BOOK: Arbitrage Theory in Continuous Time, Third Edition, Tomas Bjork 8 Questions: exercise: 4.1, 4.3, 4.4, and 4.5; exercise: 5.6, and 5.12; exercise: 7.4, and 7.5 Handwriting solution is NOT...

Ans. 4.1
a)
dz(t) = α dt + αdW(t)
whereas, Z satisfies SDE.
dZ(t) = αZ(t)dt+αZ(t)dW(t)
Z(0) = 1
Integral form,
Z(t) = 1 + α + α
After moving the expectation within the integral sign in the ds-integral and defining m by m(t) = Z(t)
Then,
m(t) = 1 + α
m(0) = 1
solving these
Z(t) = m(t) =
b)
Z(t) = W^2(t)
dZ(t)=dt+2g(t)dW(t)
Integrated form
W^2(t) = t+ 2
=
c)
dz(t) = α dt + αdW(t)
whereas, Z satisfies SDE.
dZ(t) = αZ(t)dt+αZ(t)dW(t)
Z(0) = 1
Integral form,
Z(t) = 1 + α + α
After moving the expectation within the integral sign in the ds-integral and defining m by m(t) = Z(t)
Then,
m(t) = 1 + α
m(0) = 1
solving these
Z(t) = m(t) = =
d)
we know that
X(t) = x0 + +
W is a wiener process. So rewriting the above equ.
dX(t) = µ(t)dt+σ(t)dW(t)
Where X = Stochastic differential.
X(0) = α
dz(t) = α dt + αdW(t)
dZ(t) = αZ(t)dt+αZ(t)dW(t)
dZ(t) = αZ(t)dt+αZ(t)dW(t)
Z(0) = 1
Integral form,
Z(t) = 1 + α + α
After moving the expectation within the integral sign in the ds-integral and defining m by m(t)...