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Jan 30 In-Class Exercise Thread.

Hey Everyone,

Post your solutions to the Jan 30 In-Class Exercise to this thread.

Best,

Chris

Hey Everyone, Post your solutions to the Jan 30 In-Class Exercise to this thread. Best, Chris

-- Jan 30 In-Class Exercise Thread

 Sample Space = {HH, HT, TH, TT}
 P(X) = P(HH U HT) = 0.5
 P(Y) = P(HH U TH) = 0.5
 P(Z) = P(HT U TH) = 0.5
 P(X \cap Y) = P(HH) = 0.25 = P(X)P(Y)
 P(Y \cap Z) = P(TH) = 0.25 = P(Y)P(Z)
 P(X \cap Z) = P(HT) = 0.25 = P(X)P(Z)
 They are all pair-wise independent.
 P(X \cap Y \cap Z) = P(empty set) = 0 != P(X)P(Y)P(Z)
 They are not mutually independent.

(Edited: 2019-01-30)

Sample Space = {HH, HT, TH, TT} P(X) = P(HH U HT) = 0.5 P(Y) = P(HH U TH) = 0.5 P(Z) = P(HT U TH) = 0.5 P(X \cap Y) = P(HH) = 0.25 = P(X)P(Y) P(Y \cap Z) = P(TH) = 0.25 = P(Y)P(Z) P(X \cap Z) = P(HT) = 0.25 = P(X)P(Z) They are all pair-wise independent. P(X \cap Y \cap Z) = P(empty set) = 0 != P(X)P(Y)P(Z) They are not mutually independent.

-- Jan 30 In-Class Exercise Thread

Sample space: HH,HT,TH,TT 1. X = .5 {HH,HT} 2. Y = .5 {TH,HH} 3. Z = .5 {TH,HT}

Pairwise Independence: P(X n Y) = 1/4 = P(X)P(Y) = (.5)(.5) P(X n Z) = 1/4 = P(X)P(Z) = (.5)(.5) P(Y n Z) = 1/4 = P(Y)P(Z) = (.5)(.5)

Mutual Independence: P(X n Y n Z) = 0, but P(X)P(Y)P(Z) = .5^3, thus the events are not mutually independent.

(Edited: 2019-01-30)

Sample space: HH,HT,TH,TT 1. X = .5 {HH,HT} 2. Y = .5 {TH,HH} 3. Z = .5 {TH,HT} Pairwise Independence: P(X n Y) = 1/4 = P(X)P(Y) = (.5)(.5) P(X n Z) = 1/4 = P(X)P(Z) = (.5)(.5) P(Y n Z) = 1/4 = P(Y)P(Z) = (.5)(.5) Mutual Independence: P(X n Y n Z) = 0, but P(X)P(Y)P(Z) = .5^3, thus the events are not mutually independent.

-- Jan 30 In-Class Exercise Thread

((resource:1548885245908-1008845637.jpg|Resource Description for 1548885245908-1008845637.jpg))

-- Jan 30 In-Class Exercise Thread

 Sample Space   S = {HH, HT, TH, TT}

 Pr{X} = First Head  = 1/2 (HH, HT)
 Pr{Y} = Second Head = 1/2 (HH, TH)
 Pr{Z} = Different   = 1/2 (HT, TH)

 PAIRWISE INDEPENDENCE

 Pr{XnY} = Pr{X} * Pr{Y} = 1/2 * 1/2 = 1/4
 There is a 1/4 chance the outcome is HH.

 Pr{XnZ} = Pr{X} * Pr{Z} = 1/2 * 1/2 = 1/4.
 There is a 1/4 chance the outcome is HT.

 Pr{YnZ} = Pr{Y} * Pr{Z} = 1/2 * 1/2 = 1/4.
 There is a 1/4 chance the outcome is TH.

 NOT MUTUALLY INDEPENDENT

 Pr{XnYnZ} does not equal Pr{X} * Pr{Y} * Pr{Z} = 1/2 * 1/2 * 1/2 = 1/8
 There is a 0 chance the first toss is heads, the second toss is heads, and the two flips are different.

(Edited: 2019-01-30)

Sample Space S = {HH, HT, TH, TT} Pr{X} = First Head = 1/2 (HH, HT) Pr{Y} = Second Head = 1/2 (HH, TH) Pr{Z} = Different = 1/2 (HT, TH) PAIRWISE INDEPENDENCE Pr{XnY} = Pr{X} * Pr{Y} = 1/2 * 1/2 = 1/4 There is a 1/4 chance the outcome is HH. Pr{XnZ} = Pr{X} * Pr{Z} = 1/2 * 1/2 = 1/4. There is a 1/4 chance the outcome is HT. Pr{YnZ} = Pr{Y} * Pr{Z} = 1/2 * 1/2 = 1/4. There is a 1/4 chance the outcome is TH. NOT MUTUALLY INDEPENDENT Pr{XnYnZ} does not equal Pr{X} * Pr{Y} * Pr{Z} = 1/2 * 1/2 * 1/2 = 1/8 There is a 0 chance the first toss is heads, the second toss is heads, and the two flips are different.

