Home exercise on chapters 6 and 10 & its solution - Stochastic Calculus PDF

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Home Exercises - Stochastic Calculus 2014 - Home exercises and its solutions - Stochastic Calculus 2014 merged files: Home exercises on Chapters 6 and 10.pdf - Solved exercises on Chapters 5-6 and 10.pdf...

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TMS 165/MSA350 Stochastic Calculus Home Exercises for Chapters 6 and 10 in Klebaner’s Book Througout this set of exercises B = {B(t)}t≥0 denotes Brownian motion. Task 1. Consider the CKLS SDE from Exercise 4 of Exercise session 5 dX (t) = (α + β X(t)) dt + σ X(t)γ dB (t) for t > 0,

X (0) = X0 ,

with parameters α, σ > 0, β ≥ 0 and γ > 1, and where X0 has the stationary distribution (according to Exercise 4 of Exercise session 5) so that the solution {X(t)}t≥0 is a stationary process (see Task 2 below). Show that  Z t X(s)γ dB (s) = C t E

for t ≥ 0,

0

for some strictly negative constant C < 0. Task 2. Let {X(t)}t≥0 be a time homogeneous diffusion process that has a stationary distribution and is started according to that stationary distribution at time t = 0. Prove that X is a stationary process, which is to say that     P X(t1 +h) ≤ x1 , . . . , X(tn +h) ≤ xn = P X(t1 ) ≤ x1 , . . . , X (tn ) ≤ xn for 0 < t1 < . . . < tn and h > 0.

Task 3. Find three SDE that explode, that display transcience but not explosion, and that display recurrence, respectively, but do not feature to exemplify these properties in Klebaner’s book. Also, find three SDE where the issue whether the above three mentioned properties hold depends on the starting value of the SDE.   Task 4. Find a PDE that is solved by the fair price p(x, t) = E max{X (T )−K, 0} 

X(t) = x} (for a constant K > 0) of an European call option at time t ∈ [0, T ) for an asset price {X(t)}t∈[0,T ] given by the CKLS SDE from Exercise 4 of Exercise session 5. Task 5. Can an Ornstein-Uhlenbeck process [a solution to the Langevin equation 5.12 in Klebaner’s book] become a Brownian motion by means of a change of measure? In that case, how? Otherwise, why not? Task 6. Assume that we have observed the solution {X(t)}t∈[0,T ] to the CKLS SDE in Task 1 where the γ coefficient is known. Find the estimators of the coefficients α, β and σ according to the methodolgy of Section 10.6 in Klebaner’s book.

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TMS 165/MSA350 Stochastic Calculus Solved Exercises for Chapters 5-6 and 10 in Klebaner’s book Througout this exercise session B = {B(t)}t≥0 denotes Brownian motion. Exercise 1. Show that −αt

X(t) = e



 σ  √ B(e2αt) − B(1) + x0 2α



for t ≥ 0

is an Ornstein-Uhlenbeck process in the sense that it got the same distributional properties (finite dimensional distributions) as the solution   Z t  −αt αr {X(t)}t≥0 = e x0 + σ e dB(r) 0

t≥0

to the Langevin SDE dX (t) = −α X(t) dt + σ dB(t) for t > 0,

X(0) = x0 ,

where α, σ > 0 and x0 ∈ R are constants. Solution. As both the above X processes are Gaussian they have the same finite dimensional distributions if their mean and covariance functions agree. Here we clearly have E{X(t)} = e−αtx0 for t ≥ 0 for both the X processes. Further, we have   σ 2 −α(s+t) e Cov B (e2αs ) − B (1), B (e2αt ) − B (1) 2α  σ 2 −α(s+t)  2α min{s,t} e e −1−1+1 = 2α  σ 2  −α|s−t| e − e−α(s+t) for s, t ≥ 0 = 2α

Cov{X(s), X (t)} =

for the first X process, while Theorem 4.11 in Klebaner’s book shows that Z min{s,t}  σ 2  −α|s−t| 2 −α(s+t) Cov{X(s), X (t)} = σ e e2αr dr = e − e−α(s+t) 2α 0 for s, t ≥ 0 for the second X process. Exercise 2. Use the expression for an Ornstein Uhlenbeck process expressed in terms of B from Exercise 1 to find the transition density function for the solution to the Langevin SDE (the Ornstein Uhlenbeck process).

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Solution. We have 

  σ  √ B(e2α(t+s) ) − B(1) + x0 2α     σ = e−α(t+s) x0 + √ e−α(t+s) B (e2α(t+s) ) − B (e2αs ) + B (e2αs ) − B (1) 2α  σ −α(t+s)  B (e2α(t+s) ) − B (e2αs ) + e−αtX(s), = √ e 2α

X(t+s) = e−α(t+s)

where   σ √ e−α(t+s) B(e2α(t+s) ) − B(e2αs ) 2α   2 −2αt is an N 0, (σ /(2 α)) (1−e ) -distributed random variable independent of {X(r )}r≤s .   It follows that (X(t+s)|X(s) = x) is N e−αtx, (σ 2 /(2α)) (1−e−2αt ) -distributed, so that   √ d α (y − x e−αt )2 α p(y, t + s, x, s) = exp − 2 P (y, t+s, x, s) = p σ (1 − e−2αt ) dy π (1 − e−2αt ) σ for t+s > s ≥ 0 and x, y ∈ R.

