, Y=2) + P(X=1, Y=3) + P(X=1, Y=4)
1
1
2
2
3
3 4
4
Example2.9(Fig2.10)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
Column sums: marginal PMF pX(x)
E.g., pX(1) = P(X=1) = P(X=1, Y=1) + P(X=1, Y=2) + P(X=1, Y=3) + P(X=1, Y=4) = pX,Y(1,1) + pX,Y(1,2) + pX,Y(1,3) + pX,Y(1,4) =
1
1
2
2
3
3 4
4
Example2.9(Fig2.10)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
Column sums: marginal PMF pX(x)
E.g., pX(1) = P(X=1) = P(X=1, Y=1) + P(X=1, Y=2) + P(X=1, Y=3) + P(X=1, Y=4) = pX,Y(1,1) + pX,Y(1,2) + pX,Y(1,3) + pX,Y(1,4) =
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pX ,Y (1,y)y∑
1
1
2
2
3
3 4
4
Example2.9(Fig2.10)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
Column sums: marginal PMF pX(x)
E.g., pX(1) = P(X=1) = P(X=1, Y=1) + P(X=1, Y=2) + P(X=1, Y=3) + P(X=1, Y=4) = pX,Y(1,1) + pX,Y(1,2) + pX,Y(1,3) + pX,Y(1,4) = = 1/20 + 1/20 + 1/20 = 3/20
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pX ,Y (1,y)y∑
1
1
2
2
3
3 4
4
Example2.9(Fig2.10)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
3/20 Column sums: marginal PMF pX(x)
E.g., pX(1) = P(X=1) = P(X=1, Y=1) + P(X=1, Y=2) + P(X=1, Y=3) + P(X=1, Y=4) = pX,Y(1,1) + pX,Y(1,2) + pX,Y(1,3) + pX,Y(1,4) = = 1/20 + 1/20 + 1/20 = 3/20
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pX ,Y (1,y)y∑
1
1
2
2
3
3 4
4
Example2.9(Fig2.10)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
3/20 6/20 8/20 3/20
Column sums: marginal PMF pX(x)
E.g., pX(1) = P(X=1) = P(X=1, Y=1) + P(X=1, Y=2) + P(X=1, Y=3) + P(X=1, Y=4) = pX,Y(1,1) + pX,Y(1,2) + pX,Y(1,3) + pX,Y(1,4) = = 1/20 + 1/20 + 1/20 = 3/20
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pX ,Y (1,y)y∑
1
1
2
2
3
3 4
4
Example2.9(Fig2.10)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
3/20
7/20
7/20
3/20
Row sums: marginal PMF pY(y)
3/20 6/20 8/20 3/20
Column sums: marginal PMF pX(x)
E.g., pX(1) = P(X=1) = P(X=1, Y=1) + P(X=1, Y=2) + P(X=1, Y=3) + P(X=1, Y=4) = pX,Y(1,1) + pX,Y(1,2) + pX,Y(1,3) + pX,Y(1,4) = = 1/20 + 1/20 + 1/20 = 3/20
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pX ,Y (1,y)y∑
1
1
2
2
3
3 4
4
JointPMFformorethantwodiscreterandomvariables
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pX1 ,X 2 ,…,Xn(x1,x2,…,xn ) = P(X1 = x1,X2 = x2,…,Xn = xn )
JointPMFformorethantwodiscreterandomvariables
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pX1 ,X 2 ,…,Xn(x1,x2,…,xn ) = P(X1 = x1,X2 = x2,…,Xn = xn )
The marginal PMF for each Xi can be obtained fromthe joint PMF by summing over all the x j 's other than xi,for example,
pX1(x1) = … pX1 ,X 2 ,…,Xn
(x1,x2,…,xn )xn
∑x3
∑x2
∑
FuncQonsofmanydiscreterandomvariables
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Let Y = g(X1,X2,…,Xn ). Then
pY (y) = pX1 ,X 2 ,…,Xn(x1,x2,…,xn )
{(x1 ,…,xn )|g(x1 ,…,xn )= y}∑
FuncQonsofmanydiscreterandomvariables
€
Let Y = g(X1,X2,…,Xn ). Then
pY (y) = pX1 ,X 2 ,…,Xn(x1,x2,…,xn )
{(x1 ,…,xn )|g(x1 ,…,xn )= y}∑
E[g(X1,X2,…,Xn )] = g(x1,…,xn )pX1 ,…,Xn(x1,…,xn )
x1 ,…,xn
∑
FuncQonsofmanydiscreterandomvariables
€
Let Y = g(X1,X2,…,Xn ). Then
pY (y) = pX1 ,X 2 ,…,Xn(x1,x2,…,xn )
{(x1 ,…,xn )|g(x1 ,…,xn )= y}∑
E[g(X1,X2,…,Xn )] = g(x1,…,xn )pX1 ,…,Xn(x1,…,xn )
x1 ,…,xn
∑
If g(X1,X2,…,Xn ) = a0 + a1X1 + a2X2 +…+ anXn, thenE[g(X1,X2,…,Xn )] = a0 + a1E[X1] + a2E[X2] +…+ anE[Xn ]
Example2.9(conQnued)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
3/20
7/20
7/20
3/20
3/20 6/20 8/20 3/20
1
1
2
2
3
3 4
4
Z=X+2Y (a) Find pZ(k). (b) Find E[Z].
