3
$\begingroup$

Basic question to which I surprisingly did not find an answer on here.

What's the best approximation to the break-even (with respect to stock price) for an option that was hedged fully at point of trade (and not adjusted later). (by fully I mean however many deltas the option had - not 100 deltas per option)

I have seen the approximation Premium/Delta and $\sigma * (\Delta t)^{0.5}$.

Are these model specific?

Am I incorrect to have thought that break-evens should be slightly asymmetric (the effect of charm will change the delta of the option leg, but the static stock hedge will not adjust)?

Any sources about delta hedged options are also appreciated.

|improve this question
$\endgroup$
  • $\begingroup$ I don't understand what you mean with break-even. What variable are you targeting with break-even, the stock price at expiry? $\endgroup$ – luckylwk Oct 9 '13 at 17:39
  • 1
    $\begingroup$ yes, that's what I was trying to say with "wrt. stock price" $\endgroup$ – frickskit Oct 9 '13 at 19:15
3
$\begingroup$

Lets give it a rough go then.

Two assumptions. (1) We disregard repo (to lend the stock you may want to short) or financing on your hedged position. And (2) We assume no trading of the gamma on the option.

Then I would assume the break-even is equal to the expiry should be equal to... (CALL) paidPremium/(1-hedgedDelta) + callStrike (PUT) putStrike - paidPremium/(1-hedgedDelta)

Assume you pay 2.5 USD for a ATM CALL that is hedged ATM with delta 52%. You only make money if on expiry the spot it above your strike and you then only make money on (100-52) 48% delta that you have run. So you need to make 2.5 USD with 48% delta (is 5.21 USD per 'unhedged' delta). Which puts the breakeven-spot at 105.21?

The other way around you can find it by dividing the premium by the hedged delta and take the hedge-level minus that value (for a CALL). The your short hedge has paid for the premium.

(sorry, i am not familiar with including formula's)

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.