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This is a commonly asked question and I have not been able to find a satisfactory answer to it. Let me first phrase it here. Suppose that interest rates are $0$ and consider an at the money put and an at the money call (both European) on the same stock (self-financing, hence no dividends) with the same expiry. The payoff of the put is bounded by the strike whereas the payoff of the call is unbounded. But the put-call parity dictates that they have the same price. How do you explain this "contradiction"?

The explanations I have found are all about how stock prices are lognormally distributed etc. This explanation cannot possibly be true since the put-call parity is model-independent. To make it a bit formal I consider the first theorem of asset pricing. Then for the put and the call we have

$$C = E^Q[\max(S_T - S_0,0)]$$ $$P = E^Q[\max(S_0 - S_T,0)]$$

Here $Q$ is the risk-neutral measure (or the $T$-terminal measure). Subtracting $P$ from $C$ we get $$C - P = E^Q[\max(S_T - S_0,0)] - E^Q[\max(S_0-S_T,0)]$$ Since expectation is linear and $S_T - S_ 0 = \max(S_T - S_0,0) - \max(S_0-S_T,0)$ we have $$C - P = E^Q[S_T - S_0] = E^Q[S_T] - S_0$$ According to the first theorem of asset pricing absence of arbitrage implies that discounted asset prices are martingales. The discount rate is $0$ in this example. Hence, $E^Q[S_T] = S_0$. Finally, $$C = P$$ So I arrived at this conclusion based on two things: there is no arbitrage and interest rates are $0$. Nowhere I directly or indirectly imposed an assumption on the distribution of $S_T$ under $Q$. So what is the explanation of this phenomenon? The only explanation I can think of is that it is not necessarily an arbitrage opportunity if two contracts, one with bounded payoff and the other unbounded payoff, have the same price. Another example is a bond and a stock that coincidentally have the same price. Bond has a bounded (fixed) payoff whereas the stock has unbounded payoff. Nothing weird about that. Likewise, nothing fishy about the put and the call in the example. Is there more to this?

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    $\begingroup$ This indicates that the (risk neutral) distribution of future asset prices is skewed that's all (and indeed a lognormal is skewed, but you are right this is a model independent result) $\endgroup$ – Quantuple Apr 7 '17 at 12:34
  • $\begingroup$ @Quantuple There is nothing about the distribution of prices in the derivation (except maybe for the martingale property). So $S_T$ could have a symmetric distribution with finite support and this would still be true. With finite support the unboundedness would not lead to confusion but still price distribution having skew is not an implication here I think. $\endgroup$ – Calculon Apr 7 '17 at 12:47
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    $\begingroup$ I was of course assuming continuous support over $[0,\infty[$, I indeed find it hard to claim that a stock price has a finite support (especially since the relationship you state holds $\forall T>0$). What is the question? Because there is no contradiction at all indeed. Two options that have the same prices $\iff$ their expected payout under the risk-neutral measure is the same. If the payout the former is bounded, while that of the latter is not, the equality can then only be explained by a difference in the weights => asymmetrical probability distribution (support/skewness/whatever) $\endgroup$ – Quantuple Apr 7 '17 at 13:10
  • $\begingroup$ @Quantuple There is no contradiction. I am only looking for an explanation for why the put and the call in the question have the same price and why there is nothing strange about it even though one has bounded payoff and the other not. $\endgroup$ – Calculon Apr 7 '17 at 13:17
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One thing to notice is that indeed $E^Q[S_T]=S_0$, by construction, even though the stock price can only drop down to 0, but it can go up to $2\,S_0, 3\,S_0, 100\,S_0$, ...

Thus, implicitly, there are constraints on the distribution of the stock price at time T (otherwise $S_0$ would change):

  • it could be that the actual distribution is symmetric around $S_0$ (in particular, $S_T$ could then at most go up to $2\,S_0$) - that would immediately make plausible that calls and puts are priced the same
  • it could be that price could, say, drop to 10% or rise to 1000% of $S_0$ - but then, to bring the expectation of $S_T$ back to $S_0$, we'd need a very high probability of dropping, and a very low probability of rising ($P_{down} \cdot 0.1 + P_{up} \cdot 10 = 1$).

