π Detailed Solutions to Homework Exercises: Conditional Expectation and Properties of Brownian Motion
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This page contains detailed solutions to key homework exercises from Chapter 1 (Independence, Conditional Expectation) and Chapter 2 (Brownian Motion and Its Properties) of the Stochastic Differential Equations course. All solutions are presented in collapsible boxes; click to view the detailed derivation process.
Part I: Independence and Conditional Expectation
Exercise 1
Problem: Let the probability density function of the random variable \(X\) be \(f(x) = ax(1-x), x \in (0,1)\). For all other \(x\), \(f\) is zero. (1) Find the constant \(a\); (2) Let \(Y = X^3\), find the probability density function of \(Y\).
Solution (Click to expand)
(1) Finding the constant \(a\)
By the normalization property of the probability density function, its integral over the entire space equals 1:
Solving this integral yields:
Solving gives: \(a = 6\) [cite: 1519].
(2) Finding the probability density function of \(Y=X^3\)
First, find the cumulative distribution function \(F_Y(y)\) of \(Y\). Since \(X \in (0,1)\), we have \(Y = X^3 \in (0,1)\). For \(y \in (0,1)\):
Substituting the probability density function of \(X\) and computing the integral:
Differentiating \(F_Y(y)\) yields the probability density function \(f_Y(y)\) of \(Y\) [cite: 1519]:
In summary, the probability density function of \(Y\) is:
Exercise 2
Problem: Let \(X, Y\) be independent random variables, \(f(x,y)\) be a bounded continuous function, and \(F_X\) be the probability distribution function of \(X\). Prove: (1) \(\mathbb{E}[f(X,Y)|Y] = \int f(x,Y)dF_X(x)\); (2) \(\mathbb{P}(X+Y \leqslant x | Y) = F_X(x-Y)\).
Solution (click to expand)
(1) Proof:
[cite_start]We adopt the rigorous proof method of the Standard Machine from measure theory [cite: 1520]:
Step 1 (Indicator functions): Let \(f(x,y) = I_A(x)I_B(y)\), where \(A, B\) are Borel sets. Since \(X, Y\) are independent, using properties of conditional expectation (given \(Y\), \(I_B(Y)\) can be factored out, and the conditional expectation of independent variables equals the unconditional expectation):
On the other hand, for the integral form on the right-hand side:
The two are equal, so the statement holds for indicator functions.
Step 2 (Simple functions): By the linearity of expectation, the conclusion holds for simple functions, which are finite linear combinations of indicator functions [cite: 1520].
Step 3 (Non-negative measurable functions): For any non-negative measurable function, there exists a monotone increasing sequence of simple functions approximating it. By the Monotone Convergence Theorem, the conclusion holds for non-negative measurable functions [cite: 1520].
Step 4 (Bounded continuous functions): Any bounded continuous function can be decomposed into its positive and negative parts \(f = f^+ - f^-\), and by definition it is Lebesgue integrable. Therefore, the original formula holds for all bounded continuous functions. QED [cite: 1520].
(2) Proof:
The conditional probability can be directly written in the form of conditional expectation:
Let \(g(X, Y) = I_{\{X+Y \leqslant x\}}\). Using the result proven in part (1):
Transform the integration domain equivalently to \(u \leqslant x - Y\):
QED [cite: 1521].
Exercise 3
Problem: Let \(A, B\) be two events in the measure space \((\Omega, \mathcal{F}, \mathbb{P})\). Compute the conditional expectation \(\mathbb{E}[\chi_A|\chi_B]\). (Note: \(\chi\) denotes the indicator function.)
