-- Jan 30 In-Class Exercise Thread

Sample Space = {TT, TH, HT, HH}. X = {HT,HH}, Y = {TH, HH}, Z = {TH, HT}, Pr(X) = 0.5, Pr(Y) = 0.5 Pr(Z) = 0.5 X and Y are pairwise independent if Pr(X and Y) = Pr(X) and Pr(Y) Proof: Pr(X and Y) = {HT} = 0.25 = Pr(X)*Pr(Y) X and Z are pairwise independent if Pr(X and Z) = Pr(X) and Pr(Z) Proof: Pr(X and Z) = {HT} = 0.25 = Pr(X)*Pr(Z) Y and Z are pairwise independent if Pr(Y and Z) = Pr(Y) and Pr(Z) <=> Pr(Y and Z) = {TH} = 0.25 = Pr(Y)*Pr(Z) X, Y and Z are not mutually independent if Pr(X and Y and Z) != Pr(X)*Pr(Y)*Pr(Z) Proof: Pr(X and Y and Z) = {} = 0, but Pr(X)*Pr(Y)*Pr(Z) = 0.5*0.5*0.5 = 0.5^3 != 0

-- Jan 30 In-Class Exercise Thread

Sample spaces: X: {HH, HC} Y: {HH, CH} Z: {HC, CH} Showing pairwise independence: definition: @BT@Pr(X \cap Y) = Pr(X)Pr(Y)@BT@ @BT@Pr(X \cap Y) = 0.25@BT@ ({HH} or @BT@1/2 * 1/2@BT@), Pr(X)Pr(Y) = 0.25 @BT@Pr(Y \cap Z) = 0.25@BT@ ({TH} or @BT@1/2 * 1/2@BT@), Pr(Y)Pr(Z) = 0.25 @BT@Pr(X \cap Z) = 0.25@BT@ ({HT} or @BT@1/2 * 1/2@BT@), Pr(X)Pr(Z) = 0.25 Showing mutual independence @BT@Pr(X \cap Y \cap Z) = 0@BT@ ({} since no one event occures in each set) Pr(X)Pr(Y)Pr(Z) = 1/8 or (1/2 * 1/2 * 1/2)

-- Jan 30 In-Class Exercise Thread

 Sample space: {HH, HT, TH, TT}

 X,Y,Z pairwise independent. 
 Pr(XnY) = Pr(X)Pr(Y)
 1/4 = 1/2 * 1/2
 Pr(XnZ) = Pr(X)Pr(Z)
 1/4 = 1/2 * 1/2
 Pr(YnZ) = Pr(Y)Pr(Z)
 1/4 = 1/2 * 1/2

 X,Y,Z not mutually independent. 
 Pr(XnYnZ) != Pr(X)Pr(Y)Pr(Z)
 0 != 1/2 * 1/2 * 1/2 = 1/8

Sample space: {HH, HT, TH, TT} X,Y,Z pairwise independent. Pr(XnY) = Pr(X)Pr(Y) 1/4 = 1/2 * 1/2 Pr(XnZ) = Pr(X)Pr(Z) 1/4 = 1/2 * 1/2 Pr(YnZ) = Pr(Y)Pr(Z) 1/4 = 1/2 * 1/2 X,Y,Z not mutually independent. Pr(XnYnZ) != Pr(X)Pr(Y)Pr(Z) 0 != 1/2 * 1/2 * 1/2 = 1/8

-- Jan 30 In-Class Exercise Thread

 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(Y \cap Z) = P(TH) = 0.25 
 P(X \cap Z) = P(HT) = 0.25
 They are all pair-wise independent.

 P(X \cap Y \cap Z) = P(empty set) = 0
 P(X) * P(Y) * P(Z) = 1/8

Sample Space = {HH, TT, HT, TH} 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(Y \cap Z) = P(TH) = 0.25 P(X \cap Z) = P(HT) = 0.25 They are all pair-wise independent. P(X \cap Y \cap Z) = P(empty set) = 0 P(X) * P(Y) * P(Z) = 1/8 So the three event are not mutually independent.

