# Tag Info

## New answers tagged stochastic-processes

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You are generating a price series, in time steps $p_{dt},p_{2dt},p_{3dt},...$. I assume? So if you do $\text{Median}\{p_{dt},p_{2dt},p_{3dt},..\}$ then you will get a value biased towards $0$ (for the parameters you gave, and the sample size you have). If you want to look at the distributional characteristics of the price, then you have to do so per each ...

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The Feller condition applies without modification. That is under the assumption that $v$ is square-root process with poisson-arrival jumps (as you wrote), and assuming the jump distribution is strictly positive and initial level $v_0>0$. The reason is, conditional on no jumps occuring, the process is just a square root process, for which the references ...

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If there are no arbitrage opportunities there is no dominant hedge or long position. Why would there be an arbitrage opportunity if everything was priced correctly? There may be arbitrage opportunities even if there are no dominant hedges or long positions. Put-call parity shows arbitrage opportunities of badly priced options regardless of long position ...

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If we are going to have the form \begin{align*} dr = A dt + BdW_t, \end{align*} Then both A and B are functions of $t$ and $r_t$, otherwise, $r_t$ is normal. However, note that \begin{align*} r_t = \exp\Bigg(\frac{1}{\sigma(t)}\bigg(\int_0^t \theta(s)\sigma(s) ds +\sigma(0)\ln r_0 + \int_0^t\sigma^2(s) dW_s\bigg)\Bigg). \end{align*} That is, $r_t$ is ...

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Please clarify rigorously what you mean by each term. It is not true that no dominance is a consequence of no arbitrage. Think of the put-call parity: $C-P=S-K$, assuming $r=0$ since it's inconsequential. If there is no short selling then we can have: $C-P \geq S-K$ without arbitrage but No Dominance would not hold. If you think very deeply about this, ...

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First, I don't see $S^0$ appear anywhere, so I assume it is just used somewhere else. Second, there is probably a point (i) in your question, because you included the point (ii). I'd expect that $\xi_0$ is actually defined there. If that's the case, then $\xi_t$ is predictable. Otherwise, if $\xi_0$ is not defined anywhere then the process $\xi_t$ is ...

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The exercise is really not about replicating a call with asset or nothing. It is simply about the PDE of the delta of a call. The usual derivation of the BS equation starts by considering a portfolio short the option $$\Pi_t = \delta^0_t B_t + \delta_t S_t - V(t,S_t)$$ Assuming the portfolio is selfinancing (and interest rate = 0), we get  d\Pi_t ...

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I think the title here is misleading. Let's go back to the BS world with $r=0$ to $a(S_t)=S_t \sigma.$ In that case, all you are saying is that you can replicate a call option by holding $N(d_1)$ units of stock at time $t.$ What does this have to do with the second equation? I am guessing that this is the price process of an asset of nothing option with ...

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Without getting into all the Math one thing should be clear that: Call option is equivalent to: long asset or nothing AND short cash or nothing options. You cannot replicate a call option without asset or nothing since replicating portfolio for long call requires holding N(d1) quantity of the underlying. Asset or nothing gives you this exposure directly. ...

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There is a qualitative shift in the shape of the density. When V is small it is monotone decaying. When V is large it looks more like a Gaussian. Another reason he uses two schemes is that he wants match two moments of the density. When V is small, the moment matching equations for the quadratic Gaussian are unsolvable. When V is large they are unsolvable ...

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