pricing of european put option

[Lun]

let's assume the spot price S=0
I find that longer the time to maturity (TTM), lower the option price (S is still zero, I don't change it)

However, with more time, there should be more time value, the option should be more expensive, it contridicts with the fact I stated above. There is no such a contridiction for call option.

Can anybody points out what's wrong ?

Thanks

 

[Josu]

Hi Lun.

Let's see. If the spot price is 0, then it will be always 0 (think of when it does happen spot price =0 ) so the payoff at maturity T of the option will always be K (strike).
Now, as the cash flow at maturity is known, the value of the option is just the present value of K, let's say, exp(-r*T)*K.
Then you can see what's happening there: the longer T, the smaller the discount term exp(-r*T), then , obviously, the smaller the price of the put.

Does it make sense?

 

[Lun]

Well, let's have a discussion.

If your logic is correct, try to apply it on a call option.

Keep S = 2K

the payoff at maturity T of the option will always be K (strike).

Now, as the cash flow at maturity is known, the value of the option is just the present value of K, let's say, exp(-r*T)*K.
Then you can see what's happening there: the longer T, the smaller the discount term exp(-r*T), then , obviously, the smaller the price of the CALL.

However, the fact is that the opposite happens. Can you explain ?

 

[tinoob]

Hi Lun
The other aspect of your question might be the answer.

I think it all depends on if the Put is ITM, OTM, ATM . If the put is ITM, there is very no Time Value premium and most of this is Intrinsic Value and it is highly volatile and hence high Premium. as Volatility is propotional to Premium.

thanks

 

[Lun]

Hi Tinoob,

Can you explain further ? Sorry, I can't catch what you mean.

 

[Josu]

Well, let's have a discussion.

If your logic is correct, try to apply it on a call option.

Keep S = 2K

the payoff at maturity T of the option will always be K (strike).

Now, as the cash flow at maturity is known, the value of the option is just the present value of K, let's say, exp(-r*T)*K.
Then you can see what's happening there: the longer T, the smaller the discount term exp(-r*T), then , obviously, the smaller the price of the CALL.

However, the fact is that the opposite happens. Can you explain ?

Lun, in this case is not the same logic.
For the put, a 0 spot price in the underlying is going to be 0 all the time.
However, in the case of the call option, you can't say "let's keep S=2K"; you don't know that when you purchase a call option with S=2K initially. The price will fluctuate for sure. And in that case, it's true the longer the expiry date T, the higher the call price. ( despite S(0)=K, S(T) is not necesarily K, so the payoff can be different than K; higher, lower, or 0)

 

[Lun]

if zero is such a special number, let's play a real case.

K = 100,
keep S = 1 or 5 for put
keep S = 200 for call

As TTM is larger, the price for put is lower

I'm using a numercial method to study the relationship between option price & time, I have to assume that the spot price keeps constant. I mean, I plot the curve "option price against spot" with different time, and then observe the relationship.

 

[Josu]

Ok, I understand what you say.
It all has to do with the dynamic model used for the underlying in the B-S model; the geometric Brownian motion.
dS(t)=r*S(t)dt+s*S(t)*dW(t), where s is the volatility, and r the interest rate. That's the stock dynamic SDE used in the Black-Scholes model, under the risk neutral measure.
(1 comment: 0 was special spot price because, as you can see, once S reaches 0, dS(t)=0 all the time, and S(t)=0, for all t)
Now, regarding the question: the geometric Brownian motion paths have an increasing tendency (or they have so low probability of falling under S(0), more precisely) If you draw them, you will see it clearly. As you increase t, you will see how S(t) grows , given an initial spot price S (which is always the same for you)
Then, as you increase expiry time, S(t) will be higher and then puts will cost less (payoff is decreasing when St increases; it is K-S(t) )
On the other hand, for calls, it happens the opposite.

The model of the underlying is assuming an increase of the stock (though non.determistic), so that should help to understand the logic behind put& call prices when chaging T

 

[rajanS]

if zero is such a special number, let's play a real case.

K = 100,
keep S = 1 or 5 for put
keep S = 200 for call

As TTM is larger, the price for put is lower

I'm using a numercial method to study the relationship between option price & time, I have to assume that the spot price keeps constant. I mean, I plot the curve "option price against spot" with different time, and then observe the relationship.

no. by law of one price, if the stock price is 0. it will never go up. if it does, there is automatically an arbitrage. i.e, get it for free and sell it immediately as soon as its price goes up. if you are assuming spot price to be constant, you should use a value other than 0 for it to make any sense. infact, a 0 spot price means that call price is 0 as well because N(d1) = 0. and N(2) = 0.

 

[abcdefgh]

pricing of european put option let's assume the spot price S=0
I find that longer the time to maturity (TTM), lower the option price (S is still zero, I don't change it)

However, with more time, there should be more time value, the option should be more expensive, it contridicts with the fact I stated above. There is no such a contridiction for call option.

Can anybody points out what's wrong ?

Thanks

I think Tinoob has the right approach. Its a model independent fact. In this case you have the stock at a point where the payoff is maximum. (max(K-S,0) ) is at its peak. Since S can only go up, this value has only one way to go, that is down. So more the time you leave it, more it will hurt your payoff hence you should pay a lower price for it. Less the time you leave, lower is the probability that it will move upper than 0 so you pay more than if you had more time. That is the time value of money here, unfortunately negative.

 

[Lun]

Sorry for late reply, I think I find the answer for myself. The reason for me to ask about spot price = 0 is that I'm using finite difference method, that's why I ask about zero. The fact is that it's not just the case for zero, and the answer is on ITM/ATM/OTM. When it's ITM, intrinsic value >>>> time value, so even TTM is larger, the effect of time value is small. With discounting value, larger TTM, the option price is smaller. When it's OTM, time value dominates, so larger TTM, higher the option price.

However, my next question is why this logic cannot apply to call option. I mean, if we apply the same logic, when it's ITM, intrinsic value >>>> time value, so even TTM is larger, the effect of time value is small. With discounting value, larger TTM, the option price is smaller. The fact is that, for call, the reverse happens, larger the TTM, higher the option price.

 

[koupparis]

Your stock process is a Brownian motion with drift. S goes up on average. Hence the longer the TTM the more expensive the call is (the call has more time to expire ITM) and the less expensive the put is (the put has more time to expire OTM).