black scholes - Put Call Symmetry

I want to show the Put Call Symmetry without using the explicite Black Scholes formula. In other words I want to show

Call(t, x, K, T) = Pull(t, K, x, T)

where $S_t = x$ for $t \in [0, T]$.

I got the hint to use $\mathbb{E}_x[(S_T - K)^+] = xK \mathbb{E}_{1/x}[(\frac{1}{K} - S_T)^+]$. For that reason I wanted to use Girsanov and conclude from that $\mathbb{E}_x[(S_T - K)^+] = \mathbb{E}_{K}[(x - S_T)^+]$ holds. For simplicity I assume our interest rate is zero and consider the Black Scholes model with the filtration which is generated by the Brownian motion. I also found the paper form Peter Carr (https://www.math.uchicago.edu/~rl/PCSR22.pdf) but I didn't really understand it.

Thank you for your help!

Answer

The Black-Scholes symmetry formula is valid only under Black-Scholes as its name suggests. It works only for a lognormal $S$. For other models, you can find symmetry relations but they will be different.

Here is an interpretation that will help you link the result to the distributions:

Martingale case

The symmetry relation is:

$$Call^{BS}(S_0, K, T) = Put^{BS}(K, S_0, T)$$

Because the dynamics of a lognormal variable starting at $K$ are the same as those of lognormal variable starting at $S_0$ if we multiply it by $\frac{K}{S_0}$, we can write the put price as follows:

$$Put^{BS} = \mathbb{E}\left[\left(S_0 - K \frac{S_T}{S_0}\right)^+\right] = \mathbb{E}\left[\frac{S_T}{S_0}\left(\frac{S_0^2}{S_T} - K\right)^+\right]$$

So, we can express the call-put symmetry as follows:

$$\mathbb{E}[(S_T - K)^+] = \mathbb{E}\left[\frac{S_T}{S_0}\left(\frac{S_0^2}{S_T} - K\right)^+\right]$$

More generally, for every positive function $f$: $$\mathbb{E}[f(S_T)] = \mathbb{E}\left[\frac{S_T}{S_0} f\left(\frac{S_0^2}{S_T}\right)\right]$$

Which could be interpreted as follows:

The law of $S_T$ under $\mathbb{Q}$ is the same as the law of $\frac{S_0^2}{S_T}$ under $\mathbb{Q}^S$ which is defined by its Radon-Nikodym derivative: $\frac{d\mathbb{Q}^S}{d\mathbb{Q}} = \frac{S_T}{S_0}$

This interpretation answers your question as to where to start with Girsanov.

General case

Just for reference, in the case where the drift $\mu$ is not zero, but rather $r$ or $r-q$, the idea is to use a power of S_T to get a martingale:

1. 
   

   $S_t ^ \alpha$ is lognormal. With the right value of $\alpha$ you can make it a martingale. It is eas to show that this value is $\alpha_0 = 1 - \frac{2\mu}{\sigma}$

   
   


2. 
   

   For every given positive $f$: $$\mathbb{E}[f(S_T)] = \mathbb{E}\left[\left(\frac{S_T}{S_0}\right)^{\alpha_0} f\left(\frac{S_0^2}{S_T}\right)\right]$$