# Tag Info

Accepted

Accepted

### 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 ...
Accepted

### 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 ...

### 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, ...
Accepted

### 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 ...
Accepted

### 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 ...
Accepted

### 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 ...
Accepted

### 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\...

### 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 ...
Accepted

### 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 ...

### 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 ...
Accepted

### 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 - \...
Accepted

### 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, ...
Accepted

### 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 ...
Accepted

### 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,...

### 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 ...
### 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 $\mathbb Q^{M}$,...
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 ...