Solution (click to expand)
Since \(\chi_B\) can only take the values 0 or 1, the \(\sigma\)-algebra it generates is very simple: \(\sigma(\chi_B) = \{\emptyset, \Omega, B, B^c\}\). Therefore, the conditional expectation \(\mathbb{E}[\chi_A|\chi_B]\) must be \(\sigma(\chi_B)\)-measurable, which means it must be constant on \(B\) and \(B^c\). Let us set:
According to the Radon-Nikodym derivative definition of conditional expectation, for any \(\Lambda \in \sigma(\chi_B)\), the integral equality must hold:
Case 1: Take \(\Lambda = B\)
If \(\mathbb{P}(B) > 0\), then \(c_1 = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = \mathbb{P}(A|B)\) [cite: 1523].
Case 2: Take \(\Lambda = B^c\)
If \(\mathbb{P}(B^c) > 0\), then \(c_2 = \frac{\mathbb{P}(A \cap B^c)}{\mathbb{P}(B^c)} = \mathbb{P}(A|B^c)\) [cite: 1524].
In summary, the explicit expression for the conditional expectation is:
Exercise 4
Problem: Let \(\mathcal{V}_1\) and \(\mathcal{V}_2\) be two independent \(\sigma\)-fields, and let \(X\) be an integrable random variable. Prove: \(\mathbb{E}[\mathbb{E}(X|\mathcal{V}_1)|\mathcal{V}_2] = \mathbb{E}[X]\).
Solution (click to expand)
Proof:
First, by the definition of conditional expectation, the inner expectation \(\mathbb{E}(X|\mathcal{V}_1)\) must be a \(\mathcal{V}_1\)-measurable random variable [cite: 1529].
Since the problem states that \(\mathcal{V}_1\) and \(\mathcal{V}_2\) are independent \(\sigma\)-fields, the \(\mathcal{V}_1\)-measurable random variable \(\mathbb{E}(X|\mathcal{V}_1)\) is naturally independent of the \(\sigma\)-field \(\mathcal{V}_2\).
According to the property of conditional expectation with respect to an independent \(\sigma\)-field (if a random variable \(Z\) is independent of a \(\sigma\)-field \(\mathcal{G}\), then \(\mathbb{E}[Z|\mathcal{G}] = \mathbb{E}[Z]\)), we can remove the conditioning on \(\mathcal{V}_2\) [cite: 1529]:
Finally, by the law of total expectation (i.e., the degenerate form of the Tower Property):
Combining the two equations, the original proposition is proved [cite: 1529]. \(\square\)
Exercise 5
Problem: Let \(X\) and \(\{X_n\}\) be a sequence of random variables on \((\Omega, \mathcal{F}, \mathbb{P})\), and for a fixed \(1 \leqslant p < \infty\), \(\mathbb{E}[|X_n|^p] < +\infty\). Assume \(\lim_{n\to\infty}\mathbb{E}[|X_n - X|^p] = 0\). Prove that for any sub-\(\sigma\)-field \(\mathcal{V} \subset \mathcal{F}\) of \(\mathcal{F}\), the following holds:
Solution (click to expand)
Proof:
Consider the function \(\Phi(x) = |x|^p\). Since \(p \geqslant 1\), this is a convex function. Using the linearity of conditional expectation and the conditional Jensen's inequality, we combine the conditional expectations inside the absolute value and move the absolute value inside the expectation [cite: 1530]:
Taking the unconditional expectation on both sides of the above inequality and applying the law of total expectation [cite: 1530, 1531]:
Taking the limit as \(n \to \infty\) on both sides of the inequality. Since it is given that \(\lim_{n\to\infty}\mathbb{E}[|X_n - X|^p] = 0\), and the left-hand side expectation is always non-negative, by the squeeze theorem [cite: 1532]:
The original proposition is proven. \(\square\)
Exercise 6
Problem: Let \(\Omega = \{1, 2, \cdots, 7, 8\}\), \(\mathcal{F} = 2^\Omega\) (i.e., the \(\sigma\)-algebra consisting of all subsets of \(\Omega\)). For \(i \leqslant 4\), \(\mathbb{P}(\{i\}) = 1/10\), and for \(i > 4\), \(\mathbb{P}(\{i\}) = 3/20\). Define \(X = \chi_{\{1,2,3,4\}} + 2\chi_{\{5,6,7,8\}}\), \(Y = \chi_{\{1,5\}} + 2\chi_{\{2,3,4,6,7,8\}}\). \(\mathcal{V}\) is the \(\sigma\)-algebra generated by \(\{1,2\}\) and \(\{3,4\}\), \(\mathcal{H}\) is the \(\sigma\)-algebra generated by \(\{1,2,3,4\}\). Compute: (1) \(X\mathbb{E}[Y]\); (2) \(\mathbb{E}[\mathbb{E}[XY|\mathcal{V}]|\mathcal{H}]\).