-- Jan 30 In-Class Exercise Thread

Sample space: {HH, HT, TH, TT} 
 
Pr(X) = 2/4 = 1/2, {HH, HT}
Pr(Y) = 2/4 = 1/2, {HH, TH}
Pr(Z) = 2/4 = 1/2, {HT, TH} 
 
Pairwise independent
Pr(X^Y) = Pr(X) * Pr(Y) = 1/2 * 1/2 = 1/4, {HH}
Pr(Y^Z) = Pr(Y) * Pr(Z) = 1/2 * 1/2 = 1/4, {TH}
Pr(X^Z) = Pr(X) * Pr(Z) = 1/2 * 1/2 = 1/4, {HT} 
 
Not mutually independent
Pr(X^Y^Z) = 0
Pr(X) * Pr(Y) * Pr(Z) = 1/8

(Edited: 2019-02-23)

<pre> Sample space: {HH, HT, TH, TT} Pr(X) = 2/4 = 1/2, {HH, HT} Pr(Y) = 2/4 = 1/2, {HH, TH} Pr(Z) = 2/4 = 1/2, {HT, TH} Pairwise independent Pr(X^Y) = Pr(X) * Pr(Y) = 1/2 * 1/2 = 1/4, {HH} Pr(Y^Z) = Pr(Y) * Pr(Z) = 1/2 * 1/2 = 1/4, {TH} Pr(X^Z) = Pr(X) * Pr(Z) = 1/2 * 1/2 = 1/4, {HT} Not mutually independent Pr(X^Y^Z) = 0 Pr(X) * Pr(Y) * Pr(Z) = 1/8 </pre>

-- Jan 30 In-Class Exercise Thread

`2^{2^2}`

@BT@2^{2^2}@BT@

-- Jan 30 In-Class Exercise Thread

X = Pr(A1)Pr(A2) = 0.5 * 1 = 0.5 Y = 1 * 0.5 = 0.5 Z = 0.5*0.5*2 = 0.5

-- Jan 30 In-Class Exercise Thread

 S= {HH, HT, TH, TT}
 A ={HH, HT}
 B ={HH, TH}
 C ={HT, TH}

 Pr(A) = Pr(B) = Pr(C)=0.5

`Pr (A ^^ B) = Pr({HH}) = 0.25`

 and P(A)* P(B) = 0.5 * 0.5 = 0.25 
 i.e. Pr(A and B) = P(A)* P(B)

likewise `P(A ^^ C) = P({HT}) = 0.25`

 i.e, Pr(A and C) = P(A)* P(C)

 P(C and B) = P({TH}) = 0.25 
 i.e, Pr(C and B) = P(B)* P(C)

 Since P(A and B and C) = P ({}) != P(A)*P(B)*P(C)  therefore not mutually independent

(Edited: 2019-01-30)

S= {HH, HT, TH, TT} A ={HH, HT} B ={HH, TH} C ={HT, TH} Pr(A) = Pr(B) = Pr(C)=0.5 @BT@Pr (A ^^ B) = Pr({HH}) = 0.25@BT@ and P(A)* P(B) = 0.5 * 0.5 = 0.25 i.e. Pr(A and B) = P(A)* P(B) likewise @BT@P(A ^^ C) = P({HT}) = 0.25@BT@ i.e, Pr(A and C) = P(A)* P(C) P(C and B) = P({TH}) = 0.25 i.e, Pr(C and B) = P(B)* P(C) Since P(A and B and C) = P ({}) != P(A)*P(B)*P(C) therefore not mutually independent

-- Jan 30 In-Class Exercise Thread

The sample space is{ (HT), (TH), (TT), (HH)}

P(X) = 0.5 = {(HT), (HH)}

P(Y) = 0.5 = {HH, TH}

P(Z) = 0.5 = {HT, TH}

P(XnY) = 1/4 -> (XnY)= {HH}

P(XnZ) = 1/4 -> (XnZ)={HT}

P(YnZ) = 1/4 -> (YnZ)={TH}

So they are pairwise independent

P(X n Y n Z) = 0

X n Y n Z) = {}

But P(X) P(Y) P(Z)=1/8.

So they are mutually dependent.

(Edited: 2019-02-03)

The sample space is{ (HT), (TH), (TT), (HH)} P(X) = 0.5 = {(HT), (HH)} P(Y) = 0.5 = {HH, TH} P(Z) = 0.5 = {HT, TH} P(XnY) = 1/4 -> (XnY)= {HH} P(XnZ) = 1/4 -> (XnZ)={HT} P(YnZ) = 1/4 -> (YnZ)={TH} So they are pairwise independent P(X n Y n Z) = 0 X n Y n Z) = {} But P(X) P(Y) P(Z)=1/8. So they are mutually dependent.