Time Value of Option

I am working on time value of option, and especially with dividend, and I have the following questions. First if the consider the Black Scholes models with no dividends and free interest rate $r = 0$ If the write the call price : $$\operatorname{Call Price }= \operatorname{Intrinsic value + Time value}.$$ Where intrinsic $$\operatorname{value} = \max( S_t - K,0)$$ then using the call-put parity. It can be shown that

$$\operatorname{Time Value of Call} = \operatorname{Time Value of Put}.$$

However when we have continuous dividend or free interest rate I don't manage to get the following identity. Or at least I manage to get it for the Forward $F_{t,T} = S_t e^{(r-q)(T-t)}$. As $$\frac{dS_t}{S_t} = (r-q)dt + \sigma dW_t \rightarrow \frac{dF_{t,T}}{F_{t,T}} = \sigma dW_t .$$

Then the call put parity but for the future can be rewrite :

$$F_{t,T} - K = \operatorname{Call}(F_{t,T},K,\sigma,T,t) - \operatorname{Put}(F_{t,T},K,\sigma,T,t)$$

$$F_{t,T} - K = F_{t,T} - K +\operatorname{ Time Value Call Forward} - \operatorname{Time Value Put Forward}$$

$$\operatorname{Time Value call forward} = \operatorname{Time Value Put forward}$$

Do I am missing something or we do not have the relation Intrinsic Value of Call = Intrinsic Value of Put when the free interest rate and the the continuous dividend yield are not equal to 0.

I may have another question on when to exercise call option but I still need to think about it.

In advance thank you very much.

• You wrote, "Intrinsic Value of Call = Intrinsic Value of Put", but that's not correct. E.g. if $S_t > K$ then intrinsic of call is $S_t - K > 0$ but intrinsic of put with the same strike is $0$. Could this be the source of confusion? – bcf Jul 24 '16 at 19:32
• Hello, Thanks a lot, I meant Time Value, I edit my message – ashu24 Jul 24 '16 at 19:46

$$C(T,K) - P(T,K) = ( F_{t,T} - K ) B(t,T)$$
$$C(T,K) - P(T,K) = ( F_{t,T} - K ) + \text{TimeValueCall}(T,K) - \text{TimeValuePut}(T,K)$$