# Tag Info

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): $$[Y,X]=[W_t,-\int_0^t\Theta_udW_u]=-\int_0^t\Theta_ud[W_u,W_u]=-\int_0^t\... 10 It depends on the purpose of your simulation. If you want to model the asset price path for pricing some derivative then you need the risk-neutral measure (thus you take the risk-less rate as drift). Why? Because the risk-neutral measure makes your pricing compatible with the pricing of other contracts in the market. It makes the prices consistent. If ... 10 First, let's check if these models are abritrage free. The first fundamental theorem of asset pricing says that if there exists an equivalent probability measure under which \frac{S_t}{\beta_t} = e^{-t}S_t is a martingale, then the market is arbitrage free, so we will check whether such an equivalent martingale measure exists. This is where we will use ... 9 The result you're looking for is$$ \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 ... 9 Consider the Heston (1993) model under the real world measure (\mathbb{P}) \begin{align*} \mathrm{d}S_t&=\mu^\mathbb{P} S_t\mathrm{d}t+\sqrt{v_t}S_t\mathrm{d}B_{S,t}^\mathbb{P}, \\ \mathrm{d}v_t&=\kappa^\mathbb{P} (\bar{v}^\mathbb{P}-v_t)\mathrm{d}t+\sigma^\mathbb{P}\sqrt{v_t}\mathrm{d}B_{v,t}^\mathbb{P}, \end{align*} where \mathrm{d}B_{S,t}^\... 7 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)$$for any a \in \... 6 Under the stock numeraire measure, \frac{B_t}{S_t} is a Martingale. We can compute$$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.$$So the growth rate of the ... 6 Your notations are really hard to follow as you define \mathbb{P} twice at the beginning. The notation \mathbb{P} = \mathbb{\hat{P}} and \mathbb{P} =\mathbb{\tilde{P}} is not meaningful as the probability measure \mathbb{P} is already fixed and used for the real world probability measure. I think that this is the reason why you are getting confused. ... 6 Bond Price Dynamics I do not know the source of the bond dynamics you show above but seeing how we are dealing with an affine model there is a very elegant way to derive those. Due to the model being affine the bond price is given by$$P(t,T)=A(t,T)e^{-r(t)B(t,T)}$$you can find the exact formulas for A(t,T) and B(t,T) in this document (or just read ... 6 Your mistake is actually made at the beginning: "Introducing a new process: d\tilde{W}_t = dW_t +\frac{\mu-r}{\sigma} dt " This is incorrect. Rather, d\tilde{W}_t = dW_t -\frac{\mu-r}{\sigma} dt  Otherwise, your derivation is correct. After correcting for the sign error, your final equation becomes \Phi(x)=e^{-\lambda x-\frac{1}{2}\lambda^2 t}. ... 5 In a practical manner, here is how you get to the PDE of your option: Use Girsanov theorem to go from the real-world measure to the risk-neutral measure (basically subtract the market price of risk \mathrm dW^Q_t = \mathrm d W^P_t - \frac{\mu -r}{\sigma} \mathrm dt). This will change your SDE. Discounted option price e ^{-rt} v(t, S_t) has to be a ... 5 The answer is yes. Proof: Theorem (Radon-Nikodym) Let (\Omega, \mathcal{F}) be a measurable space. Let \mathbb{P} and \widetilde{\mathbb{P}} be two \sigma-finite measures. Let \widetilde{\mathbb{P}} be absolutely continuous w.r.t. \mathbb{P} (i.e. \widetilde{\mathbb{P}}\ll\mathbb{P}). THEN: (\exists) measurable function f:\Omega\to[0;+\... 5 A very interesting topic ! Black-Scholes originally did not make use of the Girsanov theorem and arrived at the equation the way you described. Later theoretical work on arbitrage pricing uncovered the concepts of the risk-neutral measure and derivatives pricing as an expectation under that measure. That work relies on stochastic calculus far more and one ... 4 It doesn't imply$$ \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 (Bloomberg and Reuters News are fond is reporting that some name is trading at some such CDS spread, "which implies N% probability of default". They neglect to mention what recovery assumption they used, and that this is risk-neutral probability, not physical.) For corporate names, Ed Altman published the well-known paper on Z-score. His basic idea ... 4 I don't think that you can apply Girsanov's theorem in this way, noting that I don't understand your (short) argument in the comments. This is how I would proceed, which makes sense mathematically, but the economic interpretation is then a bit strange. Let us write the SDE's a bit different, noting that it still preservers the same correlation structure \... 4 Let Y_t= e^{B_t} and Z_t = B_{t}-t / 2. Then, \begin{align*} dX_t &= Z_t dY_t + Y_t dZ_t + d\langle Y, Z\rangle_t\\ &=(B_{t}-t / 2)e^{B_t}\big( dB_t + 1/2\,dt \big) + e^{B_t}\big(dB_t -1/2\, dt\big) + e^{B_t} dt\\ &=e^{B_t}(B_t-t / 2+1)dB_t + e^{B_t}(B_t/2-t / 4 -1/2+1)dt\\ &=e^{B_t}(B_t-t / 2+1)d\big(B_t+1/2t\big). \end{align*} We ... 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$$Both statements express this same idea: Moving from the old measure to new one adds a drift ct, so if you take a Wiener process (under the old measure)... 3 Let (V_t)_{t \geq 0} denote a self-financing wealth process in foreign currency units. In the absence of arbitrage, the former process should emerge as a martingale when expressed in the foreign money market numéraire i.e.$$ V_0 = \Bbb{E}^{\Bbb{Q}^f} \left[ \frac{B_0^f}{B_T^f} V_T \right] \tag{1} $$Still by absence of arbitrage, the value of that same ... 3 It might be easier to go the other way: start with$$ 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}\...

3

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 ...

3

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 $S_0$ ...

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

Nothing new in this answer - I have just consolidated what others have said in the answer and the comments, and put the explanation next to each step. I have to move the original equations so as to have one equation on each line:

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 ...

3

I assume all three models are stated under the money-market measure: then there is no arbitrage if the discounted pay-off is a martingale under the money-market Numeraire. Therefore to show no arbitrage for all three models, we would want to show that: $$\mathbb{E}\left[\frac{S_t}{\beta_t}|\mathcal{F_0}\right]=\frac{S_0}{\beta_0}$$ Model a: \frac{S_0}{\...

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, $Q$.

Only top voted, non community-wiki answers of a minimum length are eligible