# Tag Info

0

you have to be careful to distinguish between trinomial trees in a theoretical sense which do not give unique prices, and trinomial trees chosen as an approximation to the risk-neutral measure of the BS model. In the second case, they are an effective numerical method as are binomial trees. Trinomial trees are more useful when you want to ensure nodes lie ...

3

Since American style options allow early exercise, put-call parity will not hold for American options (unless they are held to expiration). In practice, there is also a difference between calls and puts for European options as well. The full description is here: What causes the call and put volatility surface to differ?

0

Consider two different strikes $K_1$ and $K_2$. Then, for $i=1$ and $2$, \begin{align*} C(K_i)-P(K_i) &= e^{-r T}E(S_T-K)\\ &=e^{-qT}S_0 - e^{-rT}K_i. \end{align*} Now, the two unknown parameters $r$ and $q$ can be solved from the two equations.

0

I think the delta-replicating of $\sigma_2$ call is just a fancy way of saying "hedging the call option bought at $\sigma_1$ volatility, with deltas based on $\sigma_2$ volatility". This is full arbitrage in case the hedging/replicating is optimal, and just a statistical arbitrage in real life. You probably do not need such sophisticated proof of why this is ...

0

Suppose you have a call/put pair with the same strike $K.$ Then a position long the call and short the put has the payoff of a forward struck at $K:$ $$C(K) - P(K) = e^{-rT} \mathbb{E}[ S(T) - K ],$$ Where $C(K)$ and $P(K)$ are the call and put price, $r$ is the interest rate, and $T$ is the time to expiry. Then by linearity of expectation and the ...

2

Pricing via characteristic functions arises naturally in models that involve Levy processes. Therefore I can see how Black's formula for swaptions can be generalized for Levy dynamics: As in Black's model take the annuity as numeraire, and define the relevant measure $Q$ Black assumes that under this measure the swap rate is martingale GBM, that is to say ...

0

It depends on how close your strike is to the forward (at expiry). Lets say you have an option which is expiring in a week, the forward will be close to the spot. Hence an out of the money strike for such as option will be closer to the at the money for an option expiring in 6 months (where the forward is pretty far from the spot).

0

Take a look here: here Option values don't always go up with maturity.

0

Generally when doing trees, people discount one step at a time. So this is the value discounted from the end to that node in the tree.

0

Quick answer, doing it with 5th grade math ^^: Assuming Forward = Spot = 50 for a 10% move = 5: Call: 55 / (1 + x) = 50 -> x = 10% Put: 45 * (1 + x) = 50 -> x = 11.11...% So the Call/Put ratio equaling (10% / 11.11...%) = .9 -> Premium -> 10% (1 - .9) ... I am only 13 years old so don't hate if I'm wrong :D

1

Let $\{F(t, T), 0 \leq t \leq T\}$ be the forward process that satisfies an SDE of the form \begin{align*} dF(t, T) = \sigma F(t, T) dW_t, \end{align*} where $\sigma$ is the constant volatility, $\{W_t, t>0\}$ is a standard Brownian motion. The payoff at time $T_1$, where $0 < T_1 \leq T$, of a vanilla European forward option is of the form ...

0

I think one way about it is maybe like below... Consider value of a call option on the forward at time $t$ and forward price $F$, and the value of a put at time $t$ and forward price $(K^2/F)$. Assume they have the same strike price $K$. Then at time $T_1$ ( option expiry ), we have $$C( F,T_1 ) = ( F - K )^+ \\ P( \frac{K^2}{F}, T_1 ) = ... 2 To keep things simple let's assume you have a perfect random number generator (i.e. I will discuss only the statistics not the numerics of the problem). I will also focus on the practical matter and gloss over some mathematical details. From a practical perspective "convergence" means that you will never get an exact answer from Monte-Carlo but ... 2 the output of an MC simulation depends on the random numbers used and if the distribution used is not too weird, after 10,000 runs you will get an answer that is distributed$$ \mu + \frac{\sigma}{\sqrt{n}} Z, with Z a standard normal. Here n=10,000. With \mu the quantity you want and \sigma the standard deviation. So you won't get precisely the ... 0 Note that \begin{align*} q= \frac{e^{r\Delta t} -d}{u-d}. \end{align*} Then, \begin{align*} u = \frac{e^{r\Delta t} -d}{q} + d. \end{align*} Therefore, \begin{align*} 1-\bar{q} &= 1-uqe^{-r\Delta t}\\ &=1- \big(e^{r\Delta t} -d\big)e^{-r\Delta t}-dqe^{-r\Delta t}\\ &=de^{-r\Delta t} -dqe^{-r\Delta t}\\ &=de^{-r\Delta t}(1-q). \end{align*} 0 You said: if the index's return is negative then the note's total payoff will be 1000 x (1 - R) When the return is positive, then the payoff is 1000 + 1000 * 2.5 * max{R - 0.1, 0}. Hence, your payout function should be as follows: v(T)= Indicator {Index(T) < 1000, Index(T); 1000 < Index(T) < 1100, 1000; Index(T) > 1100, 1000 + ... 0 I think the problem lies in the principal. Normalize S_0=1000. Essentially you get S_T \text{ if } S_{T} < 1000,  1000 \text{ if } 1000 < S_T < 1100,  2.5 (S_T - 1100) \text{ if } 1100 < S_T $$I'd write this as$$ S_T - \max(S_T-1000,0) + 2.5 \max(S_T-1100,0). Since $S_0=1000$ you'll get something close to $1000.$

1

That seems to be a nice paper but I haven't worked through it completely yet. As I understand it, the goal is to replicate the holding (by an investor) of an European option using an American option, stock and bonds in a self-financing manner. As the value of the underlying changes this requires rebalancing of the option and the bond, i.e. hedging. Since ...

0

The issue I have with these approaches is that they use the unconditional distribution to eliminate the latent volatility. However, when the volatility process has very weak mean reversion one would need a very long and clean sample to make robust parameter identification from the unconditional density. They just throw away all the information from the ...

Top 50 recent answers are included