📝 Exercise Solutions: Multivariate Itô's Formula and SDE Solving in Practice
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This page contains detailed solutions to key exercises from Chapter 5 (Multivariate Itô's Formula) and Chapter 6 (Exact Solutions of Stochastic Differential Equations) of the Stochastic Differential Equations course. The content covers the calculus rules for multi-dimensional Brownian motion, the construction and verification of martingales, Bessel processes, and advanced practical techniques for solving complex SDEs using integrating factor methods and undetermined function methods.
Part I: Chapter 5 Multidimensional Itô's Formula and Martingale Verification
Exercise 1
Problem Let \(W = (W^1, \cdots, W^n)\) be an \(n\)-dimensional Brownian motion. Prove that \(Y(t) = |W(t)|^2 - nt\) (\(t \ge 0\)) is a martingale.
Solution (click to expand)
This problem can be directly proven using the multidimensional Itô's formula, showing the drift term is zero.
First, expand the squared norm of the \(n\)-dimensional Brownian motion as the sum of squares of its components:
Apply the one-dimensional Itô's formula to a single component \((W^i(t))^2\):
Summing over all components, using the linearity of the differential:
Now consider the original process \(Y(t) = |W(t)|^2 - nt\) and compute its differential:
Writing it in integral form:
Since the right-hand side consists only of Itô integrals with respect to Brownian motion, and the integrands \(2W^i(s)\) are square-integrable on compact intervals, this integral process is a martingale. Therefore, \(Y(t)\) is also a martingale. Q.E.D.
Exercise 2
Problem Let \(W = (W^1, \cdots, W^n)^T\) be an \(n\)-dimensional Brownian motion. Denote \(R = |W|\). Prove that \(R\) satisfies the following stochastic Bessel equation:
Solution (click to expand)
Here we define the multivariate function \(f(x) = |x| = \left(\sum_{i=1}^n (x^i)^2\right)^{1/2}\). Compute its partial derivatives:
First-order partial derivatives:
Second-order partial derivatives (using the quotient rule):
The multidimensional Itô's formula requires the Laplacian (sum of all second-order pure partials). Summing them:
Substitute \(X_t = W_t\) into the \(n\)-dimensional Itô's formula \(df(W_t) = \sum_{i=1}^n f_{x^i} dW^i + \frac{1}{2} \sum_{i=1}^n f_{x^i x^i} dt\) (Note: Since Brownian motions in different dimensions are independent, cross terms \(dW^i dW^j = 0\)):
Rearranging yields the classic form of the high-dimensional Bessel process:
Q.E.D.
Exercise 3
Problem (1) Verify that \(X = (\cos W, \sin W)\) is a solution to \(dX^1 = -\frac{1}{2}X^1 dt - X^2 dW, \quad dX^2 = -\frac{1}{2}X^2 dt + X^1 dW\) (2) Prove that if \(X = (X^1, X^2)\) is a solution to the above system, then \(|X|\) is a constant independent of time.
Solution (click to expand)
(1) Verifying the solution form Given \(X^1 = \cos W(t)\), apply the one-dimensional Itô's formula:
Given \(X^2 = \sin W(t)\), similarly apply Itô's formula:
The computed results exactly match the SDE given in the problem statement. Therefore, \(X = (\cos W, \sin W)\) is a solution to the system.
(2) Proving norm conservation Consider the differential of \(|X|^2 = (X^1)^2 + (X^2)^2\). By the product rule (or Itô's formula):
Substitute the system equations, noting the quadratic variation \(\langle dX^1, dX^1 \rangle = (-X^2)^2 dt = (X^2)^2 dt\):
Similarly, for \(X^2\):
Adding the two equations:
Since \(d(|X|^2) = 0\), the squared norm does not change with time, implying \(|X|\) is a constant independent of time. Q.E.D.
Exercise 4
Problem Prove that \(X(t) = (W(t)+t)\exp\left(-W(t)-\frac{1}{2}t\right)\) is a martingale.
Solution (click to expand)
The standard approach to verify a martingale is: Let \(X(t) = u(W(t), t)\), expand using Itô's formula, and prove the drift term (the \(dt\) term) is identically zero.
