Option Price vs. Volatility question

[Uncle Max]

I was drawing some graphs for the plain vanilla european options. Here is one of the graphs I got:

EUROPEAN CALL

K=$50, T = 1 year, r=5%

OTM: S(0)=$40
ATM: S(0)=$50
ITM: S(0)=$60

The question is why ATM option has value when volatility sigma=0 ? I mean what is the financial interpretation of this fact? However, similar graph for the ATM European Put starts from 0.

EUROPEAN PUT
S(0)=$50, K=$50, T=1, r=5%

 

[naeron]

Hi Max,
I guess if you take 1 day binomial model and try to find the price of the option there will still be a value. May be this is happening due to the martingale property of the stock price.
But by this logic puts should not be zero. I am not sure .. If you find an explanation lemme know

 

[pardasani]

The question is why ATM option has value when volatility sigma=0 ? I mean what is the financial interpretation of this fact? However, similar graph for the ATM European Put starts from 0.

Probably a question with the same answer is :-
"Why is the value of a ITM option something like $12.5 at sigma ~0 " --- I would expect it to be something like 10$ (10$ discounted at 5% for 1 year) because if sigma is 0 the stop is expected to remain somewhere close to 60$...

also is it just coincidental that the ATM is off by 2.5$ --- the same by which the ITM is off

btw max what method did you use to calculate the option price :-- the answer may have to do with that...

 

[Uncle Max]

Hi Max,
I guess if you take 1 day binomial model and try to find the price of the option there will still be a value. May be this is happening due to the martingale property of the stock price.
But by this logic puts should not be zero. I am not sure .. If you find an explanation lemme know

I used Black-Scholes to price those options. I can see some problems with vol=0 when computing d1 and d2 (numerically). But I don't understand why someone would pay $2.40 for this option which will never generate any payoff (financially). And why Put gives more reasonable price?

 

[Uncle Max]

Probably a question with the same answer is :-
"Why is the value of a ITM option something like $12.5 at sigma ~0 " --- I would expect it to be something like 10$ (10$ discounted at 5% for 1 year) because if sigma is 0 the stop is expected to remain somewhere close to 60$...

If there is no volatility of underlying the price at maturity of this underlying should be the same as price at time 0.

price of the option = intrinsic value + time value

If volatility is zero, time value should be equal to zero as well. Thus, we have only intrinsic value left (the amount we would receive if option exercised right now). So in case of ITM option, this value is $10.

 

[pardasani]

I used Black-Scholes to price those options. I can see some problems with vol=0 when computing d1 and d2 (numerically).

couple of points i guess:--

1) Max do you use your own C++ code for Black Scholes? --- I just verified your interesting post with Hull's DerivaGem software...... although deriva Gem does not let you calculate the price with vol<0.25 you can graph out the price vs vol till vol=0.05% and the results were similar..... ATM coverges to about 2.5, ITM coverges to about 12.5...

2) I think you may have answered the question when you observed that there are problems in calculating d1 and d2 numerically.... For instance, when sigma = 0, then the stock price is not random at all, and the price should always be equal to the intrinsic premium....

3) another interesting thing for me is that the variation of stock price with vol is more or less linear

 

[pardasani]

So in case of ITM option, this value is $10.

completely agree with you..... However in the graph you have drawn, and also in Hull's deriva gem this value is about 12.5 !!!!... is it just coincidental that the "error" is about 2.5$???

 

[Uncle Max]

1) Max do you use your own C++ code for Black Scholes? ---
2) I think you may have answered the question when you observed that there are problems in calculating d1 and d2 numerically....
3) another interesting thing for me is that the variation of stock price with vol is more or less linear

1) I used C++ code from the summer refresher course, which uses the same approximation formulas as in Hull's book. I checked answers with a couple of Black-Scholes web-pricers.

2) Why problems with computation of d1 and d2 doesn't affect Put option?

3) Indeed, this is a very interesting observation. Now, if we admit linearity, comes next question: what is a slope of this line? How do we compute it?

 

[pardasani]

Here is another wierd case... ITM put...

s(0) = 40, K=50, r=5%, T=1yr...... When vol ~0, you would expect the price to be around 10$, but it seems that it converges to about 7.5$, with an error very conveniently being $2.5

 

[dstefan]

Volatility zero menas that the price of the underlying is known without any uncertainty at any future time. But, if iunterest rate is not zero, the risk neutral value of the underlying will increase at the rate r.

The price of the underlying at maturity is \(S(0) e^{r T} = K e^{rT}\). The value of the ATM call at maturity is then \(C(T) = S(T) - K = K e^{r T} - K\). Discount to present and get \(C(0) = K(1 - e^{-r T}) \approx K r T\).

This works out to about 2.5 for ATM with strike 50, at interest rate 5%, for a one year option.

The same argument will give the price of the put.