Tag Info

New answers tagged

0

yes if you use the BS Model for computing deltas and the same model for evolving the stock price then you should replicate the pay-off of any contract.


1

Take a look at Hull's Appendix of the Volatility Smiles chapter. (Chapter 16 in my version). It gives a method to calculate the probability density function based on option prices: $$ g(K) = e^{rT} \frac{\partial ^2 c}{\partial K^2} $$ This result comes from the Breeden Litzenberger 1978 paper.


1

If $\mu$ is large, then it is more likely for the call to finish in the money. Your and my intuitions suggest that this means that the option is more valuable. But this is wrong. A call option is an insurance policy. A call option is useful because it protects you in the case that the value of the stock goes down. That is why call options are valuable for ...


1

I would also say that the pricing of some exotic products require to compute expectations of functions of the random variable at consideration, and these functions may grow more than linearly : you need finite moments in order for the prices of these exotic derivatives to be bounded.


1

In the Black-Scholes Model or Heston Model, the American option satisfies the same PDE, but with different boundaries.For an American call option $C_A(S,\tau )$, we can therefore write \begin{align} \frac{\partial {{C}_{A}}}{\partial \tau }=+\frac{1}{2}{{\sigma }^{2}}{{S}^{2}}\frac{{{\partial }^{2}}{{C}_{A}}}{\partial {{S}^{2}}}+(r-q)S\frac{\partial ...


1

There is no closed formula for American put option. However, there is an analytic solution for perpetual American put option. The only difference is that the maturity of the perpetual American option is infinite. Why that makes such a difference? That's because we can determine the optimal stopping time (and therefore optimal exercise price) if we don't ...


1

it all comes down to how you define analytic. If you push the definition far enough there are some. An exact and explicit solution for the valuation of American put options DOI:10.1080/14697680600699811 Song-Ping Zhu pages 229-242 However, it's an infinite sum of recursively defined double integrals.


1

Yes, there is none. Quoting Higham (2004): "The mathematical problem defined by (...) is much more difficult than the BlackÔÇôScholes PDE that arose without the early exercise facility. In general, there is no closed form expression for $P^{Am}(S, t$) and we must use numerical methods to obtain approximate values." Where (...) refers to the American Option ...


0

First your equation for returns is false. Forgetting about the jump, it does not reduce to Gbm returns. The variance term from Ito's formula is missing. In the case of a jump, a similar term should appear. Secondly, the distribution is obviously not "what market things are the real probabilities": you chose to impose specific sizes for the jump, the market ...


0

EDIT I think I figured it out. Under the $\mathbb{Q}$ measure, $\begin{equation} S_t \sim LN(ln(S_0) + r - 0.5\sigma^2, \ \sigma\sqrt{t}) \end{equation}$ Under the $\mathbb{P}$ measure $\begin{equation} S_t \sim LN(ln(S_0) + \alpha - 0.5\sigma^2, \ \sigma\sqrt{t}) \end{equation}$ Suppose we simulate the following stock prices Si = 96.33, 69.04, ...


0

basically, you have very few constraints. The main other constraint you might consider is that if the real-world probability of lying in a given set is positive, the risk-neutral probability must be too. So a point mass at the futures price is not valid unless you believe that this is the case in real-world too. (see chapter 6 of my book "concepts etc" ...


1

The answer is: yes. We can consider a model that assumes there is only one jump with distribution $p$, and otherwise the stock value does not change. Then for $p$ to be a martingale measure the only condition is on expectation of $p$. Hence, any distribution with desired expectation can be a marginal of some pricing measure.


0

The other moments are not free. Suppose we are in the standard BS environment with one stock and one bond and a single brownian motion. Suppose we have a derivative that at maturities pays: $V_T=S^2_T$ and we want to price it. Under the martingale measure we know that: $E_t^Q[S_T]=S_t e^{r(T-t)}$ and $Var^Q_t(S_T)=S_t^2e^{2r(T-t)}(e^{\sigma^2(T-t)}-1)$. ...


2

Let $\{P_t \mid t \geq 0\}$ be a compound Poisson process, where \begin{align*} P_t = \sum_{i=1}^{N_t} (V_i -1), \end{align*} and $N_t$ is a Poisson process with intensity $\lambda$ and jump times $\tau_i$, $i = 1, \ldots, \infty$. Let $Y_i=\ln V_i$ and $f(x)$ be the density function. Then \begin{align*} P_t - \lambda t E(V_1) &= P_t - \lambda t ...


0

Let's start with the main idea, I hope you can finish the computations yourself. Whenever you want to derive a pricing equation, try the following approach: discounted value of portfolio/option/derivative must be a martingale for non-arbitrage reasons. Since you have a Markovian dynamics in variables $t$ and $S$, you assume that the price is some function ...


0

One solution is to calculate the annual dividend yield implied by that. $Div_{yield}=\delta=1.5/40$ and then replace the $r$ on $d_+$ by $r-\delta$. A cleaner way would be to compute it using a binomial tree.


1

$S_0$: The stock price today. $p$: The probability of a price rise. $u$:The factor by which the price rises. $d$: The factor by which the price falls. Three equations are required to be able to uniquely specify values for the three parameters of the binomial model. Two of these equations arise from the expectation that over a small period of time the ...


2

In the Heston Model we have \begin{align} C(t\,,{{S}_{t}},{{v}_{t}},K,T)={{S}_{t}}{{P}_{1}}-K\,{{e}^{-r\tau }}{{P}_{2}} \end{align} where,for $j=1,2$ \begin{align} & {{P}_{j}}({{x}_{t}}\,,\,{{v}_{t}}\,;\,\,{{x}_{T}},\ln K)=\frac{1}{2}+\frac{1}{\pi }\int\limits_{0}^{\infty }{\operatorname{Re}\left( \frac{{{e}^{-i\phi \ln K}}{{f}_{j}}(\phi ;t,x,v)}{i\phi ...


0

First Question:This derivation is a special case of a PDE for general stochastic volatility models,described in books by Lewis (2000), Musiela and Rutkowski(2011) and others. The argument is similar to the hedging argument that uses a single derivative to derive the Black-Scholes PDE. In the Black-Scholes model, a portfolio is formed with the underlying ...


2

I would argue that there is some path-dependency involved. The BS model is considered the big breakthrough and it presented the world with some kind of tractable toy model. After that people saw that you had to adjust the model to account for all kinds of stylized facts (e.g. non-constant volatility for different strikes, over time and so on). Yet finite ...



Top 50 recent answers are included