Define the function \(u(x, t) = (x+t)\exp\left(-x-\frac{1}{2}t\right)\). Compute its partial derivatives:
Derivative with respect to \(t\):
First derivative with respect to \(x\):
Second derivative with respect to \(x\):
Now substitute into the Itô drift term formula: \(drift = u_t + \frac{1}{2}u_{xx}\)
Factor out the common term \(\exp\left(-x-\frac{1}{2}t\right)\):
Since the drift term \(u_t + \frac{1}{2}u_{xx} \equiv 0\), this implies \(dX(t) = u_x dW(t)\). As there is no \(dt\) term, this integral process constitutes a martingale. Q.E.D.
Exercise 5
Problem Prove that the solution to the stochastic differential equation \(dX(t) = \frac{1}{3}X(t)^{1/3} dt + X(t)^{2/3} dW(t)\) with initial value \(X(0) = x_0 > 0\) is \(X(t) = \left( x_0^{1/3} + \frac{1}{3}W(t) \right)^3\).
Solution (click to expand)
This problem is a classic example of the "guess-and-verify" method. We directly treat the given solution \(X(t)\) as a composite function and compute its stochastic differential via Itô's formula to see if it matches the given SDE.
Define the auxiliary process \(Y(t) = x_0^{1/3} + \frac{1}{3}W(t)\). Then the proposed solution can be written as \(X(t) = Y(t)^3\).
First, the differential of the auxiliary process \(Y(t)\) is:
Its quadratic variation is:
Next, apply Itô's formula to \(f(Y) = Y^3\). Compute derivatives: \(f'(Y) = 3Y^2\), \(f''(Y) = 6Y\).
Substitute \(dY(t)\) and \((dY(t))^2\):
Finally, since \(Y(t) = X(t)^{1/3}\), substitute back to eliminate the auxiliary variable \(Y\):
This exactly matches the stochastic differential equation given in the problem! Moreover, at \(t=0\), \(X(0) = (x_0^{1/3} + 0)^3 = x_0\), satisfying the initial condition. Thus the proof is complete.
Part II: Chapter 6 Practical Solutions of Stochastic Differential Equations
Exercise 1
Problem Solve the stochastic differential equations (1) \(dX = X dt + e^{-t} dW\) (2) \(dX_1 = dt + dW_1, \quad dX_2 = X_1 dW_2\)
Solution (Click to expand)
(1) Solving \(dX_t = X_t dt + e^{-t} dW_t\) This is a linear SDE, and we can use the integrating factor method. Rearranging gives \(dX_t - X_t dt = e^{-t} dW_t\). Consider multiplying by the integrating factor \(F_t = e^{-t}\). We examine the differential of the process \(Y_t = e^{-t} X_t\):
Substituting \(dX_t\) from the original equation:
Integrating both sides:
Substituting back \(Y_t = e^{-t}X_t\), we obtain the final solution:
(2) Solving \(dX_1 = dt + dW_1, \quad dX_2 = X_1 dW_2\) This is a hierarchical SDE. First, solve the first layer. Directly integrate the first equation:
Substitute the explicit expression for \(X_1(t)\) into the second equation:
Direct integration yields the final solution:
Exercise 2
Problem Prove that \(X(t) = (a\cos W(t), b\sin W(t))^T\) (where \(a, b\) are positive constants) is a solution to the following stochastic differential equation
Solution (Click to expand)
This problem introduces matrix form into the SDE. We expand the given matrix into a system of equations. Let the matrix \(M = \begin{pmatrix} 0 & -a/b \\ b/a & 0 \end{pmatrix}\). Then \(MX = \begin{pmatrix} 0 & -a/b \\ b/a & 0 \end{pmatrix} \begin{pmatrix} X^1 \\ X^2 \end{pmatrix} = \begin{pmatrix} -\frac{a}{b}X^2 \\ \frac{b}{a}X^1 \end{pmatrix}\).
Therefore, the original SDE is equivalent to the following two scalar equations:
Now we verify the given solution \(X^1 = a\cos W, X^2 = b\sin W\). Apply Itô's formula to \(X^1\):
Substituting the definitions of \(X^1\) and \(X^2\) into the right-hand side: Since \(X^2 = b\sin W\), i.e., \(\sin W = X^2/b\), we have \(-a\sin W = -a(X^2/b) = -\frac{a}{b}X^2\).
This perfectly matches the first scalar equation!
Similarly, verify \(X^2\):
Since \(X^1 = a\cos W\), i.e., \(\cos W = X^1/a\), we have \(b\cos W = \frac{b}{a}X^1\). Substituting:
This perfectly matches the second scalar equation. Verification complete.