Solution (click to expand)
Computational Preparation: Clarifying the Probability Structure
- \(\mathbb{P}(\{1,2,3,4\}) = 4 \times \frac{1}{10} = \frac{2}{5}\)
- \(\mathbb{P}(\{5,6,7,8\}) = 4 \times \frac{3}{20} = \frac{3}{5}\)
- It is easy to verify that the two core atoms of \(\mathcal{H} = \sigma(\{1,2,3,4\})\) are precisely \(\{1,2,3,4\}\) and \(\{5,6,7,8\}\). The partition of \(\mathcal{V} = \sigma(\{1,2\}, \{3,4\})\) is finer, and it is clear that \(\mathcal{H} \subset \mathcal{V}\).
(1) Computing \(X\mathbb{E}[Y]\)
First, compute the unconditional expectation of \(Y\):
Multiply this directly by \(X\):
(2) Computing \(\mathbb{E}[\mathbb{E}[XY|\mathcal{V}]|\mathcal{H}]\)
Here we utilize the extremely elegant smoothing property (Tower Property) to directly reduce dimensions, avoiding the cumbersome calculation of double conditional expectations [cite: 1533]. Since \(\mathcal{H} \subset \mathcal{V}\), the smaller \(\sigma\)-algebra plays the decisive role on the outside:
Observing the random variable \(X\), it is constantly 1 on \(\{1,2,3,4\}\) and constantly 2 on \(\{5,6,7,8\}\), which exactly corresponds to the atomic partition of \(\mathcal{H}\). Therefore, \(X\) is \(\mathcal{H}\)-measurable! As known information, it can be factored out of the conditional expectation [cite: 1535]:
Next, we compute the conditional expectation of \(Y\) on the two atoms of \(\mathcal{H}\): * On \(\{1,2,3,4\}\): \(\mathbb{E}[Y|\{1,2,3,4\}] = \frac{1\cdot P(\{1\}) + 2\cdot P(\{2,3,4\})}{P(\{1,2,3,4\})} = \frac{1/10 + 6/10}{4/10} = \frac{7}{4}\)
- On \(\{5,6,7,8\}\): \(\mathbb{E}[Y|\{5,6,7,8\}] = \frac{1\cdot P(\{5\}) + 2\cdot P(\{6,7,8\})}{P(\{5,6,7,8\})} = \frac{3/20 + 18/20}{12/20} = \frac{7}{4}\)
It is surprisingly found that \(\mathbb{E}[Y|\mathcal{H}] = 7/4\) holds constant on both partitions. Hence:
Exercise 7
Problem: Given a probability space \((\Omega, \mathcal{F}, \mathbb{P})\) and an integrable random variable \(X\), let \(\{\mathcal{F}(t)\}_{t \geqslant 0}\) be a filtration. For \(t \geqslant 0\), define \(X(t) \doteq \mathbb{E}[X|\mathcal{F}(t)]\). Prove that \(X(t)\) is a martingale with respect to \(\mathcal{F}(t)\).
Solution (click to expand)
Proof: [cite_start]We need to strictly verify the three core conditions of a martingale [cite: 1546, 1547]:
1. Integrability: Using the contraction property of conditional expectation (the absolute value special case of Jensen's inequality):
Since the problem states that \(X\) is integrable (\(\mathbb{E}|X| < \infty\)), we have \(\mathbb{E}|X(t)| < \infty\).
