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jonasre
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Joined: June 22nd, 2015, 6:19 pm

Delta Hedging Volatility Breakeven

July 16th, 2017, 1:32 pm

Hello everybody,

I am trying to derive the breakeven move for a delta hedged (Black Scholes) portfolio and refer to http://www.codeandfinance.com/option-break-even.html.

I am wondering though what happens to the sigma squared from Ito's Lemma that is included (within the product that includes gamma) when looking at a delta hedged portfolio. At which point does this drop out?

Thank you guys in advance for help
 
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frolloos
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Location: Netherlands

Re: Delta Hedging Volatility Breakeven

July 17th, 2017, 3:46 pm

The break-even volatility is the implied volatility.

Not sure what you mean with "when it drops out"
 
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jonasre
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Joined: June 22nd, 2015, 6:19 pm

Re: Delta Hedging Volatility Breakeven

July 20th, 2017, 11:02 am

When I take the Black scholes equation and set Delta = 0 then i am left with

-Theta + 0.5 * Gamma * S^2 * sigma^2 = 0
0.5 * Gamma * S^2 * sigma^2 = Theta
0.5 * S^2 * sigma^2  = Theta / Gamma
S^2 * sigma^2 = 2 * Theta / Gamma
Sqr(2Theta / Gamma) = S * sigma

...which is equal to http://www.codeandfinance.com/option-break-even.html

I am just wondering if this is the proof for sigma * Stock price being the break even move? Sorry I have expressed myself in a very weird way earlier.
 
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frolloos
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Location: Netherlands

Re: Delta Hedging Volatility Breakeven

July 20th, 2017, 3:34 pm

Suppose you buy a call and delta hedge it (by selling the underlying) and put the proceeds of long call and short stock on a bank account.

Your initial portfolio is then

[$] \Pi = C - \Delta S + (\Delta S - C) = 0 [$]

If you only delta hedge and keep other variables/parameters constant, then the change in your portfolio value (marked-to-model) after a time dt is

[$] d \Pi = \Theta dt + \frac{1}{2} \Gamma dS^2 + r (\Delta S - C)dt [$]

Assuming you canwrite [$] dS^2 = \sigma^2 S^2 dt [$] (no jumps_, then the break-even move is that value of [$] dS^2 [$] such that [$] d\Pi =0 [$]. In other words when

[$] \Theta = - \frac{1}{2} \Gamma \sigma^2 S^2 - r (\Delta S - C) [$]

Lo and behold, this is the Black-Scholes equation... But not really, since the BS equation tells you, assuming you know the vol to start with, how to solve for your option price and greeks, and this equation tells you if you know your gamma and delta and theta (from a mis-speicfied Black Scholes model) what should be the vol / daily spot move to balance - which is obviously the vol you used initially for solving Black-Scholes. The next question should be so what happens if my daily move is not my implied volatility.

Anyway, I hope you're really confused now and decide to explore further what happens if you hedge with a mis-specified model.

To summarize, assumptions to arrive at the expression in the link (which oversimplifies things):

- you're only letting [$]S[$] vary and (hence) only delta hedging
- r = 0
 
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jonasre
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Joined: June 22nd, 2015, 6:19 pm

Re: Delta Hedging Volatility Breakeven

July 20th, 2017, 10:23 pm

Wow thank you very much for your effort!

First one technicality: Shouldn't dS^2 = S^2 * sigma^2 * dz [and not dt] - I guess what you did is assuming no drift rate?

I am aware that this is a henn-egg problem. We are solving for something that we used as input. Still, this seems to be an at least "rough" approximation (although it might not be fully correct academically)?

The interest rates I have obviously neglected. I am reading more about jump diffusion as well, so I hope I am getting closer. Do you have any other keywords so I'd have a starting point where to look to find non misspecified models?

Thank you in advance. Really appreciate your help!!
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