Put price when maturity tends to infinity

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Hi all,

I have a question on how behaves the put price when maturity tends to infinity.

• According to Hull as the time to expiration increases, options become more valuable.

• When I check this with the Black Scholes formula:

Put=Kexp(-rT)N(-d2)-SN(-d1)

It appears that N(-d2) tends to 1; N(-d1) tends to zero, exp(-rT) tends to zero;

I conclude that the put price tends to zero.

Can anyone confirm that my logic is correct and what is the economic sense of conclusion?

Thanks,
Indira

 

[ShaktiRathore]

call=SN(d1)-Kexp(-rT)N(d2)
call=S(ln(S/K)+T*(r+.5sigma^2))/sigma*root(T)-Kexp(-rT)(ln(S/K)+T*(r-.5sigma^2))/sigma*root(T)
call*sigma*root(T)=S(ln(S/K)+T*(r+.5sigma^2))-Kexp(-rT)(ln(S/K)+T*(r-.5sigma^2))
call*sigma*root(T)=S(ln(S/K)-Kexp(-rT)(ln(S/K)+T*(r+.5sigma^2))+T*(r-.5sigma^2))
call*sigma*root(T)=(ln(S/K)(S-Kexp(-rT))+2rT
call=[ln(S/K)/sigma]*(S-Kexp(-rT))+(2r/sigma)T
call=a*(S-Kexp(-rT))+bT where a,b>0 since S>X,r,sigma>0
d(call)/dT=b+aKr*exp(-rT)>0 since b,a>0 as well as Kexp(-rT)>0 which implies that d(call)/dT>0 hence call price of stock is an increasing function of T so as T increases the value of call increases.
On similar lines we can prove that price of put increases as T increases. otherwise
from put call parity p+S=c+K*exp(-rT)
p=c+K*exp(-rT)-S
holding othe values constant other than T,
p increases as c and T increases.
I hope you understand the logic.

thanks

 

[aadityafrm]

Presuming question is for an European put option. As European options can only be exercised at the expiration date and the maximum it can fetch is only K, the strike price at maturity. Hence the the maximum payoff that can happen, when discounted at present time, will be K*exp(-rT). as T->infinity, the value of put tends to zero. For American put options, it may be excercised at any time before expiration, which allows an investor to sell stock at its exercise price.
Hope this answers your query.

 

[[email protected]]

Many thanks again,

Indira

 

[ShaktiRathore]

Indira,
Your logic was not appropriate i think. you were looking at things like N(d1) and exp(-rT) in isolation that is you were considering them separately. Instead of looking inside N(d1) which also has terms in variable T whose effect you were considering in option value you just considered N(d1). You need to combine all the terms in T in the formula to see the combine effect rather than looking the effect of different T terms separately on the option value.Once you consider the combined effect you can better judge the effect of T on option value. However its not easy to see the effect of T on option price with so complex a formula as Black Scholes that is why I used the calculus of differentiation to arrive at the result.
Economically the result says that as more time to maturity is there in the option the probability of stock price showing extreme trends increases as volatility is sigma*root(T) over a period of time so if T increases the total volatility increases which means that probability of upside or downside moves from exercise price increases.which increases the value of call or put option respectively.large shift from exercise price means greater value of the option and large shift is possible only with large volatility.

thanks

 

[Ashwin FRM]

as T->infinity, exp(-rT) ->0
d1, d2 -> +infinity
N(d1)->1 and N(d2)->1
c=S N(d1) -K N(d2) exp(-rT) ->S <== so call will become equal to asset price.
N(-d1)->0 and N(-d2) -> 0 ;
p=K exp(-rT) N(-d2) - S N(-d1) -> 0 <=== so the put will tend to be zero

put call parity holds, as c - p= S - K exp(-rT) -> S - 0 = S - K * 0 or S=S.

Now interesting thing is if the stock pays dividend then value of S gets decreased by the amount of divident in time T, so S exp(-qT) ->0
hence puts and calls both for dividend paying stock become zero.

 

[David Harper CFA FRM]

I agree with missindira's initial solution and aadityafrm's intuition (which also, IMO, comports with Hull's solution to 14.17(g): which asks about a call...).
I extended our BSM in the following worksheet, each column is a different BSM (ATM, OTM, ITM, low vol, high volatility) but all have terms = 1,000 years.

Simulation: https://www.dropbox.com/s/nd2eqgzll86r778/term_equals_1000.xlsx

Please note this "simulation" shows:

• At term = 1000 year, interestingly, I think as Shakti suggests, N(d2) and N(-d2) are variant to volatility changes; i.e., switching 80% vol to 10% changes N(d2) from 0 to 1. However, the discounted strike (K*exp[-rT]) always overwhelms the term to a zero. So the simulation appears to suggest:
• With respect to the put, as T increases:
  • N(-d1) tends to 0
  • exp(-rT) tends to zero
  • N(-d2) tends to either zero or ~1.0 but this doesn't matter:
  • p = [K*exp(-rT) --> 0]*[0 | 1] - S*[N(-d1) --> 1.0] --> 0 - 0 = 0
• With respect to the call, as T increases:
  • N(d1) tends to 1.0
  • exp(-rT) tends to 0
  • N(d2) tends to ~0 or ~1.0, but this doesn't matter:
  • c = S*[N(d1) -->1] - K*[exp(-T) -->0 ]*[0 | 1] = S - 0 = S

I think @aadityafrm's intuition is quite natural: the future payoff of the put is capped at the stock price. The PV of the fixed payoff tends to zero as the term increases for any positive riskfree rate. Thanks,

I've awarded three stars, so the stars are finished for this , thank you for your efforts!!

 

[David Harper CFA FRM]

mathless, that was exactly my thinking ... but unless my BSM XLS is wrong, high volatility can tend d2 to negative high values with N(d2) --> 0. I could have a mistake, intuitively I don't understand. However, mathematically it seems plausible; e.g., if ATM option, where LN(S/K) = 0, then:
d2 = (r - sigma^2/2)*T/[sigma*SQRT(T)], where ATM has LN(S/K) = 0 for convenience,
= rT/sigma*sqrt(T) - sigma^2*T/[2*sigma*SQRT(T)]
= r*sqrt(T)/sigma- sigma*sqrt(T)/2
; i.e., high volatility can switch d2 from positive to a negative (!?), thanks,

 

[Ashwin FRM]

It seems, d2 changes sign from positive to negative
as relationship of r changes from less than to more than sigma^2/2.
so if the vol increases or r decreases the d2 will shift toward negative.
Here, it does not change the answer any. Thanks.