2. Adaptability: According to the measure-theoretic definition of conditional expectation \(\mathbb{E}[X|\mathcal{F}(t)]\), it is naturally \(\mathcal{F}(t)\)-measurable. Therefore, the process \(\{X(t)\}\) is adapted to the filtration \(\{\mathcal{F}(t)\}_{t \ge 0}\).
3. Martingale Property: For any \(0 \leqslant s \leqslant t\), since \(\{\mathcal{F}(t)\}\) is a filtration, meaning information accumulates, we must have \(\mathcal{F}(s) \subset \mathcal{F}(t)\). Using the smoothing property (Tower Property) of conditional expectation:
In summary, based on these three points, the process \(X(t)\) constructed by projecting a single integrable variable onto different information filtrations must be a martingale (this is known as a Doob martingale in stochastic analysis). \(\square\)
Part II: Brownian Motion and Its Properties
Exercise 1
Problem: Let \(W(t)\) be a one-dimensional Brownian motion. Prove that for any fixed \(s>0\), \(W(t+s)-W(s)\) is a Brownian motion; and for any positive constant \(c\), \(cW(t/c^2)\) is also a Brownian motion.
Solution (click to expand)
(1) Prove that \(B(t) = W(t+s) - W(s)\) is a Brownian motion
We verify the four defining properties of Brownian motion one by one [cite: 1551]:
-
Zero initial value: \(B(0) = W(0+s) - W(s) = 0\).
-
Independent increments: For any \(0 \le t_1 < t_2 < \cdots < t_k\), the sequence of increments is:
Since the original process \(W\) has the independent increments property, and the time intervals \([t_{j-1}+s, t_j+s]\) are non-overlapping on the time axis, these increments are mutually independent.
- Stationary normal increments: For \(t > u\), the increment distribution is:
- Path continuity: Because the sample paths of \(W(\cdot)\) are almost surely continuous, the shifted paths \(B(\cdot)\) are also almost surely continuous.
In conclusion, \(B(t)\) is a one-dimensional Brownian motion.
(2) Prove that \(U(t) = cW(t/c^2)\) is a Brownian motion
Similarly, we verify the properties [cite: 1552, 1553]:
-
Zero initial value: \(U(0) = cW(0/c^2) = cW(0) = 0\).
-
Independent increments: For any \(0 \le t_1 < t_2 < \cdots < t_k\), since \(t_1/c^2 < t_2/c^2 < \cdots < t_k/c^2\) are non-overlapping, the increments of the original Brownian motion are independent. Multiplying by the constant \(c\) preserves independence.
-
Stationary normal increments: Due to the properties of linear transformation, the mean remains \(0\), and the variance is:
Hence, \(U(t) - U(s) \sim N(0, t-s)\).
- Path continuity: Scaling does not affect the continuity of sample paths.
In conclusion, \(U(t) = cW(t/c^2)\) is also a Brownian motion (this is known as the scaling invariance/fractal property of Brownian motion).
Exercise 2
Problem: Let \(W(t)\) be a one-dimensional Brownian motion, and define \(\tilde{W}(t) = \begin{cases} tW(1/t), & t > 0, \\ 0, & t = 0. \end{cases}\) Prove: \(\tilde{W}(t) - \tilde{W}(s) \sim N(0, t-s), \forall 0 < s < t\).
Solution (click to expand)
For \(0 < s < t\), we perform an identity transformation on the increment to arrange mutually independent terms [cite: 1557, 1558]:
We separate by adding and subtracting \(sW(1/t)\):
Note that here \(1/t < 1/s\). Using the independent increments property of Brownian motion, the random variable \(W(1/t) = W(1/t) - W(0)\) is independent of the increment \(W(1/s) - W(1/t)\).
Due to independence, the variances can be directly added [cite: 1561]:
Substituting the variance formula \(Var(W(u)) = u\):
Simplifying by common denominator:
Furthermore, by the linearity of expectation, \(E[\tilde{W}(t) - \tilde{W}(s)] = 0 - 0 = 0\) [cite: 1562]. Since it is a linear combination of jointly normal random variables, it must follow a normal distribution.
