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6

Let $t_0, t_1, \ldots, t_n$ be observation dates, where $0=t_0 < \cdots < t_n = T$, and $\{S_t \mid t \geq 0\}$ be the equity price process without dividend payments. Then the realized variance is defined by \begin{align*} \sum_{i=1}^n \ln^2 \frac{S_{t_i}}{S_{t_{i-1}}}. \end{align*} Note that, for sufficiently small $x$, \begin{align*} \ln (1+x) ...


0

The option portfolio would not change its value because the dividend also has been already priced in as you can assume that option markets have the same information on the underlying as the stock market. A dividend in general decreases the value of the call options because these are foregone relative to holding the asset as soon as the market has as an ...


0

It seems to me that if it were a martingale, one could make a lot of money (in expectation) by shorting it and investing the proceeds in the risk neutral asset/bank account.


3

Well, "mean reversion trading" could mean a lot of things, I am not qualified to describe it in full generality. However, there is a simple model of mean reversion called the Ornstein Uhlenbeck process that is often seen. It has two parameters \lambda and \sigma, where lambda is the strength of the mean reversion (so one over lambda is the mean reversion ...


4

This drift comes from making the discounted stock a martingale in the risk-neutral measure $\mathbb Q$ You start with a stock in $\mathbb P$ having this form: $$ dS_t = \mu S_t dt + \sigma S_t dW_t $$ You also have a discount factor $e^{rt}$. The idea is to remove the drift of the discounted process in $\mathbb Q$ so you get (after applying Girsanov's ...


3

You only have one asset in your portfolio which means that you can only statically hedge. By the definition of self financing, $V_0=\phi_0 S_0$, $V_1=V_0+\phi_1 (S_1-S_0)$, and $V_1= \phi_1 S_1$. Putting these last two together, $V_0=\phi_1 S_0 $. Hence $\phi_1=\phi_0$ and you have a static position. Intuitively, this is because you cannot trade in ...


1

The binomial model certainly is self-financing. First, get the value at every node by working backwards using risk-neutral evaluation. Then at each step and node, you get the value in the up node and the down node from where you are. You can fit a straight line as a function of stock through the two. You hold stocks and bonds to fit this straight line with ...


1

I'd add a comment if I could but don't have enough reputation. How do you know your final equation is not equal to zero. The $f_i$ have not yet been calculated in terms of $S$. Certainly the $f_i$ that are in the final set of nodes are known since they are defined in terms of the payoff and the terminal price. The ones in intermediate steps have to be ...


2

Try this paper by Rolf Poulsen : http://colloquium.mathfinance.de/abstracts/poulsen.pdf. He derives barrier option prices in the Black-Scholes model using only reflection and Girsanov's Theorem, and then discusses extensions.


2

I'll provide an extreme counterexample. Suppose your non self-financing replicating portfolio is to do the following. At time t through T you own no stock and no bonds. Then you observe X at time T and place X dollars in the portfolio. It would be incorrect to say that the options value at time t is zero. The idea of a replicating portfolio is that at ...


1

I think that you can find the answer to this question here: http://people.stern.nyu.edu/wsilber/chuang-silber%20approx%20option%20value.pdf


1

Option pricing is all about intrinsic value and time value. The intrinsic value is the difference between the strike price and the underlying market price. A call is a right to buy the underlying. Therefore intrinsic value of a call is positive when the strike price is below the underlying market price. You can buy for less than the market offer. A put is ...


3

There is a good quick well-known approximation for at-the-money options: $$\textrm{Call,Put} = 0.4 S \sigma \sqrt{T}.$$ See further discussion at What are some useful approximations to the Black-Scholes formula?.


2

I don't know where you would have read that, but no, time value cannot be negative. Time value is option value minus intrinsic value. Intrinsic value is a model-imdependent no-arbitrage bound on option value. For an out-of-the-money payoff, intrinsic value is zero, and since the call or put payoff is non-negative this is a clear lower bound. For an ...


1

Important assumptions: - we have zero interest rate, - option is perpetual, EDIT: with probability 1, share price will hit the barrier $H$ (in fact this is a hidden assumption that price changes continuously or we can at least trade at the very moment when $S_t = H$). No, we can't assume that, because , as @q.t.f noted, it would imply arbitrage. ...


