Stochastic Exponential and Girsanov - Exercises
Exercise 1
probability space and \(\mathcal{G}\) be a \(\sigma\)-algebra such that \(\mathcal{G}\subseteq \mathcal{F}\). Let also \(Z_{\infty} \in L^1\) be a random variable such that \(Z_{\infty}>0\) almost surely and \(E[Z_{\infty}]=1\). 1. Show that \(Q:\mathcal{F}\to [0,1]\), defined as \(Q[A]=E[1_A Z_\infty]\) defines a probability measure which is equivalent to \(P\) and for which holds \(E^{Q}[ X ]=E[ Z_\infty X ]\) for every positive random variable \(X\), 2. Show that for every \(\mathcal{G}\)-measurable positive random variable \(X\), it holds \(E^{Q}[ X ]=E[ E[ Z_\infty|\mathcal{G} ] X]\). 3. Let \(\mathcal{H}\) be a \(\sigma\)-algebra such that \(\mathcal{H}\subseteq \mathcal{G}\subseteq \mathcal{F}\) and \(X\) be a \(\mathcal{G}\)-measurable positive random variable. Show that it holds
$$E^Q\left[ X|\mathcal{H} \right]=\frac{1}{E\left[ Z_\infty|\mathcal{H} \right]}E\left[ Z_\infty X |\mathcal{H} \right]$$
- Let \(\mathbb{F}=(\mathcal{F}_t)\) be a filtration. Define the process \(Z\) by \(Z_t=E[Z_{\infty}|\mathcal{F}_t]\) for every \(t=1,\ldots\). Show that a process \(X\) is a martingale with respect to the measure \(Q\) if and only if the process \(ZX\) is a martingale with respect to the measure \(P\).
Exercise 2
and \(\mathbb{F}=(\mathcal{F}_t)_{0\leq t\leq T}\) a right-continuous filtration and \(B=(B_t)_{0\leq t\leq T}\) be a Brownian motion adapted to \(\mathbb{F}\). We define the process
for some progressive process \(Z=(Z_t)_{0\leq t\leq T}\) such that
We know that \(X\) is then a continuous square integrable martingale. We suppose that
for every \(0\leq t\leq T\) and some \(\varepsilon>0\). 1. Define the process \(q=(q_t)_{0\leq t\leq T}\) as
$$q_t=\frac{Z_t}{X_t},\quad 0\leq t\leq T$$
Show that
$$E\left[ \int_{0}^{T}q_s^2 ds \right]<\infty$$
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Applying Ito's formula to the process \(Y=(Y_t)_{0\leq t\leq T}\) given by
\[Y_t=\ln\left( X_t \right), \quad 0\leq t\leq T\]show that
\[X_t=\exp\left( \int_{0}^{t}q_s dB_s -\frac{1}{2}\int_{0}^{t}q_s^2 ds \right)\]
Exercise 3
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The financial markets admits an arbitrage if an only if there exists a strategy \(\alpha\) such that
\[P\left[ \sum_{t=1}^T \alpha_s\cdot (S_s - S_{s-1}) \geq 0 \right] = 1 \quad \text{and}\quad P\left[ \sum_{t=1}^T \alpha_s\cdot (S_s - S_{s-1}) > 0 \right] >0\] -
If there exists a probability measure \(Q\sim P\) such that each stock \(S^k\) is a martingale under \(Q\), then the financial market does not admit an arbitrage. (Hint: Assume that there exists such a martingale measure \(Q\) and an arbitrage strategy \(\alpha\) and show that it is a contradiction.) The reciprocal is also true and the proof relies on Hahn Banach separation theorem, and is called the fundamental theorem of asset pricing. Fundamental Theorem of Asset Pricing: The financial market does not admit an arbitrage strategy if and only if there exists a probability measure \(Q\sim P\) such that each stock \(S^k\) is a martingale under \(Q\).
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Assume the fundamental theorem of asset pricing and consider the following market with a single stock \(S=S^1\). We consider a binomial model with iid random variables \(Y_1, \ldots Y_T\) with \(P[Y_1 = 1] = p\) and \(P[Y_1 = -1] = 1-p\) for \(0<p<1\). We specify the stock price evolution as follows
\[ S_0>0 \quad \text{and} \quad S_t = \begin{cases} S_{t-1}(1+u) & \text{if }Y_t = 1\\ S_{t-1}(1+d) & \text{if }Y_t = -1 \end{cases} \]where \(-1<d <u\) so that the stock price is always positive. In other terms the stock increases by \(u/100\) or decreases by \(d/100\) percent each time according to \(Y\). - Show that the market is arbitrage free if and only if \(d<0<u\) (assume the fundamental theorem of asset pricing above). - Furthermore, show that if it is arbitrage free, there exists a unique measure \(Q \sim P\) for which \(S\) is a martingale measure, and under this measure the returns of the stock
$$R_{t+1} = \frac{S_{t+1} - S_t}{S_t}, \quad t=0, \ldots, T-1$$ are iid with $Q[R_t = u] = -\frac{d}{u-d}$
Exercise 4 (BSDE)
the auxiliary function \(g\) is affine, that is
where \(a, b\) and \(c\) are progressively measurable processes such that each of them are uniformly bounded. 1. Show that if \(\xi\) is in \(L^2\) and \(g =0\), then the BSDE [eq:01] has a solution. 2. Consider that \(c= a = 0\), proceeding to the variable change \(\tilde{Y} = e^{\int b ds}Y\), write the resulting BSDE for \(\tilde{Y}\), solve it and derive the explicit solution for \(Y\). 3. Consider that \(b = a = 0\), make the measure change with the stochastic exponential \(\tilde{Y} = \mathcal{E}(\int c dW)Y\), write the resulting BSDE for \(\tilde{Y}\), solve it and derive the explicit solution for \(Y\). 4. Consider that \(b= c = 0\), proceed to the variable change \(\tilde{Y} = \tilde{Y} + \int ads\), write the resulting BSDE for \(\tilde{Y}\), solve it and derive the explicit solution for \(Y\). 5. From these three steps derive what is the explicit solution of the BSDE when \(g(Y,Z) = bY + cZ - a\).