Hey Everyone,

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

Best,

Chris

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.

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.

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.

<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>

2^{2^2}

X = Pr(A1)Pr(A2) = 0.5 * 1 = 0.5 Y = 1 * 0.5 = 0.5 Z = 0.50.52 = 0.5

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

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.