The conclusion is proven: \(\tilde{W}(t) - \tilde{W}(s) \sim N(0, t-s)\).
Exercise 3
Problem: Let \(W(t)\) be a Brownian motion. Prove: \(\mathbb{E}[W^{2k}(t)] = \frac{(2k)!t^k}{2^k k!}, \forall t > 0\).
Solution (click to expand)
Since \(W(t) \sim N(0, t)\), using the moment generating function (MGF) of the normal distribution \(M_W(\lambda) = \mathbb{E}[e^{\lambda W(t)}]\) [cite: 1564]:
We expand both sides of the equation as Taylor series at \(\lambda = 0\):
The left-hand side is expanded using the linearity of expectation:
The right-hand side expands the exponential function:
Comparing the coefficients of \(\lambda^{2k}\) on both sides:
Rearranging yields the proof [cite: 1571]:
(Note: For odd-order moments, since the right-hand side has no odd-power terms, comparing coefficients gives \(\mathbb{E}[W^{2k+1}(t)] = 0\).)
Exercise 4
Problem: Prove: Let \(c\) be a constant, \(0 < s < t\), then \(\mathbb{E}[\exp(c(W(s) - W(t)))] = \exp\left(\frac{1}{2}c^2(t-s)\right)\).
Solution (click to expand)
Due to the stationarity and symmetry of Brownian motion increments, \(W(s) - W(t)\) and \(W(t) - W(s)\) have the same distribution, both following \(N(0, t-s)\).
This is equivalent to finding the moment generating function of a normal random variable [cite: 1573]. Let \(Z = c(W(s) - W(t))\), then \(Z \sim N(0, c^2(t-s))\).
According to the moment generating function formula for a normal distribution \(N(\mu, \sigma^2)\), \(\mathbb{E}[e^Z] = \exp(\mu + \frac{1}{2}\sigma^2)\) [cite: 1573, 1574]:
Q.E.D.
Exercise 5
Problem: Let \(U(t) = e^{-t}W(e^{2t})\), then \(\mathbb{E}[U(t)U(s)] = e^{-|t-s|}, \forall t, s \in \mathbb{R}\).
Solution (click to expand)
Without loss of generality, we assume \(t \geqslant s\).
Substituting the definition of \(U(t)\) to compute the expectation of the cross term:
For Brownian motion, we know its covariance function is \(\mathbb{E}[W(u)W(v)] = \min(u, v)\). Since we assumed \(t \geqslant s\), naturally \(e^{2t} \geqslant e^{2s}\), therefore:
Substituting this back into the original expression [cite: 1576, 1577]:
Since \(t \geqslant s\), \(s-t = -|t-s|\). If \(s > t\), the conclusion holds symmetrically. Thus \(\forall t, s \in \mathbb{R}\), \(\mathbb{E}[U(t)U(s)] = e^{-|t-s|}\), which completes the proof.
Exercise 6
Problem: Prove that almost surely \(\lim_{m \to \infty} \frac{W(m)}{m} = 0\).
Solution (click to expand)
We can use Chebyshev's Inequality combined with the Borel-Cantelli lemma to give a rigorous measure-theoretic proof.
Step 1: Bounding the variance Consider the random sequence \(Y_m = \frac{W(m)}{m}\). Since \(W(m) \sim N(0, m)\), we have:
Step 2: Strengthening the Moment Bound (Paving the Way for Borel-Cantelli)
If we only use Chebyshev's inequality with the second moment, we get \(P(|Y_m| \ge \epsilon) \le \frac{1}{m\epsilon^2}\) [cite: 1583]. However, \(\sum_{m=1}^{\infty} \frac{1}{m}\) is the harmonic series, which does not converge, preventing the direct use of the B-C lemma. Therefore, we use the fourth moment (or exploit the exponential decay property of the normal distribution). For a standard normal variable \(Z \sim N(0,1)\), \(\mathbb{E}[Z^4] = 3\).
