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## New answers tagged black-scholes

2

I think the chain of logic should be as follows: We have put value >= intrinsic. Therefore either put value > intrinsic or put value= intrinsic. If put value > intrinsic, then it is not optimal to exercise. If put value = intrinsic , it may be optimal to exercise.. Hence there is no contradiction.

3

So, from this simple no-arbitrage argument, we see that the price of the option must always be at least its intrisic value. Yes indeed However, at this point I realized something strange: if this is true, why in the world should I exercise my put option before expiry?? The inequality seems to indicate that it would be an unwise decision to ever ...

4

Under GBM $$\frac {dS_t}{S_t} = \mu dt + \sigma dW_t$$ we get $$S_T = S_0 e^{(\mu - \frac{1}{2}\sigma^2)T + \sigma W_T}$$ suggesting that $$S_T \sim \text{ln}\mathcal {N} ( \tilde {\mu}, \tilde {\sigma})$$ where \begin{align} \tilde {\mu} &= \ln S_0 + (\mu - \frac{1}{2}\sigma^2)T \\ \tilde {\sigma} &= \sigma \sqrt {T} \end{align} Now if $X \... 2 Measure change is still the most natural approach for such problems. We assume that, under the measure$P, \begin{align*} dX_t &= \mu X_t dt + \sigma X_t dW_t^1,\\ dY_t &= \mu Y_t dt + \sigma Y_t \left(\rho dW_t^1 + \sqrt{1-\rho^2} dW_t^2 \right), \end{align*} based on the Cholesky decomposition, where\{W_t^1, t \ge 0\}$and$\{W_t^2, t \ge 0\}$... 0 It depends what type of interest rate model you are using. If rates are normally distributed, the situation should be as you describe, so there should be minimal exposure to implied volatility. If rates are lognormally distributed, the higher strike option has greater time value, and has a greater volatility exposure, than the lower strike option, hence ... 1 Relatively quick Solution If$U$and$V$be normally distributed with means$\mu_u\,,\,\mu_v$, variances$\sigma^2_u\,,\,\sigma^2_v$and correlation$\rho$then we can show ( by definition of expectation and apply joint density function ) $$\mathbb{E}\left[\left(e^U-e^V\right)^+\right]={\large{e^{\mu_u+\frac{1}{2}\sigma_u^2}}}\Phi\left(d_1\right)-{\large{e^... 3 You know$$N(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{\frac{-u^2}{2}}du$$then$$N'(x)=\frac{1}{\sqrt{2\pi}}e^{\frac{-u^2}{2}}$$therefore$$\mathcal{V}=S_t\sqrt{\tau}N'(d_1)$$0 With respect I think that this issue was associated Martingale properties AND dominated convergence theorem.(May be Wrong) Let L\in(0,K) a fixed price, we can consider the following choices for the exercise of a put option with strike price K: If S_t\le K, then we exercise contract at time t, and were delighted. O.W. we should wait until the ... 2 To elaborate on the explanation provided by @Alex, the reasoning is because when we look at the PDE we notice that the S terms appear in pairs with the \dfrac{\partial}{\partial S}, i.e. S\dfrac{\partial}{\partial S} and S^2\dfrac{\partial^2}{\partial S^2}. What this says it that if we were to try a polynomial function of S then after applying ... 5 Thanks to @Phun and @oliversm I solved the problem. So I'm posting here the solution in case someone will need it. Under Black-Scholes assets dynamics are determined by a Geometric Brownian Motion, and we can define the price of a security at time t+\Delta t as:$$S_{t+\Delta t}=S_{t}\exp\left(\left(r-\frac{1}{2}\sigma^{2}\right)\Delta t+\sigma\sqrt{\... 0 The log-return of a stock over a period$\Delta t $starting at$t=0$is defined as: $$r_{\Delta t} = \ln \left( \frac{S_{\Delta t}}{S_0} \right)$$ Thus you should compute$S_{\Delta t}$as $$S_{\Delta t} = S_0 \exp ( r_{\Delta t} )$$ when you are given the$\Delta t $-period log-return i.e. the one which you sample as you propose above. Thus no ... 2$d$is a vector that collapses the$n$-dimensional vector into a real number. In the BS case$d=1$. There is nothing to be estimated. Also not that in practice affine pricing is done through FFT (and variants) rather than the direct transform you quote. 0 You mix up several things: if you sample from Brownian motion, then $$B_{t+\Delta t} - B_t$$ is normally distributed with variance$\Delta t$. Thus if you sample a standard normal$Z$(with variance 1) then you can use $$\sqrt{\Delta t} Z$$ as sample for$B_{t+\Delta t} - B_tin order to get the correct variance. Recall that constant factors enter ... 1 It is reasonable and PDE approach is not suitable. In the Black scholes model we have $$d\ln {{S}_{T}}=\,(r-\frac{1}{2}{{\sigma }^{2}})dt+\sigma d{{W}_{t}}$$ so $$d\ln {{S}_{T}^2}=\,(2r-{{\sigma }^{2}})dt+2\sigma d{{W}_{t}}$$ as a result $$\ln {{S}_{T}^2}=\ln{{S}_{t}^2}\,+(2r-{{\sigma }^{2}})(T-t)+2\sigma (W_T-W_t)$$ let $$Y(t)=(2r-{{\sigma }^{2}})(T-t)+... 2 The PDE will be the same but because the terminal condition is different the solutions will not be the same. The different boundary condition will give different values at t=T. Then the equation is marched backwards in time in both cases using the same equation but because the terminal condition is different the solutions will not agree 1 Another approach as follow. The T-Straddle option X, i.e.$$X=\left\{ \begin{align} & K-S(T)\quad ,\quad 0<S(T)\le K \\ & S(T)-K\quad ,\quad S(T)>K \\ \end{align} \right. $$has then following contract function$$\Phi (x)=\left\{ \begin{align} & K-x\quad ,\quad 0<x\le K \\ & x-K\quad ,\quad x>K \\ \end{align} \right.... 3 just take a call and a put struck atK$and add them together. For the hedge just add the hedges together as well. 5 Starting from the Black-Scholes model that $$\dfrac{dS}{S} = \mu \:dt + \sigma\:dW_t$$ where$W_t$is a standard Brownian motion, and$\sigma$and$\mu$are constant where$\sigma > 0$. Here$W_t$is a Brownian motion under the physical measure$\mathbb{P}$. We can then use Girsanov's theorem to change the measure to risk neutral measure$\mathbb{Q}$... 3 The standard starting point with modelling a stock price process is to use the Black-Scholes model for the stock price. This simply asserts that the changes in the stock price are described by the following stochastic differential equation (SDE) $$\dfrac{\textrm{d}S}{S} = \mu\:\textrm{d}t + \sigma\:\textrm{d}W_t$$ where$W_t$is a standard Brownian motion (... 0 You can see e^(rt) =St/S0, which is continuous and can generate r when t=1 time lag r= In(St/S0). 0 At$t=0$, you have a vol surface$(T,K)\to\sigma(t=0,S_t=S_0,T,K)$the hard question is which dynamics i.e$\sigma$seen as$\sigma: (t,S_t)\to ((T,K)\to \sigma(t,S_t,T,K))$and even, if you imagine that behind this, there is a deterministic function, you still have to suppose a dependence with respect to$S_t$Here are two examples 1)$\sigma(t=0,S_t=...

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