# 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.

• Hello and welcome to SE. To maximize the chances of you getting an accurate answer, please clarify your questions: (1) which paper are you referring to (reference and link if available), (2) what result are you trying to show precisely as there are many formulations for the B&S symmetry? (3) what did you try and where did you get stuck? Thanks! – byouness May 31 '18 at 17:35
• @Tipumba, I did post an answer which might help you, but don't forget to clarify your question anyway. – byouness May 31 '18 at 18:03
• Thank you for your feedback. I tried to make my question more clear. – Tipumba May 31 '18 at 20:10
• Somehow, your link points to the question instead of the paper :) – byouness Jun 1 '18 at 11:06

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.

### 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: $\frac{d\mathbb{Q}^S}{d\mathbb{Q}} = \frac{S_T}{S_0}$

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

• Thank you for your answer! How does the following equality hold $\mathbb{E}[f(S_T)] = \mathbb{E}[\frac{S_T}{S_0} f(\frac{S_0^2}{S_T})]$ ? I don't really understand this step. – Tipumba May 31 '18 at 20:16
• I think my problem lies already in the first equation of the Put. For me it is not clear how $Put^{(BS)} = \mathbb{E}[(S_0 - K \frac{S_T}{S_0})]$ holds. Is there a mathematical way to understand it so e.g. I use Girsanov and change the measure or something like that? – Tipumba Jun 1 '18 at 8:51
• And when I use the random nikodym derivative I have for the Call option something like this $\mathbb{E}[(S_T - K)^+] = \mathbb{E}^*[\frac{S_T}{S_0}(S_T - K)^+]$ or did I forget something? – Tipumba Jun 1 '18 at 9:01
• Reposting previous comment due to formatting issues: Because the equality above it is valid for every $K>0$, it implies that the distributions of $S$ under $\mathbb{Q}$ and $\frac{S_0^2}{S_T}$ under $\mathbb{Q}^S$ are equal, and hence the equality for every positive function $f$. – byouness Jun 1 '18 at 11:02
• It is just the put price in the call-put symmetry relation. The $Put^{BS}(K, S_0, T)$ is a put price with strike $K$ and with a lognormal underlying starting at $K$. $(S_t)$ is lognormal, and starts from $S_0$, so multiplied by $\frac{K}{S_0}$ it is still lognormal with the same characteristics, but starts at $K$. Hence: $$Put^{BS}(K, S_0, T) = \mathbb{E} \left[\left(S_0 - S_T\frac{K}{S_0}\right)^+ \right]$$ – byouness Jun 1 '18 at 11:09

Put Call Symmetry was first pointed out by David Bates in the zero risk-neutral drift Black Scholes setting. However it holds beyond Black Scholes but not in all models. It is complicated to fully describe all models in which it holds, but here are a couple of simple sufficient conditions beyond zero drift Black Scholes for which PCS holds. Start with the arbitrage-free Black model for forward prices and work with forward measure for which forward prices are driftless. Note that instantaneous volatility is constant in the Black 1976 model, as opposed to deterministic. Now show that the above probabilistic argument generalizes to the case when instantaneous vol is deterministic as opposed to constant. Next, independently randomize the instantaneous volatility into any stochastic process including ones that jump. So long as conditioning on the instantaneous volatility evolution reduces the forward price evolution into the deterministic vol Black evolution of forward price, PCS continues to hold. Nonzero correlation between increments of instantaneous vol and increments of underlying price causes PCS to fail. Arbitrary dependence of drift, diffusion, or jump coefficients of instantaneous vol on the underlying price also violates PCS. Some special dependencies are allowed and you have to see Carr Lee's Math finance paper for details.

I think there's a much easier and more intuitive answer to this that's somehow being missed.

A call option pays off $\max(S_t-K,0)$ at time $t$, and a put option pays off $\max(K-S_t,0)$ at time $t$. If we buy a call and sell a put, then our payoff is $\max(S_t-K,0) - \max(K-S_t,0)$. If you think about that for more than a couple of seconds, or just draw the payoff, it's clearly linear:

$$\max(S_t-K,0) - \max(K-S_t,0) = S_t-K$$

Now we have a linear payoff, and we know that when $f(X)$ is linear, then $\mathrm{E}[f(X)]=f(\mathrm{E}[X])$ (see here). So we have long call and short put = expectation of spot at time t less the strike, where the expectation of the spot at t is the forward, where we have to note that the call and put values are discounted to present value, so we need to do the same to the eventual payoff, and multiply by $Z$ - the value of $1 at time$t\$:

$$C - P = Z(F - K)$$

*there is a caveat here, that the discounting is not correlated with the underlying.