-- Jan 30 In-Class Exercise Thread

For 2 coin toss, there are 4 possible outcomes, HH, TT, HT, TH. Thus, the sample space S = {HH, TT, HT, TH}. For X, there are 2 outcomes, HH, HT meet the requirement, thus, Pr{X} = 0.5. For Y, there are 2 outcomes, TH, HH meet the requirement, thus, Pr{Y} = 0.5. For Z, there are 2 outcomes, HT, TH meet the requirement, thus, Pr{Z} = 0.5 @BT@P(X \cap Y) = P(HH) = 1/4 = 0.25, P(X) * P(Y) = 0.5 * 0.5 = 0.25@BT@. Since @BT@P(X \cap Y) = P(X) * P(Y)@BT@, X and Y are independent. @BT@P(Y \cap Z) = P(TH) = 1/4 = 0.25, P(Y) * P(Z) = 0.5 * 0.5 = 0.25@BT@. Since @BT@P(Y \cap Z) = P(Y) * P(Z)@BT@, Y and Z are independent. @BT@P(X \cap Z) = P(HT) = 1/4 = 0.25, P(X) * P(Z) = 0.5 * 0.5 = 0.25@BT@. Since @BT@P(X \cap Z) = P(X) * P(Z)@BT@, X and Z are independent. Therefore, X, Y, and Z are pairwise independent. @BT@P(X \cap Y \cap Z) = 0@BT@, P(X) * P(Y) * P(Z) = 0.5 * 0.5 * 0.5 = 0.125. Since @BT@P(X \cap Y \cap Z)@BT@ != P(X) * P(Y) * P(Z), X, Y, and Z are not mutually independent.

-- Jan 30 In-Class Exercise Thread

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

-- Jan 30 In-Class Exercise Thread

Sample space: {HH, HT, TT, TH} X: {HH, HT} Y: {HH, TH} Z: {HT, TH} pairwise independent: P(X∩Y) = P(X) * P(Y) = 1/2 * 1/2 = 1/4 P(X∩Z) = P(X) * P(Z) = 1/2 * 1/2 = 1/4 P(Y∩Z) = P(Y) * P(Z) = 1/2 * 1/2 = 1/4 Not mutually independent: As P(X∩Y∩Z) = 0, because X, Y, Z cannot happen at the same time, event X need the first coin is Head, and event Y need the second coin is Head, and event Z need the first coin and the second coin are different. But if it’s the mutually independent, it should be: P(X∩Y∩Z) = P(X)*P(Y)*P(Z) = 1/2 * 1/2 *1/2 = 1/8 So, it is not mutually independent

-- Jan 30 In-Class Exercise Thread

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-- Jan 30 In-Class Exercise Thread

 S = {HH, HT, TH, TT}
 Pr(X) = 1/2
 Pr(Y) = 1/2
 Pr(Z) = 1/2

 Pr(X ^ Y) = Pr({HH}) = 1/4 = Pr(x).Pr(Y)
 Pr(Y ^ Z) = Pr({TH}) = 1/4 = Pr(Y).Pr(Z)
 Pr(Z ^ X) = Pr({HT}) = 1/4 = Pr(Z).Pr(X)

 This shows that events are pairwise independent.

 P(X^Y^Z) = P({}) which is not equal to P(X).P(Y).P(z)
 Hence the events are not mutually independent.

S = {HH, HT, TH, TT} Pr(X) = 1/2 Pr(Y) = 1/2 Pr(Z) = 1/2 Pr(X ^ Y) = Pr({HH}) = 1/4 = Pr(x).Pr(Y) Pr(Y ^ Z) = Pr({TH}) = 1/4 = Pr(Y).Pr(Z) Pr(Z ^ X) = Pr({HT}) = 1/4 = Pr(Z).Pr(X) This shows that events are pairwise independent. P(X^Y^Z) = P({}) which is not equal to P(X).P(Y).P(z) Hence the events are not mutually independent.

-- Jan 30 In-Class Exercise Thread

$$\text{Sample Space} = \{HH, HT, TH, HH\}$$ Consider the experiment of flipping two fair coins. Let X be the event that the first coin lands heads, Y be the event that the second coin lands heads, and Z be the event that both coins land differently. Thus, $$P(X) = P(Y) = P(Z) = \frac{1}{2}$$. $$Pr(X\cap Y) = Pr(\text{HH}) = \frac{1}{4}$$ $$Pr(X\cap Z) = Pr(\text{HT}) = \frac{1}{4}$$ $$Pr(Y\cap Z) = Pr(\text{TH}) = \frac{1}{4}$$ To show pairwise independence, we must show: $$P(X\cap Y) = P(X)P(Y)$$ $$P(X\cap Z) = P(X)P(Z)$$ $$P(Y\cap Z) = P(Y)P(Z)$$ Which is shown by the probabilities enumerated above. To show mutual independence: $$P(X\cap Y\cap Z) = P(X)P(Y)P(Z)$$ However, $$P(X\cap Y\cap Z) = P(\emptyset) =0 \ne \frac{1}{2}\frac{1}{2}\frac{1}{2}=\frac{1}{8}$$ Therefore, these three events X, Y, Z are pairwise independent, but not mutually independent. *Accidentally deleted my comment made during class on Wednesday trying to edit.