Understanding put-call parity

I'm a person with math background trying to break into quantitative finance, and there's something about put-call parity that is not making sense to me. Below I'll detail my understanding of the theorem and set the parameters such that it in my opinion produces a contradiction. Sorry if this is too basic and well-known, I've tried searching for a similar question without results.

Put-call parity. Assume we have no interest rates and no dividends. Assume there is a stock with price $S_0$ at time $0$. Then, consider a put and a call option with strikes $S_0$ and maturities at $T$. The options have the price relation $C = P$.

This can be proven easily by noticing that a portfolio where we sell a put option and buy a call option at time $0$ must have the same payoff as a forward contract on the stock with maturity at $T$ and strike $S_0$. In a world with no interest rates or dividends, the forward contract with these properties has no value and the theorem follows.

My problem. The put-call parity theorem does not make any assumptions on the model, so it needs to hold under any model as long as no arbitrage is allowed. So let us assume that the stock follows the usual Black-Scholes Ito process with $0$ drift and volatility $\sigma$:

$$dS = S\sigma dW_t$$.

Let us assume that the stock price at time $0$ is equal to $S_0 = 1$. Let us also assume that the volatility $\sigma$ is very large.

It is well-known that under the Black-Scholes model the price of the call option approaches the spot price of the stock when $\sigma \rightarrow \infty$. This can be shown directly from the Black-Scholes pricing formula.

Therefore, if $\sigma$ is very large, and $S_0 = 1$, we must have $C \approx 1$. By the put-call parity, also $P \approx 1$. But now the stock price can never be $0$ (at least the probability of that is vanishing). Therefore the payoff of the put minus its price is $(1-S)^+ -P \approx (1-S)^+-1$, which is almost surely negative.

The contradiction. Sell put options and make (almost) guaranteed profit, contradicting the no-arbitrage assumption.

To put the above into numbers, if $T=1$ year, $\sigma = 10$ we already have $P \approx 0.9999994$. That means that the maximum loss in selling the put option is $0.000000057$ in the highly unlikely event that $S_T = 0$. Contrast that with the case of buying the call option and the payoff in the "equally unlikely" event that $S_T \rightarrow \infty$.

Discussion. Mathematically the put-call parity makes sense to me, but the implication above is very unintuitive. What is the explanation to this?

Solution. If the stock follows the dynamic $dS = S\sigma dW$, then we can solve $$S_T = S_0e^{-\frac{1}{2}\sigma^2T + \sigma\sqrt{T}Z}$$, where $Z \sim N(0,1)$. Thus for large values of $\sigma$ or $T$ we have $S_T \approx 0$ with high probability. Therefore the intuition that the stock goes up equally likely as down does not hold for this model.

• Actually when $\sigma \rightarrow \infty$ the probability that $S_T \approx 0$ becomes very high (i.e. it is not a highly unlikely event). This is a known paradox about GBM trajectories. With a high $\sigma T$ a few trajectories go very high, the others go to zero. – Alex C Apr 15 '18 at 1:26
• Thanks for the answer, I think I see it now. I'll update the original post with the solution and more details. – PSL Apr 15 '18 at 9:18
• You can self-answer your question and in fact it’s encouraged in this case. – Bob Jansen Apr 15 '18 at 10:35