Example2.9(conQnued)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
1
1
2
2
3
3 4
4
Z=X+2Y (a) Find pZ(k). (b) Find E[Z].
Z = X+2Y is between 3 and 12, with probability 1
Example2.9(conQnued)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
1
1
2
2
3
3 4
4
Z=X+2Y (a) Find pZ(k). (b) Find E[Z].
k pZ(k)
3
4
5
6
7
8
9
10
11
12
Z = X+2Y is between 3 and 12, with probability 1
Example2.9(conQnued)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
1
1
2
2
3
3 4
4
Z=X+2Y (a) Find pZ(k). (b) Find E[Z].
k pZ(k)
3
4
5
6
7
8
9
10
11
12
Z = X+2Y is between 3 and 12, with probability 1 pZ(3)=P(X+2Y = 3) = P(X=1 and Y=1) = 1/20
Example2.9(conQnued)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
1
1
2
2
3
3 4
4
Z=X+2Y (a) Find pZ(k). (b) Find E[Z].
k pZ(k)
3 1/20
4
5
6
7
8
9
10
11
12
Z = X+2Y is between 3 and 12, with probability 1 pZ(3)=P(X+2Y = 3) = P(X=1 and Y=1) = 1/20
Example2.9(conQnued)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
1
1
2
2
3
3 4
4
Z=X+2Y (a) Find pZ(k). (b) Find E[Z].
k pZ(k)
3 1/20
4
5
6
7
8
9
10
11
12
Z = X+2Y is between 3 and 12, with probability 1 pZ(3)=P(X+2Y = 3) = P(X=1 and Y=1) = 1/20 pZ(4)=P(X+2Y = 4) = P(X=2 and Y=1) = 1/20
Example2.9(conQnued)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
1
1
2
2
3
3 4
4
Z=X+2Y (a) Find pZ(k). (b) Find E[Z].
k pZ(k)
3 1/20
4 1/20
5
6
7
8
9
10
11
12
Z = X+2Y is between 3 and 12, with probability 1 pZ(3)=P(X+2Y = 3) = P(X=1 and Y=1) = 1/20 pZ(4)=P(X+2Y = 4) = P(X=2 and Y=1) = 1/20
Example2.9(conQnued)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
1
1
2
2
3
3 4
4
Z=X+2Y (a) Find pZ(k). (b) Find E[Z].
k pZ(k)
3 1/20
4 1/20
5
6
7
8
9
10
11
12
Z = X+2Y is between 3 and 12, with probability 1 pZ(3)=P(X+2Y = 3) = P(X=1 and Y=1) = 1/20 pZ(4)=P(X+2Y = 4) = P(X=2 and Y=1) = 1/20 pZ(5)=P(X+2Y = 5) = P(X=1 and Y=2) + P(X=3 and Y=1) = 2/20
Example2.9(conQnued)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
1
1
2
2
3
3 4
4
Z=X+2Y (a) Find pZ(k). (b) Find E[Z].
