# Put Call Symmetry for arbitrary $t\in [0,T]$

I want to assume I am in a general Black Scholes Model with $r=0$ and $\delta=0$ and the typical filtered probability space.

I know that $Call^{BS}(0, x, K, T) = Put^{BS}(0, K, x, T)$ with $x= S_0$, which is our start value, holds. I have proven this. My question is now how can I expand it to arbitrary $t \in [0, T]$ with $x = S_t$, i.e. $Call^{BS}(t, x, K, T) = Put^{BS}(t, K, x, T)$ with $x = S_t$?

It is for me of course somehow clear in a logical way but I have problems proving it. My idea was to use the markov property of our asset and get something like $Call^{BS}(t, x, K, T) = Call^{BS}(0, x, K, T) = Put^{BS}(0, K, x, T) = Put^{BS}(t, K, x, T)$ but this is of course not true yet. How do I change the starting value?

• I have till know the following $Call(0, x, K, T) = E _x [(S_T - K)^+] = E_x [ E_x[(S_T - K)^+ | F_t] ] = E_x [ E_{S_t}[(S_{T-t} - K)^+] ]$. My idea is if I can show a dependence between the call at time 0 to the call at time t I have the wanted statement. What would be the next step? Because the inner expectation is already my call at time t with starting point S_t. – Tipumba Jun 5 '18 at 11:30
• How did you prove it for $t = 0$? I don't see why the same proof wouldn't work for $t > 0$. Also, $Call^{BS}(t,x,K,T)=Call^{BS}(0,x,K,T)$ is not correct, I guess what you meant was rather: $Call^{BS}(t,x,K,T-t)=Call^{BS}(0,x,K,T-t)$. The idea is that you depend on $t$ and $T$ only through the time to maturity $T-t$. – byouness Jun 5 '18 at 12:55
• I proved it with the hints you gave me on my other post (quant.stackexchange.com/questions/40070/put-call-symmetry). Ok but how can you show this $Call(t,x,K, T-t) = Call(0, x, K, T-t)$? – Tipumba Jun 5 '18 at 14:13
• For example by writing down the black scholes call price in both cases. They will be identical. Or by remarking that the law of $S_{T - t}$ conditional on $S_0 = x$ is the same as the law of $S_T$ conditional on $S_t = x$. – byouness Jun 5 '18 at 14:43