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# 1/2 the option price, according to BS

Discussion in 'Options' started by heech, Jun 8, 2009.

1. ### heech

Hi all,

I'm just looking to a quick short-cut from someone that might know the answer.

According to Black-Scholes...

- If an ATM option is \$X

- How far out of the money would an option need to be, for it to cost X/2?

Other than trying to tackle the PDE, does anyone have a simple spreadsheet or other analytic tool that could give me an answer?

(I assume the answer will have standard BS parameters: sigma, t... )

2. ### nazzdack

1) ?......"It depends"........on the volatility skew of what you're trading.
2) The "theta" would also have influence depending on when you initiate a position.

3. ### Martinghoul

What does this have to do with skew and theta? This is a theoretical question about Black-Scholes, which assumes constant vol.

The answer is that it depends on the inputs (sigma, r, etc). Do you have access to Excel? If so, you need not be bovvered with the PDE and will just need to implement the trivial BS formula (trust me, it's a no-brainer). Once you have done that, you can play with the inputs and answer the question yourself.

4. ### spindr0

The correct answer is, "It depends" which can also be translated into "There's no short cut."

How much of a move OTM will be necessary to halve the premium will depend on the IV (not change in IV but how high when initiated), the time remaining until expiration and yes, volatility skew, if any. You can eliminate volatility skew with an assumption of constant volatility.

IOW, all factors being equal, the higher the IV of the option, the more the underlying will have to move to halve. OTOH, if it's time you're looking at time as the variable, the closer to expiration, the less the underlying will have to move. And due to the different premium retention aspects of puts and calls, the amount will not be equidistant from the strike. And you said that you wanted a short cut for multiple moving targets??

The only thing that will be linear will be the price of the stock vis a vis price at onset. IOW, the pct of price move for a halving of premium for a \$50 stock will be the same as that for a \$100 stock, all other variables being equal.

5. ### spindr0

If the theta is different, the amount of the move needed will be different, You can verify this for yourself by just using that no-brainer trivial BS formula

6. ### Martinghoul

Huh? Theta is not an input. It's one of the partials. Last I heard, option prices in BS are not defined in terms of theta.

Thus, what the answer depends on is the exogenous inputs, i.e. time to expiry, vol etc, as I have mentioned already. Those inputs also happen to define the theta. To me saying that theta defines anything is completely wrong in the classical BS context.

7. ### spindr0

Yep, in the strict definition of BS, you're absolutely correct. Theta is a derivative and in order to have a different theta, the BS inputs must be different. Ergo, if the theta is different, the answer to the OP's question will be diffferent, eg. a shortcut number for the amount of price move needed to halve the option's premium.

8. ### Martinghoul

Yep, we agree 100%...

I am just going on the assumption that the OP's original question is sort of a Black-Scholes brainteaser, rather than a question about a real-world position. It just sounds like an interview question to me, which is why I am attempting to respond to it in the same vein.

9. ### heech

No, not an interview question, real world question. I'm trying to decide a convenient place to put on a hedge. Just trying to figure out if there's a simple back-of-envelope calculation...

But yes, I do have Excel-based BS, and I can always go at it backwards by adjusting inputs to get the right value. Thanks.

10. ### Kevin Schmit

A very crude approximation:

ATM_Strike * (1 +- (sqrt(time-to-expiry-in-years) * sigma/2))

This will slightly underestimate the call and also the put strike. Which under normal skew is ok.

#10     Jun 8, 2009
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