k pZ(k)
3 1/20
4 1/20
5 2/20
6
7
8
9
10
11
12
Z = X+2Y is between 3 and 12, with probability 1 pZ(3)=P(X+2Y = 3) = P(X=1 and Y=1) = 1/20 pZ(4)=P(X+2Y = 4) = P(X=2 and Y=1) = 1/20 pZ(5)=P(X+2Y = 5) = P(X=1 and Y=2) + P(X=3 and Y=1) = 2/20
Example2.9(conQnued)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
1
1
2
2
3
3 4
4
Z=X+2Y (a) Find pZ(k). (b) Find E[Z].
k pZ(k)
3 1/20
4 1/20
5 2/20
6 2/20
7 4/20
8 3/20
9 3/20
10 2/20
11 1/20
12 1/20
Z = X+2Y is between 3 and 12, with probability 1 pZ(3)=P(X+2Y = 3) = P(X=1 and Y=1) = 1/20 pZ(4)=P(X+2Y = 4) = P(X=2 and Y=1) = 1/20 pZ(5)=P(X+2Y = 5) = P(X=1 and Y=2) + P(X=3 and Y=1) = 2/20, etc
Example2.9(conQnued)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
1
1
2
2
3
3 4
4
Z=X+2Y (a) Find pZ(k). (b) Find E[Z].
k pZ(k)
3 1/20
4 1/20
5 2/20
6 2/20
7 4/20
8 3/20
9 3/20
10 2/20
11 1/20
12 1/20
E[Z] = 3 · (1/20) + 4 · (1/20) + 5 · (2/20) + 6 · (2/20) + 7 · (4/20) + 8 · (3/20) + 9 · (3/20) + 10 · (2/20) + 11 · (1/20) + 12 · (1/20) = 7.55
Example2.9(conQnued)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
3/20
7/20
7/20
3/20
3/20 6/20 8/20 3/20
1
1
2
2
3
3 4
4
Z=X+2Y (a) Find pZ(k). (b) Find E[Z].
k pZ(k)
3 1/20
4 1/20
5 2/20
6 2/20
7 4/20
8 3/20
9 3/20
10 2/20
11 1/20
12 1/20
Alternatively, E[Z] = E[X] + 2E[Y]
Example2.9(conQnued)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
3/20
7/20
7/20
3/20
3/20 6/20 8/20 3/20
1
1
2
2
3
3 4
4
Z=X+2Y (a) Find pZ(k). (b) Find E[Z].
k pZ(k)
3 1/20
4 1/20
5 2/20
6 2/20
7 4/20
8 3/20
9 3/20
10 2/20
11 1/20
12 1/20
Alternatively, E[Z] = E[X] + 2E[Y] E[X] = 1 · (3/20) + 2 · (6/20) + 3 · (8/20) + 4 · (3/20) = 51/20
Example2.9(conQnued)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
3/20
7/20
7/20
3/20
3/20 6/20 8/20 3/20
1
1
2
2
3
3 4
4
Z=X+2Y (a) Find pZ(k). (b) Find E[Z].
k pZ(k)
3 1/20
4 1/20
5 2/20
6 2/20
7 4/20
8 3/20
9 3/20
10 2/20
11 1/20
12 1/20
Alternatively, E[Z] = E[X] + 2E[Y] E[X] = 1 · (3/20) + 2 · (6/20) + 3 · (8/20) + 4 · (3/20) = 51/20 E[Y] = 1 · (3/20) + 2 · (7/20) + 3 · (7/20) + 4 · (3/20) = 50/20
Example2.9(conQnued)
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
3/20
7/20
7/20
3/20
3/20 6/20 8/20 3/20
1
1
2
2
3
3 4
4
Z=X+2Y (a) Find pZ(k). (b) Find E[Z].
k pZ(k)
3 1/20
4 1/20
5 2/20
6 2/20
7 4/20
8 3/20
9 3/20
10 2/20
11 1/20
12 1/20
Alternatively, E[Z] = E[X] + 2E[Y] = 51/20 + 2 · (50/20) = 7.55 E[X] = 1 · (3/20) + 2 · (6/20) + 3 · (8/20) + 4 · (3/20) = 51/20 E[Y] = 1 · (3/20) + 2 · (7/20) + 3 · (7/20) + 4 · (3/20) = 50/20
Example2.10:meanofabinomialrandomvariable
• Nstudentsinclass• Eachhasprobabilitypofge]nganA
• Allgradesareindependent• X=numberofstudentsthatgetanA
• WhatisE[X]?
