Use Ito’s formula to show that the solutions of the SDE initial value problems a . { d y = B t d t + t d B t y ( 0 ) = c b . { d y = 2 B t d B t y ( 0 ) = c are (a) y ( t ) = t B t + c (b) y ( t ) = B 2 t − t + c .

Question

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Answer 1

Textbook Question

Chapter 9.4, Problem 1E

Use Ito’s formula to show that the solutions of the SDE initial value problems a . { d y = B t d t + t d B t y ( 0 ) = c b . { d y = 2 B t d B t y ( 0 ) = c are (a) y ( t ) = t B t + c (b) y ( t ) = B 2 t − t + c .

a.

Expert Solution

To determine

To show: The SDE dy=Btdt+tdBt initial value problem has a solution y(t)=tBt+c equation with specified boundary y(0)=c condition.

Explanation of Solution

Given information:

The initial conditions that are given are,

Ito's Formula y(t)=f(t,Bt)=tBt+cdy=Btdt+tdBty(0)=cbrownian motion Bt

Calculation:

As it’s known that by apply Ito's Formula to given equation shown below.

y(t)=f(t,Bt)=tBt+c .

Note that f(t,x)=tx+c The partial derivatives are

∂f∂t=x,∂f∂x=t, and ∂2f∂x2=0.

Therefore,using Ito's Formula gives, the following step below.

dy= ∂f ∂t (t, B t )dt+ ∂f ∂x (t, B t )d B t + 1 2 ∂ 2 f ∂ x 2 (t, B t ) d B t d B t = B t dt+t d B t as required.

By using initial conditions:

y(0)=0Bt+c=c .

b.

Expert Solution

To determine

To show: The SDE dy=2BtdBt initial value problem has a solution y(t)=tBt+c equation with specified boundary y(0)=c condition.

Explanation of Solution

Given information:

The initial conditions that are given are,

Ito's Formula y(t)=f(t,Bt)=tBt+cdy=2BtdBty(0)=cbrownian motion Bt

Calculation:

As it’s known that by apply Ito's Formula

Ito's Formula (brownian motion Bt)y(t)=f(t,Bt)=tBt+cdy=2BtdBty(0)=c

and the partial derivatives of f(t,x)=x2−t+c are ∂f∂t=−1,∂f∂x=2x, and ∂2f∂x2=2. Ito's Formula gives

dy=−1dt+2Bt dBt+122 dBt dBt=−dt+2Bt dBt+dt=2Bt dBt as required.

By using initial conditions.

y(0)=02−0+c=c

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Answer 2

Textbook Question

Chapter 9.4, Problem 1E

Use Ito’s formula to show that the solutions of the SDE initial value problems a . { d y = B t d t + t d B t y ( 0 ) = c b . { d y = 2 B t d B t y ( 0 ) = c are (a) y ( t ) = t B t + c (b) y ( t ) = B 2 t − t + c .

a.

Expert Solution

To determine

To show: The SDE dy=Btdt+tdBt initial value problem has a solution y(t)=tBt+c equation with specified boundary y(0)=c condition.

Explanation of Solution

Given information:

The initial conditions that are given are,

Ito's Formula y(t)=f(t,Bt)=tBt+cdy=Btdt+tdBty(0)=cbrownian motion Bt

Calculation:

As it’s known that by apply Ito's Formula to given equation shown below.

y(t)=f(t,Bt)=tBt+c .

Note that f(t,x)=tx+c The partial derivatives are

∂f∂t=x,∂f∂x=t, and ∂2f∂x2=0.

Therefore,using Ito's Formula gives, the following step below.

dy= ∂f ∂t (t, B t )dt+ ∂f ∂x (t, B t )d B t + 1 2 ∂ 2 f ∂ x 2 (t, B t ) d B t d B t = B t dt+t d B t as required.

By using initial conditions:

y(0)=0Bt+c=c .

b.

Expert Solution

To determine

To show: The SDE dy=2BtdBt initial value problem has a solution y(t)=tBt+c equation with specified boundary y(0)=c condition.

Explanation of Solution

Given information:

The initial conditions that are given are,

Ito's Formula y(t)=f(t,Bt)=tBt+cdy=2BtdBty(0)=cbrownian motion Bt

Calculation:

As it’s known that by apply Ito's Formula

Ito's Formula (brownian motion Bt)y(t)=f(t,Bt)=tBt+cdy=2BtdBty(0)=c

and the partial derivatives of f(t,x)=x2−t+c are ∂f∂t=−1,∂f∂x=2x, and ∂2f∂x2=2. Ito's Formula gives

dy=−1dt+2Bt dBt+122 dBt dBt=−dt+2Bt dBt+dt=2Bt dBt as required.

By using initial conditions.

y(0)=02−0+c=c

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Answer 3

Textbook Question

Chapter 9.4, Problem 1E

Use Ito’s formula to show that the solutions of the SDE initial value problems a . { d y = B t d t + t d B t y ( 0 ) = c b . { d y = 2 B t d B t y ( 0 ) = c are (a) y ( t ) = t B t + c (b) y ( t ) = B 2 t − t + c .

a.

