Question

Let A1, A2 and A3 be events except with respective probabilities 1/6 , 1/5, and 1/4.Let N be the number of these events that occur.

a) Write down a formula for N in terms of indicators.

b) Find E(N).



In each of the following cases, calculate Var(N):

c) A1, A2, A3 are disjoint;

d) they are independent;

e) A1 is in A2 is in A3.

Answer 1

Answer a)

Answer b)

Since I_i are indicators R.V.s, we have that E(I₁) = 1/6 , E(I₂) = 1/5 , and E(I₃) = 1/4

By linearity of expected value:

E(N) = E(I₁ + I₂ + I₃) = E(I₁) + E(I₂) + E(I₃) = 1/6 + 1/5 + 1/4 = (10 + 12 + 15)/60 = 37/60

Answer c)

Var(N) = Var(I₁ + I₂ + I₃) (By Def. of N)

Var(N) = E[(I₁ + I₂ + I₃)²] ? [E(I₁ + I₂ + I₃)]² (Computational formula for Var)

The second term of this expression is [E(I₁ + I₂ + I₃)]² = E(N)² = (37/60)² = 1369/3600

E[(I₁ + I₂ + I₃)²] = E[I₁² + I₂² + I₃² + 2I₁I₂ + 2I₁I₃ + 2I₂I₃] (Squaring inside)

E = E(I₁²) + E(I₂²) + E(I₃²) + 2E(I₁I₂) + 2E(I₁I₃) + 2E(I₂I₃) (Linearity of E)

E = E(I₁) + E(I₂) + E(I₃) + 2E(I₁I₂) + 2E(I₁I₃) + 2E(I₂I₃)     (Idempotence of I_i)

But, I_iI_j = I_i?j , so P(I_i?j = 1) = 0 since events are mutually disjoint. So, E(I_iI_j ) = 0

E[(I₁ + I₂ + I₃)²] = E(I₁) + E(I₂) + E(I₃) = E(N) = 37/60  Combining these two expressions:

Var(N) = E(N) ? E(N)² = 37/60 - 1369/3600 = 851/3600

Answer d)

Note that Var(I₁) = pq = (1/6)*(5/6) = 5/36. Likewise, Var(I₂) = 4/25 , and Var(I₃) = 3/16.

By independence, Var(N) = Var(I₁ + I₂ + I₃) = Var(I₁) +Var(I₂) +Var(I₃) = 5/36 + 4/25 + 3/16 = (500+576+675)/3600

Var(N) = 1751/3600

Answer e)

We proceed as we did for part c).

Var(N) = Var(I₁ + I₂ + I₃) (By Def. of N)

Var(N) = E[(I₁ + I₂ + I₃)²] ? [E(I₁ + I₂ + I₃)]² (Computational formula for Var)

The second term was previously computed in c).

For the first term:

E[(I₁ + I₂ + I₃)²] = E(I₁) + E(I₂) + E(I₃) + 2E(I₁I₂) + 2E(I₁I₃) + 2E(I₂I₃) (Previous calculation)

But, I₁I₂ = I₁ since A1 ? A2. Likewise, I₁I₃ = I₁ since A1 ? A3 and I₂I₃ = I₂ since A2 ? A3. Hence

E[(I₁ + I₂ + I₃)²] = E(I₁) + E(I₂) + E(I₃) + 2E(I₁I₂) + 2E(I₁I₃) + 2E(I₂I₃)

E[(I₁ + I₂ + I₃)²] = 5E(I₁) + 3E(I₂) + E(I₃)

E[(I₁ + I₂ + I₃)²] = 5*1/6 + 3*1/5 + 1/4

E[(I₁ + I₂ + I₃)²] = (50+36+15)/60 = 101/60

Finally,

Var(N) = E[(I₁ + I₂ + I₃)²] ? E(N)² = 101/60-1369/3600 = (6060-1369)/3600

Var(N) = 4691/3600