Example2.10:meanofabinomialrandomvariable
€
Let Xi =1, if the i - th student gets an A0, otherwise
Example2.10:meanofabinomialrandomvariable
€
Let Xi =1, if the i - th student gets an A0, otherwise
X1,X2,…,XN are Bernoulli random variables with E[Xi] = p.
Example2.10:meanofabinomialrandomvariable
€
Let Xi =1, if the i - th student gets an A0, otherwise
X1,X2,…,XN are Bernoulli random variables with E[Xi] = p.X1 + X2 +…+ XN = X is the number of successes in N independent Bernoulli trials.
Example2.10:meanofabinomialrandomvariable
€
Let Xi =1, if the i - th student gets an A0, otherwise
X1,X2,…,XN are Bernoulli random variables with E[Xi] = p.X1 + X2 +…+ XN = X is the number of successes in N independent Bernoulli trials.Therefore, X is a binomial random variable, with the following PMF :
pX (k) =Nk
pk (1− p)N−k
Example2.10:meanofabinomialrandomvariable
€
Let Xi =1, if the i - th student gets an A0, otherwise
X1,X2,…,XN are Bernoulli random variables with E[Xi] = p.X1 + X2 +…+ XN = X is the number of successes in N independent Bernoulli trials.Therefore, X is a binomial random variable, with the following PMF :
pX (k) =Nk
pk (1− p)N−k
Hard way of evaluating E[X] :
E[X] = kNk
pk (1− p)N−k
k= 0
N
∑
Example2.10:meanofabinomialrandomvariable
€
Let Xi =1, if the i - th student gets an A0, otherwise
X1,X2,…,XN are Bernoulli random variables with E[Xi] = p.X1 + X2 +…+ XN = X is the number of successes in N independent Bernoulli trials.Therefore, X is a binomial random variable, with the following PMF :
pX (k) =Nk
pk (1− p)N−k
Hard way of evaluating E[X] :
E[X] = kNk
pk (1− p)N−k
k= 0
N
∑
Easy way :E[X] = E[X1] + E[X2] +… + E[XN ] = Np
Example2.10:meanofabinomialrandomvariable
€
Let Xi =1, if the i - th student gets an A0, otherwise
X1,X2,…,XN are Bernoulli random variables with E[Xi] = p.X1 + X2 +…+ XN = X is the number of successes in N independent Bernoulli trials.Therefore, X is a binomial random variable, with the following PMF :
pX (k) =Nk
pk (1− p)N−k
Hard way of evaluating E[X] :
E[X] = kNk
pk (1− p)N−k
k= 0
N
∑
Easy way :E[X] = E[X1] + E[X2] +… + E[XN ] = NpE.g., if N = 60 and p =1/5, then E[X] =12.
Example2.11
• npeoplethrowtheirhatsinabox• eachpicksuponehatatrandom
• X=numberofpeoplewhogettheirownhatback
• WhatisE[X]?
