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2

Another take on the question which uses stochastic calculus [Digression] Assume deterministic and constant rates without loss of generality. Also assume the absence of arbitrage opportunities and market completeness Let $B_t$ denote the time-$t$ value of a risk-free money market account in which 1 unit of currency $C$ has been invested at $t=0$: ...


0

A few years ago I asked a similar question on MO: http://mathoverflow.net/questions/22828/big-picture-concerning-ito-integral-stratonovich-integral-and-standard-results My take today is that you really don't need this heavy mathematical machinery for standard BS but as soon as you move on to more sophisticated (and realistic) models you surely do, so it is ...


6

Let $\tau = T-t$. Then \begin{align*} S_T = S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, Z}, \end{align*} where $Z$ is a standard normal random variable, independent of $\mathcal{F}_t$. Moreover, \begin{align*} E\left(S_T 1_{\{S_T >K\}}\mid \mathcal{F}_t \right) &= E\left(S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, ...


0

Let \begin{align*} C(S, K, t) = SN(d_1) - e^{-rt}KN(d_2) \end{align*} denote the Black-Scholes call option price with initial asset value $S$, strike $K$, and maturity $t$. Note that \begin{align*} \frac{\partial C}{\partial S} = N(d_1). \end{align*} For the above barrier option, note that \begin{align*} E_0 &= V_0 N(d_1)-e^{-rt}KN(d_2) -\bigg[V_0 ...


1

There is a logical fallacy in your argument. The price of a European call expiring 1 day before a dividend payment may well be greater than that of a call expiring after it. In other words, claiming that $$ C_E (S_0,K,t_D-1\text {day}; D, t_D) < C_E (S_0,K,T; D, t_D) $$ is not necessarily true. Try the above inequality with a huge dividend (e.g. $D ...


2

Yes it can be done. However, bear in mind that a naive explicit FD scheme is not unconditionally stable (see CFL stability condition). As far as your initial/boundary conditions issue is concerned: [Time domain] Use terminal condition $V(S,T)=h(T)$ where $h(T)$ figures the payoff of the target derivative claim at maturity $T$ (e.g. $h(T)=(S-K)^+$ for a ...


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Any position that is long the market. Eg long stocks, short puts on stocks etc, is being compensated for taking risk. Any position that is bearish eg short the market, or short calls on the market, is being penalized for taking the risk. There's no contradiction. Investors overall are long stocks, and they need to get paid to take the risk. That's what ...


4

Peter Jaeckel has written various papers on this. "by implication" and "Let's be rational" are the most recent ones. He also provides code on his website www.jaeckel.org. (Note: the question asked for literature.)


1

Look on Google for Asymptotic behavior of Implied Volatility Near Infinity you will find results like : $$I(K) \stackrel{K\to\infty}{=} \sqrt{\frac{2}{T}}\left(\sqrt{\ln \frac{K}{C(K)}}-\sqrt{\ln\frac{1}{C(K)}}\right) +\text{O}_{K\to \infty}\left(\frac{\ln\ln\frac{1}{C(K)}}{\sqrt{\ln\frac{1}{C(K)}}}\right)$$


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I guess you could use the following link for data for training approximations: http://www.scientific-consultants.com/nnbd.html


1

I think you mix up marginal law, and law of the process. Your $Z_k$ must have three values, you just have to write the value of $\ln(H_k)$ for each possible value Let $X$ taking values $(x_1,x_2,...,x_n)$ and $p_i=P(X=X_i)$, then $P(f(X)=f(x_i))=\sum_{j=1}^n p_j\mathbf{1}_{f(x_j)=f(x_i)}$ if $f$ is a one-to-one mapping, you get $f(x_j)=f(x_i)\Rightarrow ...


1

Let \begin{align*} V(t, S_t) = E\Big(e^{-r(T-t)} g(S_T)\mid \mathcal{F}_t \Big) \end{align*} be the risk-neutral value at time $t$ of the option payoff $g(S_T)$. Then $\{e^{-rt}V(t, S_t), 0 \le t \le T\}$ is a martingale. Consequently, \begin{align} -rV + \frac{\partial V}{\partial t} + (r-q)S\frac{\partial V}{\partial S_t}+\frac{1}{2}\sigma^2 S_t^2 ...


1

In your answer, you don't include dividend. I am sorry to say it is wrong. Payoff function is $$ g(S_T) = (S_T - K_1)_+ - 2(S_T - \frac{K_1+K_2}{2})_+ + (S_T - K_2)_+ $$ BS pricing formula with dividend gives $$ V(t=0,S) = e^{-r}E(g(\tilde{d}S_T)) = \tilde{d} \left(BS_{call}\left(\frac{K_1}{\tilde{d}}\right) - 2BS_{call}\left(\frac{K_1+K_2}{2 ...


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This is an interesting and not so easy question. Here's my 2 cents: First, you should distinguish between mathematical models for the dynamics of an underlying asset (Black-Scholes, Merton, Heston etc.) and numerical methods designed to calculate financial instruments' prices under given modelling assumptions (lattices, Fourier inversion techniques etc.). ...


0

It's clear that over time $t\rightarrow t+dt$ the (expected) implied variance in the BS world (under the risk-neutral measure $\mathbb{Q}$) is $\sigma^2 dt$. The realized variance of an arbitrary single log stock path (under the physical measure $\mathbb{P}$) is $$\sum_{i=1}^\infty \sigma^2(W_{i+1}-W_i)^2 \approx \sum_{i=1}^n ...



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