10

Shreve's theorem also called "Girsanov II" indeed represents a special case of the general "Girsanov I" from Wiki above, with $$Y_t:=W_t,$$$$X_t:=-\int_0^t\Theta_udW_u$$ We can show: $$[Y,X]=-\int_0^t\Theta_udu$$ by using general Stochastic Calculus rules (e.g. p.37, 6.6 here): $$[Y,X]=[W_t,-\int_0^t\Theta_udW_u]=-\int_0^t\Theta_ud[W_u,W_u]=-\int_0^t\...


10

It depends on the purpose of your simulation. If you want to model the asset price path for pricing some derivative then you need the risk-neutral measure (thus you take the risk-less rate as drift). Why? Because the risk-neutral measure makes your pricing compatible with the pricing of other contracts in the market. It makes the prices consistent. If ...


8

The result you're looking for is $$ \left. \frac{d\Bbb{P}}{d\Bbb{Q}}\right\vert_{\mathcal{F}_t} = \left( \left. \frac{d\Bbb{Q}}{d\Bbb{P}}\right\vert_{\mathcal{F}_t} \right)^{-1} $$ This is a result from measure theory but since you mention it, let's see how we can show it based on Girsanov theorem. Starting from the definitions you provide and introducing ...


6

IMHO the problem isn't stated correctly indeed, in the sense that the Radon-Nikodym derivative provided as the "solution" is not the unique way to define a measure $\mathbb{Q}$ equivalent to $\mathbb{P}$ and under which $X_t$ is a martingale. Just take $$\frac {d\mathbb{Q}}{d\mathbb{P}} =\mathcal{E}\left(-\int_0^t \cos(s) dW_s + a\right)$$ for any $a \in \...


6

Your notations are really hard to follow as you define $\mathbb{P}$ twice at the beginning. The notation $\mathbb{P} = \mathbb{\hat{P}}$ and $\mathbb{P} =\mathbb{\tilde{P}}$ is not meaningful as the probability measure $\mathbb{P}$ is already fixed and used for the real world probability measure. I think that this is the reason why you are getting confused. ...


6

Under the stock numeraire measure, $\frac{B_t}{S_t}$ is a Martingale. We can compute $$d\frac{B_t}{S_t}= \frac{1}{S_t}dB_t -\frac{1}{S_t^2}B_tdS_t+\frac{1}{S_t^3}B_t\sigma^2S_t^2dt\\=\frac{B_t}{S_t}\left(rdt -\mu dt -\sigma dW_t +\sigma^2dt\right)$$ so the growth rate $\mu$ that makes this a Martingale is $$ \mu = r+\sigma^2.$$ So the growth rate of the ...


6

Bond Price Dynamics I do not know the source of the bond dynamics you show above but seeing how we are dealing with an affine model there is a very elegant way to derive those. Due to the model being affine the bond price is given by $$P(t,T)=A(t,T)e^{-r(t)B(t,T)}$$ you can find the exact formulas for $A(t,T)$ and $B(t,T)$ in this document (or just read ...


6

Your mistake is actually made at the beginning: "Introducing a new process: $d\tilde{W}_t = dW_t +\frac{\mu-r}{\sigma} dt $" This is incorrect. Rather, $d\tilde{W}_t = dW_t -\frac{\mu-r}{\sigma} dt $ Otherwise, your derivation is correct. After correcting for the sign error, your final equation becomes $\Phi(x)=e^{-\lambda x-\frac{1}{2}\lambda^2 t}$. ...


5

In a practical manner, here is how you get to the PDE of your option: Use Girsanov theorem to go from the real-world measure to the risk-neutral measure (basically subtract the market price of risk $\mathrm dW^Q_t = \mathrm d W^P_t - \frac{\mu -r}{\sigma} \mathrm dt$). This will change your SDE. Discounted option price $e ^{-rt} v(t, S_t)$ has to be a ...


