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 Mar25 comment What is the fair price of this option? qtf: let me think about your no-arb argument. I don't get why you wrote that the "Black-Scholes model with zero rates and positive volatility gives C=1/H". This is an American binary call option, not a European one. Mar25 comment What is the fair price of this option? qtf: the OP wanted to find out the price of the option while using model-free arguments. Mar25 comment What is the fair price of this option? I don't think your answer $C \le 1/H$ is correct. As many pointed out, the stock will hit the level $H$ when given enough time. Dec9 comment Why is this stochastic integral a martingale? emcor: because of the independent increments property of $W_t$. At time $u$, $S_u$ is known but $dW^*_u \equiv W^*_{u+du}-W^*_u$ is not. Dec9 comment Why is this stochastic integral a martingale? Gordon: correct. I added a section to address the issue. Dec9 comment Why is this stochastic integral a martingale? While the OP answered his/her own question, I wonder if he/she just copied the answer from a book. In the OP's answer, it says that it "is possible to show" the lemma that $\int_0^t f(s) dW(s)$ is a martingale, but it "is omitted here to keep things short". What? That's exactly what we are trying to prove! Dec9 comment Why is this stochastic integral a martingale? StudentT: exactly! It works out only because $dW_u$ is defined as a forward difference in the Ito integral. Dec8 comment Why is this stochastic integral a martingale? Wow my answer is at -1 vote. Folks please go take a Stochastic Calculus class, as this result is fairly basic. Look at slide #12 at idsc.ethz.ch/Courses/stochasticsystems/SDE.pdf; "The expectation of stochastic integrals is zero. This is what we would expect anyway." Dec8 comment Why is this stochastic integral a martingale? fushsialatitude: good question. To move the expectation operator inside the integration, we need to satisfy Fubini's theorem. If $S(u)$ is "well-behavior", the theorem should be satisfied. Jul19 comment How to compute $\mathbb{E} \left[ (W_s + W_t - 2W_0)^2 \right]$? Richard: thanks for pointing out the typo. If $X \sim \mathcal{N}(\mu, \sigma^2)$, $\mathbb{E}\left[e^X\right] = e^{\mu + \frac{1}{2} \sigma^2}$. Jun7 comment Modelling driftless stock price with geometric Brownian motion user7056: you should complain to stack exchange as they won't let me edit my comments! Jun2 comment Modelling driftless stock price with geometric Brownian motion The superscript $Q$ indicates that the expectation is taken over the risk-neutral probability measure [en.wikipedia.org/wiki/Risk-neutral_measure]. To compute $\mathbb{E}^Q\left[ e^{\sigma W_t} \right]$, note that, for $X \sim \mathcal{N}(\mu, \sigma^2)$, $\mathbb{E}\left[ e^X\right] = e^{\mu+\sigma^2/2}.$ Apr21 comment How to calculate the expected value of a function of a standard brownian motion (Wiener process) If the OP is not comfortable with using $\cos x = \Re \{ e^{i x} \}$, let $\cos x = \frac{e^{i x} + e^{-i x}}{2}$ and proceed from there. Jan27 comment Time-zero price of two specific contingent claims @user8: i see, but then I think your answer is wrong, since $\frac{V_0}{B_0} = \mathbb{E}^Q \left[ \frac{V_T}{B_T} \middle\vert \mathcal{F}_0\right] = \mathbb{E}^Q \left[ \frac{\int_0^T S_u \; du}{B_T} \middle\vert \mathcal{F}_0\right]$. In other words, the factor $e^{r u}$ appears in the integrand; it's not canceled out by $B_T = e^{r T}$. Jan27 comment Time-zero price of two specific contingent claims @user8: Although your answer agrees with mine if we take the limit of $r \rightarrow 0$ in my expression, why are you assuming that $r=0$? Jan26 comment Time-zero price of two specific contingent claims I think your expression for $S_t =S_0 e^{\sigma W^Q_t-\frac{1}{2}\sigma^2t}$ is wrong. Under the risk-neutral measure $Q$, $S_t= S_0 e^{(r-\sigma^2/2)t + \sigma W_t}$. Jul27 comment Deterministic interpretation of stochastic differential equation You missed the important random variable $\phi$ when proceeding from the 2nd-last equation to the last equation, which makes the last equation not deterministic. Mar26 comment Does implied vol vary for calls vs puts? cf16: 1. I agree with the conclusion but do not follow your reason, as I said. To derive put-call parity, we use model-free arbitrage arguments (e.g., without referring to any implied volatility, etc.). Then you suddenly claim that put-call parity "means that volatility of call...is the same as volatility of put". 2. In addition, the version of put-call parity you referenced does not hold true for American options [ math.nyu.edu/~cai/Courses/Derivatives/lecture8.pdf ]. One needs to use a version with inequalities: $$S_0 - K \leq C- K \leq S_0 - K e^{-r t}$$ Mar26 comment Does implied vol vary for calls vs puts? I do not follow your reasoning at all. Invoking put-call parity does not lead one to say that the "volatility of call...is the same as volatility of put", as put-call parity can be derived using model-free arguments only (i.e., without using the Black-Scholes model, etc.). Mar13 comment How to implement a long-term trade on oil? Freddy: No one asked you if or how crude oil would go down, but you rattled off a bunch of scenarios regarding Iran, world peace, etc.