Example2.11
€
Let Xi =1, if the i - th person gets his own hat0, otherwise
Example2.11
€
Let Xi =1, if the i - th person gets his own hat0, otherwise
pXi(k) =
1/n, if k =1 1−1/n, if k = 00, otherwise
Example2.11
€
Let Xi =1, if the i - th person gets his own hat0, otherwise
pXi(k) =
1/n, if k =1 1−1/n, if k = 00, otherwise
E[Xi] =1⋅ 1n
+ 0 ⋅ 1− 1n
=
1n
Example2.11
€
Let Xi =1, if the i - th person gets his own hat0, otherwise
pXi(k) =
1/n, if k =1 1−1/n, if k = 00, otherwise
E[Xi] =1⋅ 1n
+ 0 ⋅ 1− 1n
=
1n
X = X1 + X2 +…+ Xn
Example2.11
€
Let Xi =1, if the i - th person gets his own hat0, otherwise
pXi(k) =
1/n, if k =1 1−1/n, if k = 00, otherwise
E[Xi] =1⋅ 1n
+ 0 ⋅ 1− 1n
=
1n
X = X1 + X2 +…+ Xn
E[X] = E[X1] + E[X2] +…+ E[Xn ] = n ⋅ 1n
=1
CondiQoningarandomvariableonanevent
• CondiQonalPMFofarandomvariableX,condiQonedonaneventAwithP(A)>0:
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pX |A (x) = P(X = x | A) =P( X = x{ }∩ A)
P(A)
CondiQoningarandomvariableonanevent
• CondiQonalPMFofarandomvariableX,condiQonedonaneventAwithP(A)>0:
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pX |A (x) = P(X = x | A) =P( X = x{ }∩ A)
P(A)Note :
P(A) = P( X = x{ }∩ A)x∑ , and therefore
pX |A (x) =1x∑
CondiQoningarandomvariableonanotherrandomvariable
• IfXandYarerandomvariables,condiQonalPMFpX|YofXgivenYisdefined,forally>0,as:
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pX |Y (x | y) = P(X = x |Y = y) =P X = x{ }∩ Y = y{ }( )
P Y = y{ }( )=pX ,Y (x,y)pY (y)
CondiQonalPMF:examples
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
1
1
2
2
3
3 4
4
x 2 3 4
pX|Y(x|4) 1/3 1/3 1/3
CondiQonalPMF:examples
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
1
1
2
2
3
3 4
4
x 2 3 4
pX|Y(x|4) 1/3 1/3 1/3
x 1 2 3 4
pX|Y(x|3) 1/7 2/7 3/7 1/7
CondiQonalPMF:examples
0 1/20 1/20 1/20
1/20 2/20 3/20 1/20
1/20 2/20 3/20 1/20
1/20 1/20 1/20 0
y
x
Joint PMF pX,Y(x,y)
1
1
2
2
3
3 4
4
x 2 3 4
pX|Y(x|4) 1/3 1/3 1/3
x 1 2 3 4
pX|Y(x|3) 1/7 2/7 3/7 1/7
y pY|X(y|3)
1 1/8
2 3/8
3 3/8
4 1/8
Example:memorylesspropertyofageometricrandomvariable
• X=thenumberofindependentcoinflipsunQlfirstH,withP(H)=p.
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pX (k) =(1− p)k−1 p, if k =1,2,…0, otherwise
Example:memorylesspropertyofageometricrandomvariable
• X=thenumberofindependentcoinflipsunQlfirstH,withP(H)=p.
• GiventhatthefirsttrialisaT,whatisthecondiQonalPMFofX–1?
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pX (k) =(1− p)k−1 p, if k =1,2,…0, otherwise
Example:memorylesspropertyofageometricrandomvariable
• X=thenumberofindependentcoinflipsunQlfirstH,withP(H)=p.
• GiventhatthefirsttrialisaT,whatisthecondiQonalPMFofX–1?
€
pX (k) =(1− p)k−1 p, if k =1,2,…0, otherwise
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Y = X −1A = X ≥ 2{ }pY |A (m) = ?
Example:memorylesspropertyofageometricrandomvariable
• X=thenumberofindependentcoinflipsunQlfirstH,withP(H)=p.
• GiventhatthefirsttrialisaT,whatisthecondiQonalPMFofX–1?
• IntuiQon:sinceflipsareindependent,theoutcomeofthefirstflipdoesnotinfluencethefuture.
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pX (k) =(1− p)k−1 p, if k =1,2,…0, otherwise
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Y = X −1A = X ≥ 2{ }pY |A (m) = ?
Example:memorylesspropertyofageometricrandomvariable
• X=thenumberofindependentcoinflipsunQlfirstH,withP(H)=p.
• GiventhatthefirsttrialisaT,whatisthecondiQonalPMFofX–1?
• IntuiQon:sinceflipsareindependent,theoutcomeofthefirstflipdoesnotinfluencethefuture.Therefore,theremainingflipsaierthefirstoneshouldhavethesamecondiQonaldistribuQonaspX.Inotherwords,theanswermustbepY|A=pX.
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pX (k) =(1− p)k−1 p, if k =1,2,…0, otherwise
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Y = X −1A = X ≥ 2{ }pY |A (m) = ?
Example:memorylesspropertyofageometricrandomvariable
• X=thenumberofindependentcoinflipsunQlfirstH,withP(H)=p.
