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akki
Topic Author
Posts: 21
Joined: September 10th, 2008, 3:05 am

### Price of a one day straddle

Hi,Suppose the stock is S today and tomorrow it can go to only 2 states. In state 1, it is S(1+ \sigma) and in state 2, it is S(1 - \sigma). So then one would expect the price of a one day straddle (assuming r = 0) to be S*(\sigma) but it turns out to be less than that, it is actually .8*S*(\sigma). Why ? I guess it related to Jensen's inequality but I would appreciate if someone can throw some light on boththe mathematical aspect and the economic reasons. Tks!

Alkmene
Posts: 301
Joined: January 18th, 2007, 10:19 pm

### Price of a one day straddle

What?

daveangel
Posts: 17031
Joined: October 20th, 2003, 4:05 pm

### Price of a one day straddle

i dont think that's right. the straddle in a world where the stock can either go to S(1+h) or S(1-h) from S is worht hS. You must be thinking of a continuum of possible prices between S(1-h) and S(1+h).
knowledge comes, wisdom lingers

DocToc
Posts: 239
Joined: January 20th, 2010, 9:32 am

### Price of a one day straddle

the 0.8 comes from the normality assumption, sqrt(2/pi) is approx 0.8, ill give you some time to think about where this comes from before spilling the beans - its nothing to do with Jensens inequality. Think in the line of Taylor Series f(x+h) = f(x) + f'(x)dh + 0.5*f''(x) (dh)^2+... where f'(x) is the first derivative of f wrt x and f''(x) is the 2nd deriv wrt x.(And i think Dave Angel is right - so in the case where you have a continuous price process then the above will hold given normality).
Last edited by DocToc on January 9th, 2011, 11:00 pm, edited 1 time in total.

akki
Topic Author
Posts: 21
Joined: September 10th, 2008, 3:05 am

### Price of a one day straddle

Tks Dev Angle and DocToc. I see how a taylor expansion of the normal cdf gives a factor of 0.8. Yes, Dave Angel is correct, it has to be a continuous price process. If you have a continuum of possible prices between S(1-h) and S(1+h), What would the expected payoff be? I was expecting it to be S x h x sqrt(T), but since the price turns out to have a factor of 0.8, there seems to be some disparity between the payoff and the price.

daveangel
Posts: 17031
Joined: October 20th, 2003, 4:05 pm

### Price of a one day straddle

i dont think the expected payoff is S h sqrt(T)... this is just the one sd move in the underlying. the expected payoff is the value of the option.
knowledge comes, wisdom lingers

DocToc
Posts: 239
Joined: January 20th, 2010, 9:32 am

### Price of a one day straddle

i haven't done this for a very long time BUT:From what I remember about risk neutral probability distns, we should have (assuming r = 0) p(up move) = 1/2 and p(down move) = 1/2...Assume K = S then we have that the call option is atm. Therefore in the discrete case:E(Payoff_Call) = 0.5*[ S(1 + h) - S] = 0.5*hE(Payoff_Put) = 0.5* [ S - S (1 - h) ] = 0.5*hTherefore today the price of the straddle is 'h' - which is what you'd expect i.e. the volatility as the price of a straddle atm should be more or less linear in volatility - its a vol trade. Thats my 2cents anyway - might be wrong did some of this stuff too long ago.DocToc
Last edited by DocToc on January 11th, 2011, 11:00 pm, edited 1 time in total.

DocToc
Posts: 239
Joined: January 20th, 2010, 9:32 am

### Price of a one day straddle

By assuming that the diffusion is between S(1+h) and S(1-h) you're making things quite complicated to do mathematically. In the black scholes world you'd assume S is with [0,\infty]