Expert Solution

To determine

To show: The SDE dy=Btdt+tdBt initial value problem has a solution y(t)=tBt+c equation with specified boundary y(0)=c condition.

Explanation of Solution

Given information:

The initial conditions that are given are,

Ito's Formula y(t)=f(t,Bt)=tBt+cdy=Btdt+tdBty(0)=cbrownian motion Bt

Calculation:

As it’s known that by apply Ito's Formula to given equation shown below.

y(t)=f(t,Bt)=tBt+c .

Note that f(t,x)=tx+c The partial derivatives are

∂f∂t=x,∂f∂x=t, and ∂2f∂x2=0.

Therefore,using Ito's Formula gives, the following step below.

dy= ∂f ∂t (t, B t )dt+ ∂f ∂x (t, B t )d B t + 1 2 ∂ 2 f ∂ x 2 (t, B t ) d B t d B t = B t dt+t d B t as required.

By using initial conditions:

y(0)=0Bt+c=c .

b.

Expert Solution

To determine

To show: The SDE dy=2BtdBt initial value problem has a solution y(t)=tBt+c equation with specified boundary y(0)=c condition.

Explanation of Solution

Given information:

The initial conditions that are given are,

Ito's Formula y(t)=f(t,Bt)=tBt+cdy=2BtdBty(0)=cbrownian motion Bt

Calculation:

As it’s known that by apply Ito's Formula

Ito's Formula (brownian motion Bt)y(t)=f(t,Bt)=tBt+cdy=2BtdBty(0)=c

and the partial derivatives of f(t,x)=x2−t+c are ∂f∂t=−1,∂f∂x=2x, and ∂2f∂x2=2. Ito's Formula gives

dy=−1dt+2Bt dBt+122 dBt dBt=−dt+2Bt dBt+dt=2Bt dBt as required.

By using initial conditions.

y(0)=02−0+c=c

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Answer 4

Textbook Question

Chapter 9.4, Problem 1E

Use Ito’s formula to show that the solutions of the SDE initial value problems a . { d y = B t d t + t d B t y ( 0 ) = c b . { d y = 2 B t d B t y ( 0 ) = c are (a) y ( t ) = t B t + c (b) y ( t ) = B 2 t − t + c .

a.

Expert Solution

To determine

To show: The SDE dy=Btdt+tdBt initial value problem has a solution y(t)=tBt+c equation with specified boundary y(0)=c condition.

Explanation of Solution

Given information:

The initial conditions that are given are,

Ito's Formula y(t)=f(t,Bt)=tBt+cdy=Btdt+tdBty(0)=cbrownian motion Bt

Calculation:

As it’s known that by apply Ito's Formula to given equation shown below.

y(t)=f(t,Bt)=tBt+c .

Note that f(t,x)=tx+c The partial derivatives are

∂f∂t=x,∂f∂x=t, and ∂2f∂x2=0.

Therefore,using Ito's Formula gives, the following step below.

dy= ∂f ∂t (t, B t )dt+ ∂f ∂x (t, B t )d B t + 1 2 ∂ 2 f ∂ x 2 (t, B t ) d B t d B t = B t dt+t d B t as required.

By using initial conditions:

y(0)=0Bt+c=c .

b.

Expert Solution

To determine

To show: The SDE dy=2BtdBt initial value problem has a solution y(t)=tBt+c equation with specified boundary y(0)=c condition.

Explanation of Solution

Given information:

The initial conditions that are given are,

Ito's Formula y(t)=f(t,Bt)=tBt+cdy=2BtdBty(0)=cbrownian motion Bt

Calculation:

As it’s known that by apply Ito's Formula

Ito's Formula (brownian motion Bt)y(t)=f(t,Bt)=tBt+cdy=2BtdBty(0)=c

and the partial derivatives of f(t,x)=x2−t+c are ∂f∂t=−1,∂f∂x=2x, and ∂2f∂x2=2. Ito's Formula gives

dy=−1dt+2Bt dBt+122 dBt dBt=−dt+2Bt dBt+dt=2Bt dBt as required.

By using initial conditions.

y(0)=02−0+c=c

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

To show: The SDE dy=Btdt+tdBt initial value problem has a solution y(t)=tBt+c equation with specified boundary y(0)=c condition.

Explanation of Solution

Given information:

The initial conditions that are given are,

Ito's Formula y(t)=f(t,Bt)=tBt+cdy=Btdt+tdBty(0)=cbrownian motion Bt

Calculation:

As it’s known that by apply Ito's Formula to given equation shown below.

y(t)=f(t,Bt)=tBt+c .

Note that f(t,x)=tx+c The partial derivatives are

∂f∂t=x,∂f∂x=t, and ∂2f∂x2=0.

Therefore,using Ito's Formula gives, the following step below.

dy= ∂f ∂t (t, B t )dt+ ∂f ∂x (t, B t )d B t + 1 2 ∂ 2 f ∂ x 2 (t, B t ) d B t d B t = B t dt+t d B t as required.

By using initial conditions:

y(0)=0Bt+c=c .

b.

