# William S. Wong

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 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. Mar6 comment How does the “risk-neutral pricing framework” work? This answer does not mention the change of measure from the real-world probability measure to the risk-neutral one, which I think is needed in order to understand the (first) fundamental theorem of asset pricing [ en.wikipedia.org/wiki/Fundamental_theorem_of_asset_pricing ]. Mar6 comment How to implement a long-term trade on oil? Freddy: the OP stated that "I believe that one of the most compelling case of long-term trade is the long position on oil." The issue here is not "if" the OP should go long crude, but "how" to do it. Mar6 comment How to implement a long-term trade on oil? Darren: I'm glad you agreed with me. The front-month WTI crude is trading at \$91. If someone is interested to go long crude for the long-term, it seems silly NOT to take advantage of the backwardation to buy some 3 years out at \$86 or 5 years out at \$83. Dec6 comment Version of Girsanov theorem with changing volatility Agreed with Jase' comment that $$E_Q\left[ F(X) \right] = E_P\left[ F(X) \frac{dQ(x)}{dP(x)} \right]$$ . Dec1 comment What are the main limitations of Black Scholes? I don't think stock prices is a Wiener process, otherwise they might become negative. Instead, the relative change of stock prices is a Wiener process. That is$dS_t/S_t \sim \cal{N}(r \, dt, \sigma^2 \, dt)$, where$S_t$is the price of a stock at time$t$. Nov11 comment Probability of touching This question is entitled the "probability of touching", but the OP was asking for the probability of an option expiring in the money. The two are not the same. Note how many folks mentioned "stopping time" and "barriers" in their answers. Nov6 comment How to create a Stochastic Process through pre specified points? Since the OP wanted to "generate intra-day prices with the same OHLC properties", maybe one way to account for the high-of-the-day (HOD) is to note the time of the actual HOD, and add an additional fixed point to the Brownian bridge. Specifically, if the process is$X_t$, and let$X_{t_O} = x_O$,$X_{t_H} = x_H$,$X_{t_C} = x_C$, and$t_O \leq t_H \leq t_C$. Build a Brownian bridge that goes through the three points. We can handle the low-of-the-day similarly. Oct31 comment Is the stock price process a martingale or a Markov process? I think there is a typo in the 2nd integral$X_t = (\int_0^t X_s ds) dW_t$. Did you mean$dX_t = (\int_0^t X_s ds) dW_t$, or$X_t = \int_0^t (\int_0^r X_s ds) dW_r\$?