10 votes
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Girsanov Theorem and Quadratic Variation

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^...
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10 votes

Heston stochastic volatility, Girsanov theorem

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&...
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10 votes
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Change of measure and Girsanov's Theorem: Do the following models admit arbitrage and are they complete?

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^{...
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9 votes
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Girsanov Theorem, Radon-Nikodym Derivative backward

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 ...
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7 votes
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Given $\mathbb Q$ and $X_t$ is $\mathbb Q$-Brownian, find $\frac{d\mathbb Q}{d\mathbb P}$ / Uniqueness of Brownian or Radon-Nikodym derivative

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{...
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6 votes

Use of Girsanov's theorem in bond pricing

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 ...
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6 votes
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Radon-Nikodym derivative and risk natural measure

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 ...
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  • 1,319
6 votes
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Change-of-measure: Dynamics of $\log(S_t)$ with $S_t$ as numeraire

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}\...
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6 votes
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How to use the Girsanov theorem to prove $\hat{W_t}$ is a $\hat{\mathbb P}$-Brownian motion?

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 ...
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  • 110
6 votes
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Are all change of measure operations between equivalent probability measures Doléans-Dade exponentials?

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{\...
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  • 510
5 votes

What is the easiest way to learn Option pricing with PDE?

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 $\...
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5 votes

Black Scholes model without using Girsanov's theorem? It might happen?

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 ...
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  • 1,346
5 votes

Heston stochastic volatility, Girsanov theorem

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, ...
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  • 131
4 votes
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Why do we need $dS_t=r S_tdt+\sigma S_tdW_t^Q$?

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 ...
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4 votes
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theoretical reason for which we can use monte carlo simulation for option pricing

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 ...
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4 votes
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Objective probability of default from CDS spread

(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 ...
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4 votes
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Change of measure for a stochastic process to be a martingale

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\...
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3 votes

Bayes Theorem with change of measure

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 ...
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3 votes
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Change of Numeraire to price European swaptions

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 ...
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3 votes

Girsanov Transform and Likelihood Process Domestic to Foreign

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 ...
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3 votes
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Question about the Cameron-Martin-Girsanov (CMG) theorem

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 - ...
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3 votes

Girsanov Theorem application to Geometric Brownian Motion

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 ...
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3 votes

Girsanov theorem in CMS convexity derivation

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, ...
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3 votes

Ito, Stochastic Exponential and Girsanov

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 - \...
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3 votes
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Ito, Stochastic Exponential and Girsanov

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, ...
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3 votes
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true or false: the risk-neutral measure is useless in this situation

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 ...
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3 votes
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On Girsanov Theorem to switch from Risk-Neutral to Stock Numeraire

(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,...
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  • 4,963
3 votes

Change of measure and Girsanov's Theorem: Do the following models admit arbitrage and are they complete?

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 ...
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  • 4,961
3 votes
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Where does the term $\gamma$ come from when moving from measure $\mathbb Q^{N}$ to $\mathbb Q^{M}$?

From your assumption that \begin{align*} dX(t)\cdot d\frac{M(t)}{N(t)} &= \frac{M(t)}{N(t)}X(t)\gamma(t)dt,\\ dX(t) &= X(t)\big(\mu(t)dt+\sigma(t)dW(t)\big), \end{align*} under $\mathbb Q^{M}$,...
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2 votes

Radon-Nikodym derivative and risk natural measure

Five years late to the party, but let me put my two cents in for intuition. Let $W^P_t \equiv W_t$, and $W^Q_t \equiv \tilde{W_t}$ to easier remember under what measure either one is a standard ...
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