Expert Solution

To determine

To show: The SDE dy=2BtdBt initial value problem has a solution y(t)=tBt+c equation with specified boundary y(0)=c condition.

Explanation of Solution

Given information:

The initial conditions that are given are,

Ito's Formula y(t)=f(t,Bt)=tBt+cdy=2BtdBty(0)=cbrownian motion Bt

Calculation:

As it’s known that by apply Ito's Formula

Ito's Formula (brownian motion Bt)y(t)=f(t,Bt)=tBt+cdy=2BtdBty(0)=c

and the partial derivatives of f(t,x)=x2−t+c are ∂f∂t=−1,∂f∂x=2x, and ∂2f∂x2=2. Ito's Formula gives

dy=−1dt+2Bt dBt+122 dBt dBt=−dt+2Bt dBt+dt=2Bt dBt as required.

By using initial conditions.

y(0)=02−0+c=c

Answer 8

a.

Expert Solution

To determine

To show: The SDE dy=Btdt+tdBt initial value problem has a solution y(t)=tBt+c equation with specified boundary y(0)=c condition.

Explanation of Solution

Given information:

The initial conditions that are given are,

Ito's Formula y(t)=f(t,Bt)=tBt+cdy=Btdt+tdBty(0)=cbrownian motion Bt

Calculation:

As it’s known that by apply Ito's Formula to given equation shown below.

y(t)=f(t,Bt)=tBt+c .

Note that f(t,x)=tx+c The partial derivatives are

∂f∂t=x,∂f∂x=t, and ∂2f∂x2=0.

Therefore,using Ito's Formula gives, the following step below.

dy= ∂f ∂t (t, B t )dt+ ∂f ∂x (t, B t )d B t + 1 2 ∂ 2 f ∂ x 2 (t, B t ) d B t d B t = B t dt+t d B t as required.

By using initial conditions:

y(0)=0Bt+c=c .

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To show: The SDE dy=Btdt+tdBt initial value problem has a solution y(t)=tBt+c equation with specified boundary y(0)=c condition.

Answer 13

Explanation of Solution

Given information:

The initial conditions that are given are,

Ito's Formula y(t)=f(t,Bt)=tBt+cdy=Btdt+tdBty(0)=cbrownian motion Bt

Calculation:

As it’s known that by apply Ito's Formula to given equation shown below.

y(t)=f(t,Bt)=tBt+c .

Note that f(t,x)=tx+c The partial derivatives are

∂f∂t=x,∂f∂x=t, and ∂2f∂x2=0.

Therefore,using Ito's Formula gives, the following step below.

dy= ∂f ∂t (t, B t )dt+ ∂f ∂x (t, B t )d B t + 1 2 ∂ 2 f ∂ x 2 (t, B t ) d B t d B t = B t dt+t d B t as required.

By using initial conditions:

y(0)=0Bt+c=c .

Answer 14

b.

Expert Solution

To determine

To show: The SDE dy=2BtdBt initial value problem has a solution y(t)=tBt+c equation with specified boundary y(0)=c condition.

Explanation of Solution

Given information:

The initial conditions that are given are,

Ito's Formula y(t)=f(t,Bt)=tBt+cdy=2BtdBty(0)=cbrownian motion Bt

Calculation:

As it’s known that by apply Ito's Formula

Ito's Formula (brownian motion Bt)y(t)=f(t,Bt)=tBt+cdy=2BtdBty(0)=c

and the partial derivatives of f(t,x)=x2−t+c are ∂f∂t=−1,∂f∂x=2x, and ∂2f∂x2=2. Ito's Formula gives

dy=−1dt+2Bt dBt+122 dBt dBt=−dt+2Bt dBt+dt=2Bt dBt as required.

By using initial conditions.

y(0)=02−0+c=c

Answer 15

b.

Expert Solution

Answer 16

b.

Expert Solution

Answer 17

Expert Solution

Answer 18

To determine

To show: The SDE dy=2BtdBt initial value problem has a solution y(t)=tBt+c equation with specified boundary y(0)=c condition.

Answer 19

Explanation of Solution

Given information:

The initial conditions that are given are,

Ito's Formula y(t)=f(t,Bt)=tBt+cdy=2BtdBty(0)=cbrownian motion Bt

Calculation:

As it’s known that by apply Ito's Formula

Ito's Formula (brownian motion Bt)y(t)=f(t,Bt)=tBt+cdy=2BtdBty(0)=c

and the partial derivatives of f(t,x)=x2−t+c are ∂f∂t=−1,∂f∂x=2x, and ∂2f∂x2=2. Ito's Formula gives

dy=−1dt+2Bt dBt+122 dBt dBt=−dt+2Bt dBt+dt=2Bt dBt as required.

By using initial conditions.

y(0)=02−0+c=c

Answer 20

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Use Ito’s formula to show that the solutions of the SDE initial value problems a . { d y = B t d t + t d B t y ( 0 ) = c b . { d y = 2 B t d B t y ( 0 ) = c are (a) y ( t ) = t B t + c (b) y ( t ) = B 2 t − t + c .

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