New answers tagged

2

First, remember that only anyone who sells a naked call or put will get market to market and have to put up cash as it moves. The market makers who do the bulk of the trading will never sell an outright, naked option. They will always hedge with the actual stock. So you might be buying a put, but the person selling it to you is going to sell a share of ...


2

Chapter 15 in Arbitrage Theory in Continuous Time by Tomas Björk.


1

C: Is there any other way? - yes, borrow the shares and go short. Does such a scenario [short squeeze] damage the market? It is something bad that shouldn't exist? - how do you define that? re B - your options could be cash-settled. Doesn't necessarily affect the seller's hedging strategy, just pointing it out.


-2

Recently because of some personal reason, I tried to price Snowball Autocall using MC and PDE, assuming single underlying. 12 months Snowball, Monthly autocall observations, Daily Put Down & In. Payoff: if Autocall, then 100% principal + autocall coupon if Knock in and No Autocall, then client lose because of short put if No autocall and No Knock in, ...


1

The 50 value you compute is the payoff of your structure if the underlying is worth 100 at maturity. The initial value is the expected payoff of your trade, then, given the decomposition in call and puts, and given the replication principle, it should be: $V_t = Put_0(100,8) + Put_0(150,8) + Call_0(200,8)$, with $Put_0(k,T)$ and $Call_0(k,T)$ the price at ...


0

I was not able to reproduce @MainCom's example on a spreadsheet, hence I gave it a shot myself: $$ \begin{align} C(r)&:=e^{-r\Delta t}\left(\frac{e^{r\Delta t}-D}{U-D}\left(U-K\right)^++\frac{U-e^{r\Delta t}}{U-D}\left(D-K\right)^+\right)\\ \Rightarrow C(r+\rho)&=e^{-r\Delta t-\rho\Delta t}\left(\frac{e^{r\Delta t + \rho\Delta t}-D}{U-D}\left(U-K\...


1

A theoretical answer to your question is provided by Black-Scholes but remember that actual option prices are set by supply and demand (with guidance from models for different market participants.) You could also try to fit a model to the actual data to solve this problem, though that would be a much more complex (and costly) solution.


4

At time $T_0$, the strike price becomes known and the option turns into a ``normal'' put option, i.e. \begin{align*} V(T_0,S_{T_0}) &= S_{T_0}e^{-r(T-T_0)}\Phi(-d_2)-S_{T_0}e^{-q(T-T_0)}\Phi(-d_1) \\ &= S_{T_0}\underbrace{\left(e^{-r(T-T_0)}\Phi(-d_2)-e^{-q(T-T_0)}\Phi(-d_1)\right),}_{=:p} \end{align*} where $p$ is indeed independent of the stock ...


1

Assume that the price of the next step are $u,d$, with probability $p,1-p$. The the discounted payoff is $e^{-r}(p(u - K)^+ + (1-p)(d-K)^+)$. Now suppose the interest rate $r$ is increased by $\Delta r$. Then the new discounted payoff would be $e^{-r - \Delta r}(p(e^{\Delta r}u - K)^+ + (1-p)(e^{\Delta r}d-K)^+) = e^{-r}(p(u - e^{-\Delta r}K)^+ + (1-p)(d-e^{-...


2

Old and golden question, and maybe a new perspective: As the previous answers have pointed out, distinction needs to made between "skewness" and "skew". The former is the third moment of returns, and the latter is what volatility traders/portfolio managers usually associate with the difference between two implied volatilities straddling ...


0

My understanding is that technically, B-S uses the 'short rate' which is the instantaneous rate of borrow/lending for term T, denoted $r_t(T)$. I.e. at time $t=0$, if you invest £1 risk-free for term T, at T your investment will be worth $1\times e^{r_{t=0}(T)*T}$. Now, to obtain values for $r_{t}(T)$ you need to construct a yield curve for varying T. Note ...


5

I answer from a general discrete time/discrete state model point of view. This includes the binomial tree model as a special case. In finite dimensions, you can interpret asset payoffs and returns as vectors and retreat to linear algebra. Suppose you have $N$ states of nature and $J$ assets. Your payoff matrix is \begin{align*} A=\begin{pmatrix} X_1(\omega_1)...


2

Numéraire Change The time-$t$ price of a zero-coupon bond maturing at time $T$ is $$P(t,T)=\mathbb{E}^\mathbb{Q}_t\left[\exp\left(-\int_t^T r_s\text{d}s\right)\right].$$ Let $\mathbb{Q}$ be our standard risk-neutral probability measure which uses a locally risk-free bank account, $\text dB_t=r_tB_t\text dt$, as numéraire. From Geman et al. (1995), we know \...


1

Benoit Mandelbrot argued that a cauchy distribution was a closer fit for stock prices in several papers and in his book The (Mis)behavior of markets. Surprised no one has mentioned him yet.


4

what's the point if your average trader simply uses Black-Scholes (BS) and fudges all those real-wold problems like widening spreads, suddenly significant borrow rates, etc. etc. into the "risk-free rate" (with which we all can borrow and lend any amount, sure cough), the "volatility of the stock" (yeah, we all know about that one.) and ...


3

as it was stated correctly in the question all long butterfly options have to have a non-negative premium in order for No-arbitrage to hold. So we can say that: No-Arbitrage holds implies All Butterfly spreads have a non-negative premium. However, the reverse is not true. Just because all butterfly spreads have non-negative premiums does not mean that ...


4

I am not sure why the question you link to does not provide an answer. I’ll try to answer it but it is really similar to what has already been said there. Bottom line is: if the value $K$ is reachable by the underlying asset $S$, that is $K$ belongs to the domain of process $S$, then the butterfly should be strictly positive. First note that the butterfly is ...


2

As @Kermittfrog said in the comment, in Black formula for options on futures price you need to insert the futures price $F$: $$C = e^{-rT}[FN(d_1) - KN(d_2)]$$ where $r$ is the discounting rate. Here, $d_1$ depends only on $F$ (no rate involved). For Black-Scholes formula for options on spot price (assume asset pays no dividend to keep it clean), we have: $$...


1

Black Scholes uses a continuously compounded rate $r$. To go from a $T$-year annually compounded rate $\hat{r}$ to a $T$-year continuously compounded $r$ you use the formula $e^{rT} = (1+\hat{r})^T$ So to solve for the Black-Scholes continuously compounded rate you take logs and simplify which gives $r = \ln(1+\hat{r})$. This is what Damodoran quotes on page ...


1

In case this is useful for anyone else who comes across this, the issue was that I had set up my vector of terminal values the wrong way round (ie from smallest to largest rather than largest to smallest). The code to set up the terminal vector should read as follows: # set up terminal vector and prob vector c_n = np.zeros(steps + 1) c_n[0] = s * (u ** ...


0

I think the author is just saying that \begin{equation} \begin{split} \frac{S - e^{-r}S_d}{S_u - S_d} S_u + \frac{e^{-r}S_u - S}{S_u - S_d} S_d &= \frac{S S_u - e^{-r}S_d S_u + e^{-r}S_uS_d - SS_d}{S_u - S_d} \\ &= \frac{S (S_u - S_d)}{S_u - S_d} \\ &= S \end{split} \end{equation} and that \begin{equation} \begin{split} \frac{S - e^{-r}S_d}{...


Top 50 recent answers are included