# Tag Info

## New answers tagged black-scholes

1

There's no best method. The question is : what is the behavior of the volatility structure (atm and skew) when the underlying moves? Each method assumes something different. In the real market, one method might work well for a period of time (in the sense that it minimizes residual p/l), but then another method might take over as best. Practitioners ...

0

Thr second sum should be: $$\sum_{i=6}^{10}u_i = 0.10039773$$ This gives a mean of $0.0067648$ and a standard deviation of $\sigma=.028836$. To avoid these errors you should use something to automate your calculations. Something like a spreadsheet.

2

Let's talk about your first equation: If you exercised your option early, you got this payoff. But if you are a rational investor you'd realize that this is less than what you would get if you would just sell your option itself. i.e. the payoff at time t will be more than S(t)-K because the option is worth more than that as it also has some time value. so ...

0

Given the assumptions of the Black Scholes model (continuous hedging, stock is driven by one dimensional GBM with all parameters known) the stock drift does not matter: all information about the expected return on the option is already contained in the current stock price (which reflects investor's preferences with respect to the known stock volatility and ...

0

The payoff can be decomposed as \begin{align*} \phi(S) &= 100 \, I_{50 \le S_T < 100}\\ &= 100 \, \big(I_{S_T \ge 50} - I_{S_T \geq 100}\big). \end{align*} Note that, under the risk-neutral measure $P$, \begin{align*} E(I_{S_T \ge K} \mid \mathcal{F}_t) &= P(S_T \ge K \mid \mathcal{F}_t)\\ &= N(d_2), \end{align*} where \begin{align*} d_2 = ...

1

Certainly, you must agree that $$C_{T}-P_{T}=\left(S_{T}-K\right)^{+}-\left(K-S_{T}\right)^{+}=S_{T}-K.$$ Therefore, since $$C_{t}=e^{-r\left(T-t\right)}E_{Q}\left[C_{T}\right]\text{ and }P_{t}=e^{-r\left(T-t\right)}E_{Q}\left[P_{T}\right]$$ it follows by the linearity of $E$ that $$C_{t}-P_{t}=e^{-r\left(T-t\right)}E_{Q}\left[C_{T}-P_{T}\mid ... 1 Yes, you are right. It appears to be a trivial typographical error in the book. I checked the formulas on Wikipedia https://en.wikipedia.org/wiki/Greeks_%28finance%29 and they agree with yours. The signs are obvious also since N(.) is between 0 and 1, i.e. non-negative. Now, about the reasoning starting with "from a logical point of view". Are you familiar ... 0 No. Under BS you get rewarded for the risk you bear, and the risk premium (ie the instantaneous excess return over the risk free rate) is \frac{\mu-r}{\sigma}\ \sigma^*_t\ dt where \sigma^*_t is the instantaneous vol of your position. For a simple stock position the instantaneous vol is constant, \sigma^*_t=\sigma therefore you have a constant excess ... 1 To get the answer, you should know the difference between forward and futures. If all options in your strategy will not be really settled, instead, just an P&L is marked, then you will find in the long run your return is 0. This is similar to a forward contract. However, if these options are settled, you will get a realized P&L which is MtM weekly. ... 1 This is entirely true. The basic pricing formula that is intended to work for all assets (including options) is $$P=E[m*X]$$ where P is the price, m the discount factor, and X the payoff. This can also be rewrite $$1=E[m*R]$$ with R the return. This is know as the Euler Equation. In the ... 2 BS does not require this. The real-world drift of the stock can be greater than the risk-free rate so on average you make money. If this seems weird, forget the option and consider the stock. If you buy it and hold it for a week then you ought to make money on average about the risk-free rate since you get a risk compensation for the fact you can lose ... 1 I actually discuss this question at length in chapter 1 of More Mathematical Finance. The essential point is that if you can write$$ X=YZ  with $Y,Z$ independent $E(Z)=1$ and $Z>0$ then $X$ is more uncertain than $Y.$ It then follows from Jensen's inequality that the price of an option on $X$ that has a convex pay-off will be at least as high as the ...

4

We consider the case $S\leq K$ only. In this case, the intrinsic value is zero. Note that, \begin{align*} \frac{\partial C}{\partial S} = N(d_1) >0. \end{align*} That is, $C$ is a strictly increasing function of the spot level $S$. Moreover, \begin{align*} \lim_{S\rightarrow 0} d_{1, 2} = -\infty. \end{align*} Then, \begin{align*} \lim_{S\rightarrow 0} C ...

0

I think I have part of it. Assume zero interest rate and T = 1. Then the call price C is C = S.N(d1) – K.N(d2) where S is underlying price, K is strike, and d1 = ln(S/K)/V+V/2 d2 = d1 – V/2 d1 and d2 roughly represent the moneyness in terms of standard deviation, including the term V/2 which is added in d1, and subtracted in d2. Nd1 and Nd2 ...

2

Per @SKRX's suggestion, another solution is provided below. For simplicity, we assume that the stock price process $\{S_t \mid t \geq 0\}$ follows an SDE, under the risk-neutral measure $\mathbb{Q}$, of the form \begin{align*} \frac{dS_t}{S_t} = r dt + \sigma dW_t, \end{align*} where $r$ is the constant interest rate, $\sigma$ is the constant volatility, ...

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