Exercise 3. Solve the Stratanovich SDE dX (t) = −α dt + σ X(t) ∂B (t)

for t > 0,

X(0) = x0 ,

where α, σ > 0 and x0 ∈ R are constants. Solution. By Theorem 5.20 in Klebaner’s book the above SDE is equivalent to the Itˆ o SDE dX(t) =

1

2

 σ 2 X(t) − α dt + σ X (t) dB (t) for t > 0,

X(0) = x0 .

This in turn is a rather simple form of the linear SDE treated in Section 5.3 in Klebaner’s book, with a solution given by  Z X(t) = U (t) x0 − α

0

t

ds U (s)



where U (t) = eσB (t) ,

which is to say that X(t) = x0 eσB (t) − α eσB (t)

Z

0

t

e−σB(s) ds for t ≥ 0.

Exercise 4. The CKLS (Chan-Koralyi-Longstaff-Sanders) SDE is given by dX (t) = (α + β X(t)) dt + σ X(t)γ dB(t) 2

for t > 0,

X(0) = x0 ,

where α, σ, γ, x0 > 0 and β ∈ R are constants. This SDE is used in contemporary mathematical finance research as a model for, e.g., interest rates and/or deseasonalized eletricity prices, and is famous for being very hard to do inference for and very hard to simulate when γ > 1. Determine the stationary distribution for this SDE when it exists. Solution. First note that the fact that α, x0 > 0 ensures that the solution is strictly positive when it exists. From Equation 6.69 in Klebaner’s book we further see that the stationary probability density function is given by  Z x 2 (α + β y ) 1 dy exp π(x) = σ 2 y2γ C x2γ 1

for x > 0,

whenever this function can be normalized to become a density, that is, whenever  Z x Z ∞ 2 (α + β y ) 1 C= dy dx < ∞. exp 2γ σ 2 y2γ 1 0 x The issue whether C is finite or not in turn clearly boils down to check the integrability properties of the function Z x  2 (α + β y ) 1 f(x) = 2γ exp dy x σ 2 y2γ 1 as x ↓ 0 and as x ↑ ∞. Now, as x ↓ 0 we see that    C1 x−2γ for γ ∈ (0, 1/2),    2 f(x) ∼ C2 x2α/σ −1 for γ = 1/2,     C x−2γ exp{−(2α/(σ 2 (2γ −1))) x−(2γ −1) } for γ > 1/2, 3

where C1 , C2 , C3 > 0 are constants. This is to say that we always have the integrability

required as x ↓ 0. When x ↑ ∞ we further see that   C4 x−2γ       2  C5 x−2+2β/σ     f(x) ∼ C6 x−2γ exp{(β/(σ 2 (1−γ))) x2−2γ }     2   C7 x2α/σ −1 exp{(2β/σ 2 ) x}      C x−2γ exp{(β/(σ 2 (1−γ))) x2−2γ + (2α/(σ 2 (1 − 2γ))) x1−2γ } 8

for γ > 1, for γ = 1, for γ ∈ (1/2, 1), for γ = 1/2, for γ ∈ (0, 1/2),

where C4 , . . . , C8 > 0 are constants. This is to say that we have the integrability required

when γ >1

and γ = 1, 2β < σ 2

and γ ∈ (1/2, 1), β ≤ 0 3

and γ ∈ (0, 1/2], β < 0.

Exercise 5. Exercise 6.10 in Klebaner’s book. Solution. See the solution at the end of Klebaner’s book. Exercise 6. Let X be a standard normal distributed random variable. Show how X can be made to have any given probability density function f : R → [0, ∞) by means of a change of probability measure. Also, if X has probability density function f : R → [0, ∞), is it possible to make X have standard normal distributed by a change of probability measure? Solution. Clearly X has probability density function f under the probability measure Q(A) =

Z

√ 2 f(X) 2π eX /2 dP

A

for A ∈ F,

as this gives   Q{X ∈ B} = EQ I{X∈B} √   2 = EP I{X∈B} f(X) 2π eX /2 Z √ 2 2 1 = IB (x) f(x) 2π ex /2 √ e−x /2 dx 2π ZR = f(x) dx for B ⊆ R. B

If X instead has a strictly positive probability density function f : R → (0, ∞) from the beginning, then X is standard normal distributed under the probability measure Q(A) =

Z

A

1 2 1 √ e−X /2 dP f(X) 2π

for A ∈ F,

as this gives   Q{X ∈ B} = EQ I{X∈B}   1 −X 2 /2 1 = EP I{X∈B } √ e f(X) 2π Z 1 2 1 = f(x) dx IB (x) √ e−x /2 f(x) 2π R Z 1 2 √ e−x /2 dx for B ⊆ R. = 2π B If f is not strictly positive, then it is not possible to make X standard normal distributed by means of this approach, as we then have 4

Q{Ω} = EQ {1}   1 −X 2 /2 1 = EP √ e f(X) 2π Z 1 −x2 /2 1 √ e = f(x) dx f(x) 2π {x∈R:f (x)>0} Z 1 2 √ e−x /2 dx = 2π {x∈R:f (x)>0} < 1, so that Q is no longer a probability measure.

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