5

The answer is yes. Proof: Theorem (Radon-Nikodym) Let $(\Omega, \mathcal{F})$ be a measurable space. Let $\mathbb{P}$ and $\widetilde{\mathbb{P}}$ be two $\sigma$-finite measures. Let $\widetilde{\mathbb{P}}$ be absolutely continuous w.r.t. $\mathbb{P}$ (i.e. $\widetilde{\mathbb{P}}\ll\mathbb{P}$). THEN: $(\exists)$ measurable function $f:\Omega\to[0;+\...


4

It doesn't imply $$ \ln S_T=\ln S_0+rT+σW^Q_T,$$ it implies $$ \ln S_T=\ln S_0+(r-0.5\sigma^2)T+σW^Q_T.$$ Look up Ito's lemma. This is covered in just about any book on financial maths including my own Concepts etc.


4

Simplest explanation is Feynman-Kac theorem https://en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula Blackscholes is a parabolic PDE Solution can be written as a conditional expectation over an integration term. Conditional expectation means you need to simulate it using some distribution which leads to monte-carlo


4

A very interesting topic ! Black-Scholes originally did not make use of the Girsanov theorem and arrived at the equation the way you described. Later theoretical work on arbitrage pricing uncovered the concepts of the risk-neutral measure and derivatives pricing as an expectation under that measure. That work relies on stochastic calculus far more and one ...


3

Let $(V_t)_{t \geq 0}$ denote a self-financing wealth process in foreign currency units. In the absence of arbitrage, the former process should emerge as a martingale when expressed in the foreign money market numéraire i.e. $$ V_0 = \Bbb{E}^{\Bbb{Q}^f} \left[ \frac{B_0^f}{B_T^f} V_T \right] \tag{1} $$ Still by absence of arbitrage, the value of that same ...


3

Your (1) is incorrect because the annuity $A(T)$ is stochastic (it depends on discount rates on expiry) and therefore cannot be taken out of the expectation $E_Q[]$. This is why one resorts to pricing under the annuity measure.


3

According to CMG theorem, if $W_t$ is a Wiener process under the old measure, then under the new measure $\tilde{W_t}$ is a Wiener process, where: $$\tilde{W_t} = W_t + \langle cW, W \rangle_t = W_t - ct$$ Both statements express this same idea: Moving from the old measure to new one adds a drift $ct$, so if you take a Wiener process (under the old measure)...


3

Note that $$\frac{dQ_{T_p}}{dQ}|_{T_0} = \frac{P(T_0, T_p)}{P(0, T_p)}\frac{A(0, T_0, T_n)}{A(T_0, T_0, T_n)}$$. Then $$E^{Q_{T_p}}\big(S(T_0, T_n)\big) = E^Q\bigg(S(T_0, T_n) \frac{P(T_0, T_p)}{P(0, T_p)}\frac{A(0, T_0, T_n)}{A(T_0, T_0, T_n)}\bigg) \\ = \frac{A(0, T_0, T_n)}{P(0, T_p)} E^Q\bigg(S(T_0, T_n) \frac{P(T_0, T_p)}{A(T_0, T_0, T_n)}\bigg).$$ That ...


3

Risk-neutral pricing is to help with relative value type questions: If I know the value of this what should the value of that be if it depends in some way on this. It doesn't help with absolute value type questions: Should I buy this or that, is the implied volatility too low or high etc. Those are generally "real world measure" type questions.


3

(I might not be answering your question, but I feel this clarification is needed.) A random variable $X$ of $(\Omega, \mathcal{F})$ is a $\mathcal{F}$-measurable function $X : \Omega → \mathbf{R}$. So, $X$ depends on $\Omega$ and $\mathcal{F}$, but does not depend on the probability measure put on $(\Omega, \mathcal{F})$. It is the distribution of $X$ that ...


