IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32.

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1)

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B)

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1)

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32.

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n

events

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n

events we can use this formula for the first term, thus for this term we get

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n

events we can use this formula for the first term, thus for this term we get

P (B)

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n

events we can use this formula for the first term, thus for this term we get

P (B) = P (∪ni=1Ai)

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n

events we can use this formula for the first term, thus for this term we get

P (B) = P (∪ni=1Ai) =

∑

i

P (Ai) −∑

i<j

P (Ai ∩ Aj) +∑

i<j<k

P (Ai ∩ Aj ∩ Ak) −

· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n

events we can use this formula for the first term, thus for this term we get

P (B) = P (∪ni=1Ai) =

∑

i

P (Ai) −∑

i<j

P (Ai ∩ Aj) +∑

i<j<k

P (Ai ∩ Aj ∩ Ak) −

· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .

For the third term on the right hand side of Equation (2) we introduce theevents

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n

events we can use this formula for the first term, thus for this term we get

P (B) = P (∪ni=1Ai) =

∑

i

P (Ai) −∑

i<j

P (Ai ∩ Aj) +∑

i<j<k

P (Ai ∩ Aj ∩ Ak) −

· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .

For the third term on the right hand side of Equation (2) we introduce theevents Bi

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n

events we can use this formula for the first term, thus for this term we get

P (B) = P (∪ni=1Ai) =

∑

i

P (Ai) −∑

i<j

P (Ai ∩ Aj) +∑

i<j<k

P (Ai ∩ Aj ∩ Ak) −

· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .

For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n

events we can use this formula for the first term, thus for this term we get

P (B) = P (∪ni=1Ai) =

∑

i

P (Ai) −∑

i<j

P (Ai ∩ Aj) +∑

i<j<k

P (Ai ∩ Aj ∩ Ak) −

· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .

For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n

events we can use this formula for the first term, thus for this term we get

P (B) = P (∪ni=1Ai) =

∑

i

P (Ai) −∑

i<j

P (Ai ∩ Aj) +∑

i<j<k

P (Ai ∩ Aj ∩ Ak) −

· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .

For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as

P (B ∩ An+1)

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n

events we can use this formula for the first term, thus for this term we get

P (B) = P (∪ni=1Ai) =

∑

i

P (Ai) −∑

i<j

P (Ai ∩ Aj) +∑

i<j<k

P (Ai ∩ Aj ∩ Ak) −

· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .

For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as

P (B ∩ An+1) = P (∪ni=1 (Ai ∩ An+1))

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n

events we can use this formula for the first term, thus for this term we get

P (B) = P (∪ni=1Ai) =

∑

i

P (Ai) −∑

i<j

P (Ai ∩ Aj) +∑

i<j<k

P (Ai ∩ Aj ∩ Ak) −

· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .

For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as

P (B ∩ An+1) = P (∪ni=1 (Ai ∩ An+1)) = P (∪n

i=1Bi) ,

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n

events we can use this formula for the first term, thus for this term we get

P (B) = P (∪ni=1Ai) =

∑

i

P (Ai) −∑

i<j

P (Ai ∩ Aj) +∑

i<j<k

P (Ai ∩ Aj ∩ Ak) −

· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .

For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as

P (B ∩ An+1) = P (∪ni=1 (Ai ∩ An+1)) = P (∪n

i=1Bi) ,

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n

events we can use this formula for the first term, thus for this term we get

P (B) = P (∪ni=1Ai) =

∑

i

P (Ai) −∑

i<j

P (Ai ∩ Aj) +∑

i<j<k

P (Ai ∩ Aj ∩ Ak) −

· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .

For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as

P (B ∩ An+1) = P (∪ni=1 (Ai ∩ An+1)) = P (∪n

i=1Bi) ,

implying that we can use the inclusion-exclusion formula for this term too.

IMM - DTU 02405 Probability

2007-2-8BFN/bfn

Solution for exercise 1.3.12 in Pitman

We first recall that a proof by mathematical induction includes two steps

1. Prove that the stated formula is true for some n0

2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.

We know from exercise 1.3.11 that the formula is valid for n = 3 and consider

P(

∪n+1

i=1 Ai

)

= P ((∪ni=1Ai) ∪ An+1) . (1)

Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪

ni=1Ai.Now with the expression

for B inserted in (1) we get

P(

∪n+1

i=1 Ai

)

= P (B ∪ An+1) .

With exclusion-inclusion for two events (Page 22) we get

P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)

As

B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪

ni=1 (Ai ∩ An+1)

we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n

events we can use this formula for the first term, thus for this term we get

P (B) = P (∪ni=1Ai) =

∑

i

P (Ai) −∑

i<j

P (Ai ∩ Aj) +∑

i<j<k

P (Ai ∩ Aj ∩ Ak) −

· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .

For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as

P (B ∩ An+1) = P (∪ni=1 (Ai ∩ An+1)) = P (∪n

i=1Bi) ,

implying that we can use the inclusion-exclusion formula for this term too.The proof is completed by writing down the expansion explicitly.

1