• GiventhatthefirsttrialisaT,whatisthecondiQonalPMFofX–1?
€
pX (k) =(1− p)k−1 p, if k =1,2,…0, otherwise
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Y = X −1; A = X ≥ 2{ }
pY |A (m) =P(Y = m,X ≥ 2)P(X ≥ 2)
Example:memorylesspropertyofageometricrandomvariable
• X=thenumberofindependentcoinflipsunQlfirstH,withP(H)=p.
• GiventhatthefirsttrialisaT,whatisthecondiQonalPMFofX–1?
€
pX (k) =(1− p)k−1 p, if k =1,2,…0, otherwise
€
Y = X −1; A = X ≥ 2{ }
pY |A (m) =P(Y = m,X ≥ 2)P(X ≥ 2)
=P(X = m +1,X ≥ 2)
P(X ≥ 2)
Example:memorylesspropertyofageometricrandomvariable
• X=thenumberofindependentcoinflipsunQlfirstH,withP(H)=p.
• GiventhatthefirsttrialisaT,whatisthecondiQonalPMFofX–1?
€
pX (k) =(1− p)k−1 p, if k =1,2,…0, otherwise
€
Y = X −1; A = X ≥ 2{ }
pY |A (m) =P(Y = m,X ≥ 2)P(X ≥ 2)
=P(X = m +1,X ≥ 2)
P(X ≥ 2)=P(X = m +1)P(X ≥ 2)
, if m =1,2,…
0, otherwise
Example:memorylesspropertyofageometricrandomvariable
• X=thenumberofindependentcoinflipsunQlfirstH,withP(H)=p.
• GiventhatthefirsttrialisaT,whatisthecondiQonalPMFofX–1?
€
pX (k) =(1− p)k−1 p, if k =1,2,…0, otherwise
€
Y = X −1; A = X ≥ 2{ }
pY |A (m) =P(Y = m,X ≥ 2)P(X ≥ 2)
=P(X = m +1,X ≥ 2)
P(X ≥ 2)=P(X = m +1)P(X ≥ 2)
, if m =1,2,…
0, otherwise
=(1− p)m p
1− p, if m =1,2,…
0, otherwise
Example:memorylesspropertyofageometricrandomvariable
• X=thenumberofindependentcoinflipsunQlfirstH,withP(H)=p.
• GiventhatthefirsttrialisaT,whatisthecondiQonalPMFofX–1?
€
pX (k) =(1− p)k−1 p, if k =1,2,…0, otherwise
€
Y = X −1; A = X ≥ 2{ }
pY |A (m) =P(Y = m,X ≥ 2)P(X ≥ 2)
=P(X = m +1,X ≥ 2)
P(X ≥ 2)=P(X = m +1)P(X ≥ 2)
, if m =1,2,…
0, otherwise
=(1− p)m p
1− p, if m =1,2,…
0, otherwise =
(1− p)m−1 p, if m =1,2,…0, otherwise
= pX (m)
Example:memorylesspropertyofageometricrandomvariable
• X=thenumberofindependentcoinflipsunQlfirstH,withP(H)=p.
• GiventhatthefirstNtrialsareallT’s,findcondiQonalPMFofX–N.
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pX (k) =(1− p)k−1 p, if k =1,2,…0, otherwise
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YN = X − N; AN = X ≥ N +1{ }
Example:memorylesspropertyofageometricrandomvariable
• X=thenumberofindependentcoinflipsunQlfirstH,withP(H)=p.
• GiventhatthefirstNtrialsareallT’s,findcondiQonalPMFofX–N.
€
pX (k) =(1− p)k−1 p, if k =1,2,…0, otherwise
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YN = X − N; AN = X ≥ N +1{ }
pYN |AN(m) =
(1− p)m−1 p, if m =1,2,…0, otherwise
= pX (m)
Example
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
Let A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s).
Example
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
Let A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s).
€
pS (s) =
12 /45, if s =1 ?, if s = 2?, if s = 3
0, otherwise
Example
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
Let A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s).
€
pS (s) =
12 /45, if s =1 18 /45, if s = 2
?, if s = 30, otherwise
Example
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
Let A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s).
