New answers tagged black-scholes
1
So to expand a bit further on what Brian had mentioned, you're going to get a different vol surface given american vs european. So this is something Brian already pointed out, but one very simple and practical way that you can prove this to yourself is just to think about how the implied forwards are generated.
In the European case we use the entire strip ...
3
For a standard American exercise option expiring at $T>0$, price is still monotically increasing in volatility under the Black-Scholes model (though obviously it is not strictly monotonic, due to early exercise rendering price insensitive to volatility in some regions of parameter space).
To see this, you can use one of three techniques:
Investigate ...
2
As Christian notes, under the Black-Scholes model standard european options have prices that are monotonic in volatility.
You can see that binary options do not share this property but I suspect you are correct about convex payoffs.
4
If you modify your question to "European Call and Put under a Black-Scholes Model" the answer is: yes.
It's trivial to verify it from the formula $S e^{d_1} \sqrt{T-t}$.
For a general payoff the question is more difficult to answer. In general vega will not be positive.
I believe that you can derive some conditions on the payoff assuming a Black-Scholes ...
0
the convention for most market makers of options is to use calender time. it is also the convention in data one gets from options data vendors. another way to see this is that market makers will fade their bids near the close on a 'normal' friday for fear of holding inventory ahead of the weekend, since it is a crap shoot whether the pick-up in vols on ...
1
You can refer to Shreve's book, Volume II, Section 4.4.3 .
Assume that we have a generalized geometric Brownian motion
$$dX_t = \sigma_t dW_t + (\alpha_t - \frac{1}{2} \sigma_t^2) dt ,$$
where the drift coefficient and the volatility are functions of $t$ also. $(dX_t)^2 = \sigma_t^2 dt + \mathcal{O}(dt^{3/2})$ .
Assume that the asset price is
$$ S_t = S_0 ...
-2
under the REAL WORLD probability (measure), it is as you think: the log is μ△t
under the RISK NEUTRAL measure, the mean is changed, so the stock price is a martingale, and the mean is (μ−1/2 σ^2)△t
2
SRXX has talked about Intrest Rate Collar. Since it is not clear if you are looking for IR or equity here is my explaination of equity Collar
Equity Collar :-
Structure
:- Buy Underlying Asset (e.g. Stock) and Buy an out of money put and write out of money call
Payoff daigram
Replication COLLAR = long stock + long put (K1) + short call (K2)
As ...
1
Yep, there is one and it's leagues better than jquantlib.
https://code.google.com/p/maygard/
2
Your question lacks a bit of background to make sure that you are using the right terminology.
In short, you buy an interest rate collar to hedge exposure in rates when they get out of a zone. As you can see on the wiki page when you buy a collar, you essentially:
Buy an interest rate cap with strike price $K_c$
Sell an interest rate floor with strike ...
3
I remember this discussion here: http://www.wilmott.com/messageview.cfm?catid=3&threadid=62227
You should absolutely match your convention for time to expiration to the convention you used for calculating volatility. There seems to be other ways to proceed, as modifying the volatility to match your convention but I don't really see the point in using ...
2
Another way of seeing it is that the $-\frac12\sigma^2$ is just a correction term that comes from Jensen's inequality.
You need this when switching from supposedly symmetric returns (normal distribution) to the skewed price process (log-normal distribution).
3
The term 1/2 * sigma-squared arises through the application of Ito's Lemma. Keep in mind that the assumption is of a stock price that follows geometric BM with a constant drift and volatility. If you set up a delta-hedge portfolio and apply Ito calculus you will end up with an adjustment in the distribution by exactly above term. Another way of interpreting ...
6
So we have the BS-Model
$$dS_t=S_t(\mu dt +\sigma dW_t)$$
W.l.o.g we assume $S_0=1$. Itô's lemma implies that
$$S_t=\exp{(\sigma W_t+(\mu-\frac{1}{2}\sigma^2)t)}$$
We know that $W_t$ is normally distributed with mean $0$ and variance $t$. Now have a look at the r.v.
$$X_t=\sigma W_t+(\mu-\frac{1}{2}\sigma^2)t$$
$\sigma W_t$ is the random part and ...
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