So, I think the crucial insight here is that the payoff of the call is unbounded, yes, but so is the stock price at $S_T$ - yet the stock price now is only $S_0$, which imposes constraints that imply $C=P$.

Related puzzle, by the way: Suppose you have an upper digital (paying off if $S_T$ > $S_0$). What's the value as vol goes to infinity?

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  • $\begingroup$ Nice answer, thanks. For your puzzle, I am guessing the answer is $1$ since in the limit case it becomes a certainty that the option will end up in the money. Didn't check the math though. $\endgroup$ – Calculon Apr 7 '17 at 19:42
  • $\begingroup$ @Calculon: Glad you liked it. As for the puzzle: Consider again the second bullet point above. We fix $S_0$, and increase the vol - so possible values get further and further away on the upside (say, $50 S_0$), but they're limited on the downside (say, $0.02 S_0$). To keep the expectation in check (fixed at $S_0$), we need to reduce the probability mass "above" and increase it "below". Thus, if I'm not mistaken, the upper digital goes to 0, and the lower digital to 1, as we increase vol (while keeping the forward constant). $\endgroup$ – Fab Jul 8 '17 at 6:59
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Let me rewrite what you have a little bit. First, put-call parity does indeed demand that for forward value $F = E^Q[S_T]$, and for

$$ \{\hat{C}, \hat{P} \}= \{C, P\} \cdot \int_0^T r(s) ds $$

$$ F = K + \hat{C} - \hat{P} $$

This is true for any strike $K$. Now, when carry costs $c(t)$ and short rate $r(t)$ are identically zero, then by arbitrage arguments $F=S_0$ and we also get that

$$ F = S_0 = K + \hat{C} - \hat{P}= K + {C} - {P}. $$

If we choose $K=S_0$ with zero $r$ and $c$ we necessarily obtain $C-P=0$ or

$$ C=P $$

regardless of the underlying model.

So, yes, there is nothing fishy going on. Just from boundary conditions, any continuous model would have clearly have some strike $\tilde{K}$ at which $C_{\tilde{K}}=P_{\tilde{K}}$. Our arbitrage arguments just put us into the risk-neutral measure where it necessarily emerges that $\tilde{K}=S_0$.

If we were in real-world probability space $\Theta$ with zero rates, we would find that (generically) for $K_0 := S_0$

$$ E^\Theta\left[ (S_T-K_0)^+ \right] \neq E^\Theta\left[ (K_0-S_T)^+ \right] $$

even if we were using geometric brownian motion.

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Put Call Parity is purely based on no arbitrage opportunity strategy, independent of the distribution of stock price. One can always construct a put from a call, a risk free bond and a stock on the underlying asset.

Long put = Short Stock + Long Call + Long Bond

Hence, the reason why an bounded option like put is priced same as the unbounded call is fundamentally because the relation also includes an unbounded stock and a bounded bond. Edit as suggested by Fab in comments: The probability distribution of the future stock price is not irrelevant for put and call prices, but it is constrained (by the current stock price)

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  • $\begingroup$ Yeah but the unbounded stock and the unbounded call "cancel out". $\endgroup$ – Calculon Apr 7 '17 at 14:20
  • $\begingroup$ Yeah, that 'cancelling out' leaves the bounded bond and bounded put. The key factor to understand that derivatives are priced based on their underlying assets. So, the probability distribution of the future stock price simply becomes irrelevant when comparing put and call prices. $\endgroup$ – kasa Apr 7 '17 at 18:01
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    $\begingroup$ The probability distribution of the future stock price is not irrelevant for put and call prices, but it is constrained (by the current stock price). $\endgroup$ – Fab Apr 7 '17 at 19:41

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