1

Unfortunately I cannot upvote user2142's answer because I lack the reputation, but his reasoning makes sense to me: the price is $\$1/H$ because as the seller of the option you buy $1/H$ shares for the premium. You sell them when the $S_t$ hits $H$ to obtain the $\$1$ you have to pay to the option buyer. I think the price is model free for any model with ...


1

Let $T= \inf\{t>0: S_t = H\}$. Then the option payoff is given by $\mathbb{1}_{\{T < \infty\}}$, and the value of the option is given by $\mathbb{P}(T< \infty)$. We assume that the stock price process is a geometric Brownian motion, that is, for $t>0$ $$ S_t = \exp\big(-\frac{1}{2}\sigma^2 t + \sigma W_t\big),$$ where $\{W_t, t \geq 0\}$ is a ...


0

The differential equation guarantees no arbitrage. There is no need to list each one individually.


2

this is probably the most asked question in quantitative finance... There are many answers. One nice example to consider is what if the calls were struck at zero. The call then pays the stock price at time $T$ and so it's value today must the stock price today since we can replicate by holding one unit of stock. This will be true regardless of the drift of ...


0

Option pricing theory and interest rate theory are used within life insurance mathematics. See for example the articles of Thomas Møller: Local risk-minimization with survivor bonds (with L. Henriksen). To appear in Applied Stochastic Models in Business and Industry, 2014. On systematic mortality risk and risk-minimization with survivor swaps (with M. Dahl ...


0

You may bet on stock 1 by buying a call option on stock 1, and drive up the option price. But some arbitrageurs will immediately short the option and hedge with stock 1, pocketing the profit. These arbitrages will force the call option back to normal. Discrete or continuous-time, the logic is the same.


2

The claim payoff you describe, $g(M)$, looks to me like a tight butterfly spread that pays off only in one state of the world. Can't you just replicate that by short two calls with strike $K_0$ and long two calls, with strikes one either side at $K_0\pm 1$? Then the price of your option would be $C(K_0+1)+C(K_0-1)-2\cdot C(K_0)$. This is effectively the ...


0

Is the stock so easy to manipulate? How much does she need to spend to drive the price against her, far enough to pass the strike and make a profit? For a stock that is that thinly traded to be manipulated, is there sufficient liquidity in the options market? What about the hedging positions of the guys that sold options to her? It is also most probably ...


1

The classical connection is the http://en.m.wikipedia.org/wiki/Esscher_transform developed for actuaries in 1932 which essentially transforms the objective probability measure into the risk neutral one used in quant finance.


3

Actuarial science traditionally focuses on estimation of joint probabilities using real data where math finance is on valuation of contracts under an arbitrary distribution. It means the first one deals with methods of estimation of future distributions (the number of accidents of a given kind, the probability of someone with a given profile to have a ...


0

To a FX trader, "considerably out-of-the-money" means "low delta" and "close to at-the-money" means "close to 50% delta." That is, moneyness is measured in terms of delta. A useful way to understand this is that delta measures the probability of finishing in the money[1]. A 10% delta option has 10% chance of finishing in the money; a 50% delta option has ...


1

If you know the stock will finish above the strike, then the call option becomes a forward contract since it will always be exercised. We therefore price it as a forward. Its value at maturity is $$S_T - 60.$$ We can synthesize $S_T$ with one unit of stock costing $70.$ We can synthesize $60$ with $60$ ZC bonds which costs $60/1.055$ since the yield is ...


1

I don't know of any libraries for this. There is a pretty good literature on the problem you mention though. I suggest https://cs.uwaterloo.ca/~paforsyt/numuncert.pdf as a good paper to follow; they study numerical techniques, document pitfalls, and even prove something about convergence of their preferred approach.


0

User9403 nails it! Intuitively higher expected rate of return => higher stock price => higher (call) option price!


-2

"Why the expected return rate of a stock has nothing to do with its option price?" It has everything to do with the option price! The option price is a function of the stock price. If the expected rate of return on the stock price declines, the stock price will decline as will the option price. "Suppose I have two stocks A and B, the price is the same ...


9

$\theta$ is the "mean" for this process. If $X_t > \theta \implies (\theta - X_t) < 0 $, which means that the drift for the process is negative and tends towards $\theta$. The opposite case can be made for $X_t < \theta$ ; the process will have positive drift when $X_t$ is below $\theta$. Therefore we can consider $\kappa$ to be the "speed" of mean ...



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