Exercise 3
Problem Prove that \(X_t = e^{W_t}\) is a solution to the following stochastic differential equation:
Solution (Click to expand)
This problem is the simplest degenerate version of the Black-Scholes geometric Brownian motion model.
Let the function \(f(x) = e^x\). Substituting its derivatives \(f'(x) = e^x, f''(x) = e^x\) and \(W_t\) into Itô's formula:
Since the quadratic variation \((dW_t)^2 = dt\), and substituting the exponential function:
Replacing the coefficients on the right-hand side with the definition \(X_t = e^{W_t}\):
Rearranging the order gives \(dX_t = \frac{1}{2}X_t dt + X_t dW_t\), which perfectly matches the equation in the problem. Q.E.D.
Exercise 4
Problem Solve the stochastic differential equation
Solution (Click to expand)
Since the coefficients of the equation explicitly contain time \(t\) and Brownian motion \(W_t\), we use the Guess and Verify method. Assume the solution has the form \(X_t = f(t, W_t)\). Expand Itô's formula for the multivariate function \(f\):
Compare it term-by-term with the given SDE:
1. Match the coefficient of \(dW_t\):
Partially integrate with respect to \(w\) to obtain the structure of the function \(f\):
where \(g(t)\) is an undetermined integration constant function depending only on time.
2. Match the coefficient of \(dt\): First, compute the other partial derivatives of the assumed function \(f\): \(f_t(t, w) = e^t w^2 + g'(t)\) \(f_{ww}(t, w) = 2e^t\)
Substitute them into the drift term, which must equal the \(dt\) coefficient of the original equation:
The \(dt\) coefficient of the original equation is \(e^t(1+w^2)\). Equating the two:
This implies \(g(t) = C\) (a constant).
3. Apply the initial condition: The solution form is determined as \(X_t = W_t + e^t W_t^2 + C\). Given \(X_0 = 0\), and \(W_0 = 0\):
Therefore, the exact solution to this stochastic differential equation is:
Exercise 5
Problem Solve the stochastic differential equation
Solution (Click to expand)
This is a very challenging nonlinear SDE. Since the equation contains \(1/X_t\), it suggests we first use a square transformation to eliminate the denominator.
Step 1: Change of variables (Linearization) Let \(Y_t = X_t^2\). Apply Itô's formula to it:
Substitute the original equation \(dX_t = \frac{1}{X_t}dt + \alpha X_t dW_t\), and note the quadratic variation \((dX_t)^2 = \alpha^2 X_t^2 dt = \alpha^2 Y_t dt\):
Rearranging yields a linear SDE for \(Y_t\):
Step 2: Construct an integrating factor to solve the linear SDE To eliminate the proportional term in \(Y_t\), we construct an integrating factor \(F_t\) satisfying \(dF_t = F_t ( -\alpha^2 dt - 2\alpha dW_t )\). Based on knowledge of geometric Brownian motion, this integrating factor is:
(Note: Expanding via Itô's formula verifies \(dF_t = F_t (3\alpha^2 dt - 2\alpha dW_t)\)). We now compute the differential of the product \(d(Y_t F_t)\) using the product rule \(d(YF) = Y dF + F dY + \langle dY, dF \rangle\): The cross-variation term \(\langle dY_t, dF_t \rangle = (2\alpha Y_t)(-2\alpha F_t) dt = -4\alpha^2 Y_t F_t dt\).
Expanding the calculation:
Observe the coefficient of \(dW_t\): \(Y_t F_t(-2\alpha) + F_t(2\alpha Y_t) = 0\) (perfect cancellation). Observe the coefficient of \(dt\): \(Y_t F_t(3\alpha^2) + F_t(2 + \alpha^2 Y_t) - 4\alpha^2 Y_t F_t = Y_t F_t (3\alpha^2 + \alpha^2 - 4\alpha^2) + 2F_t = 2F_t\) (perfect cancellation of the \(Y_t\) term).
Therefore, it elegantly simplifies to:
Step 3: Integration and recovery Integrate the above from \(0\) to \(t\):
Substituting the initial conditions \(Y_0 = x_0^2\) and \(F_0 = \exp(0) = 1\), and rearranging to solve for \(Y_t\):
Substituting the explicit expressions for \(F_t\) and \(F_s\):
Finally, take the square root to recover \(X_t\) (since the problem specifies \(X_0 > 0\) and the integrand is always positive, we take the positive root):