# Tag Info

5

If you want to know what Greeks the market assigns to an option, i.e. the market implied Greeks, then you would use the implied volatility. And that is what traders like to look at.

4

Trinomial trees give incomplete markets so there is a range of possible risk neutral prices. So you have to find the possible probabilities that make the tree risk-neutral and see what prices you get. You have the correct expressions. Now just have to parametrize the set of solutions. It is one-dimensional and all the probabilities are positive so you need ...

3

First, we have $P(t)+S(t)=C(t)+B(t,T)\cdot K$, Then, $\frac{\partial P(t)}{\partial S(t)} + \frac{\partial S(t)}{\partial S(t)} = \Delta^{\text{put}}_{t}+1$ and $\frac{\partial C(t)}{\partial S(t)} + \frac{\partial [B(t,T)\cdot K]}{\partial S(t)} = \Delta^{\text{call}}_{t}+0$. Finaly, $\Delta^{\text{call}}_{t}-\Delta^{\text{put}}_{t}=1$. This relationship ...

3

Using the answer from: Chris Taylor, on math stackexchange (link): Let the price of an option at strike $K$ be given by $V(K)$. To say that the price is convex in the strike means that $$V(K-\delta) + V(K+\delta) > 2 V(K)$$ for all $K>0$ and $\delta>0$. Let's assume that the opposite is true, i.e. that there exist tradeable option contracts ...

3

For the case where $\sum_{i=1}^n \lambda_i =1$, you need only note that the payoff \begin{align*} (x-K)^+ \end{align*} is a convex function in $x$. That is, \begin{align*} \Big(\sum_{i=1}^n \lambda_i S_i -K\Big)^+ \le \sum_{i=1}^n\lambda_i(S_i-K)^+. \end{align*} Then \begin{align*} e^{-rT}E\left(\Big(\sum_{i=1}^n \lambda_i S_i -K\Big)^+\right) \le ...

3

For the US market nearly all options on securities are american whereas the options on indexes are european. What you can do is to use a database such as OptionMetrics which adjusts the stock american options to european options by taking into account the early exercise premium.

3

you just add in any auxiliary variables accumulated along the path that determine the pay-off to the regression variables. So path-dependence is not a problem. If you have previous decisions, you may need to do different regressions based on their possible values or make them into a continuous variables that can be used for regression.

3

Options are in zero net supply (like futures and other derivatives), so for evry long there is a short and for every short there is a long. The open interest is the sum of the longs which also equals the sum of the short positions.

2

You should not use the Feller condition as a constraint. In many cases its violation will be required for a good fit to the market data.

2

You can guesstimate by vega weighted implied vol. This is why: Say that you have a portfolio of options with prices $P_j$. Each one of them has a different pricing function $f_j$ (as function of vol) and a different implied vol $\sigma_j$. For each option $f_j(\sigma_j)=P_j$. Now you put them together in a single product. If the implied vol of the product ...

2

The easiest way to think of this is as follows: Settlement Price - Price at which the exchange margins all accounts for those options. Closing Price - Mid/Bid/Ask of Active Market at the exchanges last trade time. E.g. for TY Contracts this is at 5pm EST vs. a Settle Time of 3pm EST. Last Trade Price - Not all options trade every day. This is the price ...

1

For question a). From the assumptions, in particular, that $R=0$, \begin{align*} \pi_l + \pi_m + \pi_u &=1\\ \frac{1}{2}\pi_l + \pi_m + 2\pi_u&=1. \end{align*} Set $\pi_m=x$, and solve for $\pi_l$ and $\pi_u$, \begin{align*} \pi_l &= \frac{2}{3}(1-x)\\ \pi_m &= x\\ \pi_u &= \frac{1}{3}(1-x), \end{align*} where $0<x<1$. The option ...

1

The expected value of the option at maturity is simply $$\mathbb{E}[(S_T-K)^+]$$ Note that this is under the real world measure. In a B-S framework this value is given by $$e^{rT}C(\alpha;S_0, K, \sigma, T)$$ Where $C(r; S_0, K, \sigma, T)$ is the B-S call option price. Hence the expected growth rate (using a simple return) is \frac{e^{rT}C(\alpha;S_0, ...

1

OP is absolutely right in his approach and this is the underlying idea behind risk neutral valuation or even BS model. If Black-Scholes model assumptions hold, then a derivative payoff can always be replicated in such a way it would never provide return more than risk free interest rate, otherwise it will lead to arbitrage opportunities. But assumption never ...

1

it is not a "newbie question". actually it is very deep question, especially if you are trying to trade options in some way. all Greeks are about change in premium of the option. Premium consists of two components: time value + intrinsic. Greeks cannot tell you anything about this decomposition, they deal only with total sum (i.e. with overall premium). So ...

1

Note that, for $K_1 < K < K_2$, \begin{align*} -(K_2-K_1) \le (S_T-K_2)^+ - (S_T-K_1)^+ \le 0. \end{align*} Taking the conditional expectation with respect to information set $\mathcal{F}_t$, \begin{align*} -(K_2-K_1)B_t(T) \le C(T, K_2, S, t) - C(T, K_1, S, t) \le 0. \end{align*} That is, \begin{align*} -B_t(T) \le \frac{C(T, K_2, S, t) - C(T, K_1, ...

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