Can't figure out a basic question

[LucasX]

when the underlying is exactly between two strikes, using the same implied volatility for all strikes produces an OTM call has greater value than OTM puts with equal proximity to the money.

Why?

Is there any assumption on the underlying distribution associate with this argument like normal dist?

Thanks guys!

 

[spindr0]

If the assumption is that price is normally distributed then the options equidistant from strike should have the same value. I believe that Black Scholes assume a lognormal distribution so higher equidistant strikes are priced higher than the lower ones.

 

[LucasX]

Quote from spindr0:

If the assumption is that price is normally distributed then the options equidistant from strike should have the same value. I believe that Black Scholes assume a lognormal distribution so higher equidistant strikes are priced higher than the lower ones.

I think you are right. Could you further explain why higher strike valid a higher option value? It seems to me with lognormal dist, there is a lower chance for underlying rallying to the higher strike than to the lower one. Can't convince myself.

 

[wayneL]

There would be a component for cost of carry as well

 

[Martinghoul]

Quote from LucasX:

when the underlying is exactly between two strikes, using the same implied volatility for all strikes produces an OTM call has greater value than OTM puts with equal proximity to the money.

Why?

Is there any assumption on the underlying distribution associate with this argument like normal dist?

Thanks guys!

If I am not mistaken, this is a classic brainteaser question...

Is there a cost of carry in this example of yours?

EDIT: Double post with wayne.

 

[spindr0]

Quote from LucasX:

I think you are right. Could you further explain why higher strike valid a higher option value? It seems to me with lognormal dist, there is a lower chance for underlying rallying to the higher strike than to the lower one. Can't convince myself.

If the cost of carry is zero and it's a normal distribution, equidistant OTM strikes will have equal values.

If the cost of carry is zero and it's a lognormal distribution, equidistant OTM strikes will NOT have equal values. Lognormal permits greater upside than downside. You've seen the pricing. Accept it.

Can you guess what will happen if there's a cost of carry?

 

[LucasX]

Quote from spindr0:

If the cost of carry is zero and it's a normal distribution, equidistant OTM strikes will have equal values.

If the cost of carry is zero and it's a lognormal distribution, equidistant OTM strikes will NOT have equal values. Lognormal permits greater upside than downside. You've seen the pricing. Accept it.

Can you guess what will happen if there's a cost of carry?

A: Calls are higher and puts are lower.

Thanks spindr0. Your explanation is very clear and makes sense to me.

 

[nitro]

Quote from LucasX:

when the underlying is exactly between two strikes, using the same implied volatility for all strikes produces an OTM call has greater value than OTM puts with equal proximity to the money.

Why?

Is there any assumption on the underlying distribution associate with this argument like normal dist?

Thanks guys!

It's called skew by traders. That there should be a skew or a smile can be argued mathematically. People get confused because there is only one underlying, and if IV is an (mean) estimate of realized vol at expiration, then it doesn't make sense because there is only one underlying and all options expiring on the same day should have the same estimate of realized vol. But if you think of it from a risk point of view and go to the options side, it makes sense that different options expiring on the same day have different IV.

Why the skew takes the shape or the level it does, no one really knows.

 

[spindr0]

Quote from nitro:

It's called skew by traders. That there should be a skew or a smile can be argued mathematically. People get confused because there is only one underlying, and if IV is an (mean) estimate of realized vol at expiration, then it doesn't make sense because there is only one underlying and all options expiring on the same day should have the same estimate of realized vol. But if you think of it from a risk point of view and go to the options side, it makes sense that different options expiring on the same day have different IV. Why the skew takes the shape or the level it does, no one really knows.

Isn't that a function of the real world versus the thoeretical? I inferred from the OP's question that he was looking at a pricing model.

 

[Martinghoul]

Quote from nitro:

It's called skew by traders. That there should be a skew or a smile can be argued mathematically. People get confused because there is only one underlying, and if IV is an (mean) estimate of realized vol at expiration, then it doesn't make sense because there is only one underlying and all options expiring on the same day should have the same estimate of realized vol. But if you think of it from a risk point of view and go to the options side, it makes sense that different options expiring on the same day have different IV.

Why the skew takes the shape or the level it does, no one really knows.

The OP's question has NOTHING to do with skew, as it's a question about option pricing in a B-S world, where vol is constant, by construction. It has everything to do with the cost of carry.