Tag Info

New answers tagged

0

to answer my own question, there's no popular model for the question, that $dS/S=\mu(t)dt+\sigma(t)dW$, and $\sigma(t)$ is correlated with $\mu(t)$. the general framework should be stochastic volatility model, but need do the extension on my own.


3

A key property of Brownian motion is independent increments. So if $x-1 > y$, then $$ \mathbb{E}[\Delta W_x \Delta W_y] = 0 $$ because the time intervals [x-1,x] and [y-1,y] do not overlap. If they do overlap, i.e. $x-1 \leq y < x$, then \begin{align} \mathbb{E}[\Delta W_x \Delta W_y] =&\ \mathbb{E}[(W_x - W_{x-1}) (W_y-W_{y-1})] \\ =&\ ...


0

In general, there cannot be a closed-form solution of a random coefficients VG model. The reason is the drift-restriction that needs to be imposed to ensure that the discounted price process is a martingale under the risk-neutral measure. Using the bank account as numeraire, the restriction is $$ \frac{1}{\beta} > \theta + \frac{\sigma^2}{2} $$ where ...


1

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 ...


0

adam I still think that your question is a bit vague but perhaps the following will be of some help to you. First of all Itô's theorem is a tool. It will never give you the price by itself. While working out the concrete formula one might end up using it in one context or another. In case of a european option, a borel measurable function $h$ and $X_t$ ...


3

I will try to answer this a bit differently. The rigorous answer: because Ito calculus tells us that we need the second order term. Look at $$ S_t = S_0\exp(\mu t + \sigma B_t). $$ Assume that $S_0$ is known and fixed and look at by Ito's formula $$ d(S_t/S_0) = \mu dt + \sigma B_t + \frac{\sigma^2}{2} dt. $$ Then with some abuse of notation: $$ ...


4

The convexity of the exponential function of the stochastic variable $W$ makes its expectation greater than the exponentiation of the expectation of $W$. This is an example of Jensen's inequality, $E[e^{\sigma W}]> e^{\sigma E[W]}=1$. $\sigma$ can be interpreted as the magnitude of the convexity of the exponential function. This can be seen by Taylor ...


2

$$S_t = S_0\exp((r-\frac{\sigma^2}{2})t+\sigma W_t)$$ is not yet a martingale for it is not dirftless. From a probabilistic point of vew the "drift adjustment" comes into play so that the expected value of $S_t$ will be $e^{rt}$ rathern than $e^{(r+0.5\sigma^2)t}$. For the expected value of a log-normaly distributed variable with mean $\mu$ and vol ...


4

Orthogonality and independence are different concepts. The concepts are the same for Wiener processes because in the context of normal random variables, independence is equivalent to orthogonality (i.e. uncorrelatedness) Independence is the standard definition for probability. Let $\mathcal{F}, \mathcal{G}$ be the sigma algebras generated by two ...


1

you hypothesize that your data is generated by the following process: $y_t=\phi_0+\sum_{k=1}^P\phi_ky_{t-k}+\varepsilon_t$, where $\phi_k$ are your autocorrelation coefficients, and $\varepsilon_T$ - random errors. Next, you estimate your $\phi_t$ using one of the methods of estimation of autoregressive processes AR(P) of order P, e.g. see AR(P), there's no ...



Top 50 recent answers are included