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The image contains exercises related to stochastic processes, including Brownian motion, Itô calculus, and stochastic differential equations.

Solution

The uploaded image contains mathematical exercises related to stochastic processes, focusing on topics like Brownian motion, Itô calculus, and stochastic differential equations (SDEs). Let's go through the exercises step by step:


Exercise I

Calculate d(eαBt)d(e^{\alpha B_t}) and E(eαBt)E(e^{\alpha B_t}).

This involves applying Itô's Lemma to the exponential of a Brownian motion BtB_t:

  1. Itô's Lemma for f(t,Bt)=eαBtf(t, B_t) = e^{\alpha B_t} gives: df(Bt)=ftdt+fxdBt+122fx2(dBt)2.df(B_t) = \frac{\partial f}{\partial t} dt + \frac{\partial f}{\partial x} dB_t + \frac{1}{2} \frac{\partial^2 f}{\partial x^2} (dB_t)^2.

    Substituting f(Bt)=eαBtf(B_t) = e^{\alpha B_t}: d(eαBt)=αeαBtdBt+12α2eαBtdt.d(e^{\alpha B_t}) = \alpha e^{\alpha B_t} dB_t + \frac{1}{2} \alpha^2 e^{\alpha B_t} dt.

  2. For the expectation E(eαBt)E(e^{\alpha B_t}):

    • The result follows from the Gaussian distribution of BtB_t: BtN(0,t)B_t \sim N(0, t).
    • Using E(eαBt)=e12α2tE(e^{\alpha B_t}) = e^{\frac{1}{2} \alpha^2 t}.

Exercise II

Calculate dYtdY_t for Yt=e2Bt0tcos(s)dsY_t = e^{2B_t} \int_0^t \cos(s) \, ds in two ways.

  1. Direct Differentiation: Apply the product rule: dYt=d(e2Bt)0tcos(s)ds+e2Btd(0tcos(s)ds).dY_t = d\left(e^{2B_t}\right) \int_0^t \cos(s) \, ds + e^{2B_t} d\left(\int_0^t \cos(s) \, ds\right). Use d(e2Bt)d(e^{2B_t}) from Itô's Lemma and d(0tcos(s)ds)=cos(t)dtd\left(\int_0^t \cos(s) \, ds\right) = \cos(t) dt.

  2. Itô's Lemma on YtY_t: Rewrite Yt=e2Btg(t)Y_t = e^{2B_t}g(t), with g(t)=0tcos(s)dsg(t) = \int_0^t \cos(s) \, ds. Then: dYt=Yttdt+YtBtdBt+122YtBt2(dBt)2.dY_t = \frac{\partial Y_t}{\partial t} dt + \frac{\partial Y_t}{\partial B_t} dB_t + \frac{1}{2} \frac{\partial^2 Y_t}{\partial B_t^2} (dB_t)^2.


Exercise III

Solve the SDE: dXt=a(t)Xtdt+b(t)dt+c(t)dBt.dX_t = a(t) X_t dt + b(t) dt + c(t) dB_t.

1. Homogeneous Solution:

Set b=c=0b = c = 0 and solve dXt=a(t)XtdtdX_t = a(t)X_t dt. The solution is: Xt=X0e0ta(s)ds.X_t = X_0 e^{\int_0^t a(s) ds}.

2. Transformation:

Let Y(t)=e0ta(s)dsXtY(t) = e^{-\int_0^t a(s) ds} X_t, simplifying the SDE using Itô's Lemma.

3. Integral Form:

Derive the integral form of YtY_t.

4. Solve dXt=Xt1tdt+dBtdX_t = \frac{-X_t}{1-t} dt + dB_t:

  • This is a specific case of the general SDE.
  • Use Itô's Lemma and calculate the variance of YtY_t.

Exercise IV

For stopping times SS and TT of the stochastic process (Xn)(X_n): Prove that S+TS + T, STS \wedge T, and STS \vee T are stopping times.

  1. Use the definition of stopping times, considering the information up to time tt.
  2. For each operation, verify whether it satisfies the property of stopping times.

Would you like me to solve any specific part in detail? Here are related questions to explore:

  1. What are the key applications of Itô's Lemma in stochastic calculus?
  2. How does the solution of an SDE depend on the choice of drift and diffusion terms?
  3. Why is the expectation E(eαBt)=e12α2tE(e^{\alpha B_t}) = e^{\frac{1}{2}\alpha^2t}?
  4. How can stopping times be combined to create new stopping times?
  5. What is the significance of solving SDEs in finance and physics?

Tip: When working with SDEs, always confirm if the coefficients are adapted processes to ensure the solution's validity.

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Math Problem Analysis

Mathematical Concepts

Stochastic Processes
Brownian Motion
Itô Calculus
Stochastic Differential Equations
Stopping Times

Formulas

Itô's Lemma: df(t, B_t) = (∂f/∂t) dt + (∂f/∂x) dB_t + (1/2)(∂²f/∂x²) (dB_t)²
Exponential of Brownian motion: E(e^{αB_t}) = e^{(1/2)α²t}
General SDE: dX_t = a(t)X_t dt + b(t) dt + c(t) dB_t

Theorems

Itô's Lemma
Properties of Stopping Times in Stochastic Processes

Suitable Grade Level

Graduate-Level Mathematics