Applying Markov's inequality based on the fourth moment:
Step 3: Borel-Cantelli Lemma For any fixed \(\epsilon > 0\), since \(\sum_{m=1}^{\infty} \frac{1}{m^2} < \infty\), the series converges absolutely:
By the first part of the Borel-Cantelli lemma, the probability that the event \(\left\{ \left|\frac{W(m)}{m}\right| \geqslant \epsilon \right\}\) occurs infinitely often is 0. This is equivalent to: almost every sample path satisfies \(\lim_{m \to \infty} \frac{W(m)}{m} = 0\), completing the proof.
Exercise 7
Problem: Prove that \(W(t)^2 - t\) and \(\exp\left(\lambda W_t - \frac{1}{2}\lambda^2 t\right)\) \((\lambda \in \mathbb{R})\) are both martingales with respect to the history \(\mathcal{F}(t)\) of \(W(t)\).
Solution (click to expand)
Proof (1): \(W(t)^2 - t\) is a Martingale
Let \(0 \leqslant s < t\). We need to prove \(\mathbb{E}[W(t)^2 - t | \mathcal{F}(s)] = W(s)^2 - s\). Using the identity transformation \(W(t) = (W(t) - W(s)) + W(s)\) to expand the square term:
- [cite_start]First term: Due to the independent increments of Brownian motion, \(W(t)-W(s)\) is independent of \(\mathcal{F}(s)\), so the conditional expectation equals the unconditional expectation \(\mathbb{E}[(W(t)-W(s))^2] = t-s\) [cite: 1585, 1586].
- Second term: \(W(s)\) is \(\mathcal{F}(s)\)-measurable and can be factored out. The remaining increment expectation is 0.
- Third term: \(W(s)^2\) is \(\mathcal{F}(s)\)-measurable, so it equals itself.
Therefore:
Subtract \(t\):
The martingale property is proved.
Proof (2): \(\exp\left(\lambda W(t) - \frac{1}{2}\lambda^2 t\right)\) is a martingale (geometric Brownian martingale)
Again, let \(0 \leqslant s < t\). Using the exponential split:
Since \(\exp(\lambda W(s))\) is \(\mathcal{F}(s)\)-measurable, factor it out as a constant:
Because the increment \(W(t)-W(s)\) is independent of \(\mathcal{F}(s)\) and follows a \(N(0, t-s)\) distribution, using the moment generating function of the normal distribution \(\mathbb{E}[e^{\lambda Z}] = e^{\lambda^2 \sigma^2 / 2}\):
Substituting the above expectation into the expression to be proved [cite: 1587]:
Combine the exponents:
Finally, we obtain:
The martingale property is proven.
Exercise 8
Problem: Set \(X(t) = \int_0^t W(s)ds\). Prove: \(\mathbb{E}[X^2(t)] = \frac{t^3}{3}, \forall t > 0\).
Solution (click to expand)
[cite_start]This is a highly representative stochastic process integral calculation, with the core being the use of Fubini's theorem to interchange the order of expectation and integration [cite: 1588].
First, write the squared term as a double Riemann integral:
Take the expectation on both sides, and use Fubini's theorem (since the integration domain is bounded and the function is absolutely integrable) to move the expectation inside the integrals:
Since the covariance function of Brownian motion is \(\mathbb{E}[W(u)W(v)] = \min(u, v)\), substitute it into the above equation:
Because \(\min(u,v)\) is not differentiable on the diagonal \(u=v\), we partition the integration region \([0,t] \times [0,t]\) along the diagonal into two parts: the region \(u \leqslant v\) and the region \(u > v\) [cite: 1589, 1590]:
Since these two regions are completely symmetric, we can compute one and multiply by 2:
The conclusion is proven: \(\mathbb{E}[X^2(t)] = \frac{t^3}{3}\).