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stilyo
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Vol and T coupling in Derivative Price

August 6th, 2019, 4:39 am

Hi,

Suppose that a stock price follows standard GBM process with constant risk-free rate and constant vol. Suppose we also have some derivative with unknown payoff P(.) at expiration T which depends in some general way on the stock price process (it could be just ending stock price or path dependent or anything). Does vol and T always appear “coupled” as vol*sqrt(T) in the derivative price (which may or may not exist in an analytical closed-form)? Said differently, if I reduce vol by 0.5x and increase T by 4x will I always get the same derivative price and if not, under what conditions about the payoff function? Thanks for your input.
 
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bearish
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Re: Vol and T coupling in Derivative Price

August 6th, 2019, 11:16 am

That’s an interesting question. The simple answer is no. Consider e.g. a range accrual type payoff where you get $1 per day until maturity that the stock price is greater than K. That can clearly be worth more in your long T scenario. For payoffs defined strictly by some function of S(T) you should be good up to discounting, which would still make T matter. But in this case I believe the time T forward value would be the same. Not sure how much more general we can get.
 
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bearish
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Re: Vol and T coupling in Derivative Price

August 6th, 2019, 1:19 pm

Actually, on a little further reflection, there is always going to be an issue with a carry term and/or discounting causing differences in the present value of a claim, even if (as will often be the case) you have [$] \sigma [$] only appearing together with [$] \sqrt{T} [$].
 
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stilyo
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Re: Vol and T coupling in Derivative Price

August 6th, 2019, 4:53 pm

Agreed and thanks for pointing that out, I had overlooked the discounting so that clearly doesn't hold in the examples you provided or Black-Scholes where you have the K*exp(-r*T) term.

But if the payoff doesn't depend on the money market account and can be expressed only in terms of the stochastic forward price F(t), then that "coupling" between vol and sqrt(T) should hold no matter how complicated the payoff might be, right? That's my intuition but not sure if I can formally prove it to fully convince myself. 
 
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Alan
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Re: Vol and T coupling in Derivative Price

August 6th, 2019, 7:07 pm

If the derivative value is just the expectation of an arbitrary terminal payoff, then 'yes'. That follows from writing the value as an integral over a normal density with variance [$]\sigma^2 T[$] (times the payoff, of course). In other words, it's exactly the same normal density you would use to value vanilla puts/calls (with another factor of [$]\sigma^2 T[$] in the mean).
 
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stilyo
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Re: Vol and T coupling in Derivative Price

August 6th, 2019, 11:59 pm

Not sure I follow entirely, Alan - if we are looking at a lookback option for instance with payoff Max[F(0),...,F(T)]-F(T), then the derivative depends on the joint distribution of the maximum and the forward price. Is it obvious that the derivative price depends on vol*sqrt(T) (which is the case since there is a closed-form solution)? Same with geometric asian option as well.
 
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Alan
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Re: Vol and T coupling in Derivative Price

August 7th, 2019, 3:50 am

I will make a guess. If the payoffs are level invariant as in your examples, and you have 'continuous monitoring', and no cost-of-carry parameters, then what you want might follow simply from dimensional analysis of the value function. After all, you need dimensionless factors. But once you introduce discrete-time monitoring at spacings [$]\Delta t[$], then potentially dimensionless factors like [$]\sigma^2 \Delta t[$] and [$]\sigma^2 T[$] could both appear, destroying the argument. Like I said -- a guess.  
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