2

No, that's the point of Girsanov's theorem. If $Q$ is equal to $P_1$, then nothing has changed. In order to make $B_1(t)$ a standard BM we need to transition to a new Law. Namely, $Q$.


2

The process $L$ is strictly positive. This implies that $Q$ is equivalent to $P$ and not merely absolutely continuous.


2

Denote $B_t=e^{rt}$ the discount factor. Requiring $S_t/B_t$ to be a martingale it would mean the equation $S_0/B_0=E[S_t/B_t]$ hold. Therefore we can calculate the price of an option by discounting the expectation value at the maturity. If $S_t/B_t$ is not a Q-martingale, then we cannot discount the expectation value, which make calculation of $S_0$ ...


2

Suppose I give you objective probabilities $\mathbb{P}(S_T \geq K)$ of an equity finishing above a certain level $K$ at a future time $T$ (or in your case a survival probability in the form of default rates). Can you convert these to risk-neutral probabilities $\mathbb{Q}(S_T \geq K)$ ? Not immediately. First, I need to give you a model for the behaviour ...


2

It might be easier to go the other way: start with $$ d\mathcal{E}_t = \mathcal{E}_t dX_t $$ apply Ito to the $\log$ function $$ d\log(\mathcal{E})_t = \frac{1}{\mathcal{E}_t}d\mathcal{E}_t - \frac{1}{2} \frac{1}{\mathcal{E}_t^2}d\langle\mathcal{E},\mathcal{E}\rangle_t = \frac{1}{\mathcal{E}_t}\mathcal{E}_tdX_t - \frac{1}{2} \frac{1}{\mathcal{E}_t^2}\...


2

For question I, the identity \begin{align*} \rho_t = \exp\big(-\lambda_t W_t - \frac{1}{2} \lambda_t^2t\big) \end{align*} does not appear correct, unless $\lambda_t$ is a constant. For question II, yes. If $X_t = -\int_0^t \lambda_s dW_s$, then $\langle X \rangle_t = \int_0^t \lambda_s^2 ds$. For question III, you need to note that \begin{align*} \langle X ...


2

We assume that, under the $T_j$-forward probability measure $P_{T_j}$, \begin{align*} \frac{dP(t, T_j)}{P(t, T_j)} = \mu_P(t, T_j) dt + \sigma_P(t, T_j) dW_t^{T_j}, \end{align*} where $\mu_P(t, T_j)$ and $\sigma_P(t, T_j)$ are the respective drift and volatility functions. Let $Q$ be the risk-neutral probability measure. Then \begin{align*} \frac{dQ}{dP_{...


2

Notations $S_T$ and $K$ are expressed in EUR; $D^{CCY}(t,T) = \frac{\beta^{CCY}_t}{\beta^{CCY}_T}$ where $\beta^{CCY}$ is the money market account in currency $CCY$). In other words, it is the (stochastic) discount factor from $t$ to $T$ in the currency $CCY$; $X_t$ is the value of 1 EUR in COP. Answer The expression of the Radon-Nikodym derivative ...


1

Basically the argument is that we have arrow-debreu securities (instrument that pays 1 if you arrive in a certain state). In the absence of arbitrage the price of this arrow-debreu security should be the same under both measures. But the price of an arrow-debreu security is simply the probability of that event happening. Hence both measures must be the same. ...


1

The first result you are alluding to is known as the martingale representation theorem. More specifically, what you say holds for continuous paths processes. For jump processes, there can and will a $dt$ term in their martingale representation (compensator). Girsanov theorem is about change of probability measures as you correctly mention too. To me, the ...


1

If two assets have the same source of risk $W^P$ then the no arbitrage opportunity condition implies that their Sharpe ratios are the same, i.e. $\frac{\mu_1-r_f}{\sigma_1} = \frac{\mu_2-r_f}{\sigma_2}$. This is a textbook exercise and is easily proven by building a risk free portfolio long one share of the first asset and short $\frac{S_1 \sigma_1}{S_2 \...


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