# Why are put and call options worth the same despite that put has no upside whereas call has unlimited upsides?

The following is an interview question.

All Black-Scholes assumptions hold. Assume no dividends. Consider a standard European call and a standard European put on the same stock. Assume that each option has the same maturity, and is struck-at-the-money (i.e. strike equals current spot). For the sake of simplicity, assume that the interest rate is zero, Draw the payoff diagrams for each option (i.e. terminal payoff to option versus level of underlying).

This part of question is easy. Just the usual kinked payoff diagram.

However, the second part of the question throws me off.

The put has limited downside potential and no upside; the call has unlimited upside and no downside. Given the random direction of the stock price movements between now and expiration, the disparity in potential payoffs seems to suggest that the call should be worth more than the put. However, put-call parity says that this is not so. Verify the put-call parity implications and reconcile them with the seemingly disparate potential payoffs.

I have a feeling that it is due to nature of lognormal distribution as stock price follows a lognormal distribution. But I can't pinpoint this concretely.

• Yes it is the lognormality. In logmoneyness terms, strike 0 is equidistant from ATM as strike infinity is. – Frido Rolloos Dec 5 '19 at 14:14

## 2 Answers

There are two ways (or shall i say at least two ways) to look at this.

1) The option price does not depend on the promise of the pay-offs alone, but also on the probability of those payoffs. As you alluded to, if you look at the probability distribution, you will see the unlimited payoffs is so unlikely as to be irrelevant.

2) The put call parity gives you a way to analyse the relationship between the pay-offs of the two, which you can use to reconcile and justify the relationship between the put and call prices. You can say the same thing about the stock price as well: the downside is limited to the price you paid, the upside is unlimited, but then the price of the stock reflects these outcomes.

• For $1),$ is there a mathematical proof? – Idonknow Dec 5 '19 at 13:37
• You said the Black Scholes assumption hold right? Then it is just integral of payoff with respect to probability of stock, which like you said is log normal, and you know the log normal density tails off. 2) does not depend on black scholes assumption. – Magic is in the chain Dec 5 '19 at 18:28

I think it is simpler than looking at the distribution of returns. Since options can be perfectly hedged using synthetic positions, the distribution should not matter.

Let's keep it in the put-call-parity universe. To make a synthetic call, you need to buy a put, buy a stock, and borrow the money at rate r. You're paying that back over time. To make a synthetic put, you need to buy a call, sell the stock, and invest the proceeds. When creating the put, you get to receive the risk free rate r. Therefore, since you have to pay r for calls, but you get r for puts, the difference in their price reflects the difference between paying vs. receiving the risk free rate