€
pS (s) =
12 /45, if s =1 18 /45, if s = 215 /45, if s = 30, otherwise
Example
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
Let A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s).
€
pS (s) =
12 /45, if s =1 18 /45, if s = 215 /45, if s = 30, otherwise
€
P(A) =12 /45 +18 /45 = 30 /45
Example
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
Let A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s).
€
pS (s) =
12 /45, if s =1 18 /45, if s = 215 /45, if s = 30, otherwise
€
P(A) =12 /45 +18 /45 = 30 /45
pS |A (s) =
12 /4530 /45
=1230
, if s =1
?, if s = 20, otherwise
Example
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
Let A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s).
€
pS (s) =
12 /45, if s =1 18 /45, if s = 215 /45, if s = 30, otherwise
€
P(A) =12 /45 +18 /45 = 30 /45
pS |A (s) =
12 /4530 /45
=1230
, if s =1
18 /4530 /45
=1830
, if s = 2
0, otherwise
Example
y
r
Joint PMF pR,Y(r,y)
1
0
2
1
2
3
Let A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s). (b) Find pR,Y(r,y).
4/45
6/45
6/45
-1
-2
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
Example
y
r
Joint PMF pR,Y(r,y)
1
0
2
1
2
3
Let A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s). (b) Find pR,Y(r,y).
6/45
4/45 9/45
6/45 9/45
6/45
-1
-2
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
Example
y
r
Joint PMF pR,Y(r,y)
1
0
2
1
2
3
Let A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s). (b) Find pR,Y(r,y).
0 0 2/45
0 6/45 3/45
4/45 9/45 0
6/45 9/45 0
6/45 0 0
-1
-2
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
ExampleLet A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s). (b) Find pR,Y(r,y). (c) Find pX|A(x).
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
Event A P(A)=30/45
ExampleLet A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s). (b) Find pR,Y(r,y). (c) Find pX|A(x).
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
X=2
Event A P(A)=30/45
ExampleLet A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s). (b) Find pR,Y(r,y). (c) Find pX|A(x).
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
X=2 X=3
Event A P(A)=30/45
ExampleLet A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s). (b) Find pR,Y(r,y). (c) Find pX|A(x).
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
X=2 X=3 X=4
Event A P(A)=30/45
ExampleLet A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s). (b) Find pR,Y(r,y). (c) Find pX|A(x).
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
X=2 X=3 X=4 X=5
Event A P(A)=30/45
ExampleLet A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s). (b) Find pR,Y(r,y). (c) Find pX|A(x).
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
X=2 X=3 X=4 X=5
€
pX |A (x) =
4 /4530 /45
=430
, if x = 2
?, if x = 3 ?, if x = 4?, if x = 50, otherwise
Event A P(A)=30/45
ExampleLet A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s). (b) Find pR,Y(r,y). (c) Find pX|A(x).
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
X=2 X=3 X=4 X=5
€
pX |A (x) =
4 /4530 /45
=430
, if x = 2
6 /45 + 6 /4530 /45
=1230
, if x = 3
?, if x = 4?, if x = 50, otherwise
Event A P(A)=30/45
ExampleLet A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s). (b) Find pR,Y(r,y). (c) Find pX|A(x).
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
X=2 X=3 X=4 X=5
€
pX |A (x) =
4 /4530 /45
=430
, if x = 2
6 /45 + 6 /4530 /45
=1230
, if x = 3
2 /45 + 9 /4530 /45
=1130
, if x = 4
?, if x = 50, otherwise
Event A P(A)=30/45
ExampleLet A = {S ≠ 3}, Y = R − S, X = R + S (a) Find pS(s) and pS|A(s). (b) Find pR,Y(r,y). (c) Find pX|A(x).
6/45 9/45 0
6/45 9/45 3/45
4/45 6/45 2/45
s
r
Joint PMF pR,S(r,s)
1 1
2
2
3
3
X=2 X=3 X=4 X=5
€
pX |A (x) =
4 /4530 /45
=430
, if x = 2
6 /45 + 6 /4530 /45
=1230
, if x = 3
2 /45 + 9 /4530 /45
=1130
, if x = 4
3/4530 /45
=3
30, if x = 5
0, otherwise
Event A P(A)=30/45