# How to calculate the relationship between a long put and a long call with the same strike price

I have noticed that, a lot of the time, long put and call options have different prices even if they are the same distance from the strike price (lower and higher respectively). Is there a name for this, and if so, is there a formula for analyzing this effect?

I assume that the variables will be --

• Price of long put and distance from strike

• Price of long call and distance from strike

• Strike Price

• Underlying Price

EDIT:

I appreciate all these answers. But none really answer my question about how to make a ratio of puts versus calls.

EDIT 2:

I understand the ratios. However, if the stock is expected to go up, the call would be up by some degree and vice-versa. How do I calculate that ratio?

• Can you clarify the question? "a long put and a long call with the same strike price" ... "they are the same distance from the strike price" ... Do they have the same strike price or not? Are you confusing strike price with underlying price? – nanoman Sep 10 '19 at 17:47
• Your original question made no mention of ratios. You have now edited the question, asking how to calculate "that ratio" but you haven't defined what ratio you are now talking about. Got details so we don't have to make random guesses about what you're after? – Bob Baerker Sep 13 '19 at 18:26

Volatility skew is the relationship between the implied volatility of high strike options vs low. Typically options with low strikes have higher implied volatilities.

However, there are other reasons why the options pricing is different. Consider the case where stock price is 100, the call strike is 200 and the put strike is 0. Although unlikely depending on the time to expiration, it is possible that someday the stock price will be above 200. It is impossible for the stock to go below 0. There is an asymmetry in the range of potential stock prices.

Edit: Adjusted strikes to reflect feedback.

• @BobBaerker I see no mention of different expirations in Charles's answer. – nanoman Sep 10 '19 at 17:50
• @nanoman - Sorry, I meant different strikes, leading to volatility smile. etc. I think that the OP is comparing same series options... but he'll have to verify that. – Bob Baerker Sep 10 '19 at 18:06
• "Consider the case where stock price, call strike, and put strike are all 100." By put-call parity, the two option prices differ only by carry-cost terms (interest and dividends), which are typically negligible close to expiration. "It is not safe to assume the future distribution of stock price is symmetric." Arbitrage assures that despite this possible asymmetry, the options still obey the parity relationship. After all, if the unlimited gain above 100 were perceived as dominant over the limited loss below 100, why would the underlying be trading at 100 (and not more) in the first place? – nanoman Sep 10 '19 at 18:16
• @nanoman, fair point regarding the at the money. However, if the stock price is 100, a call option with 200 strike has some small value while a put with a strike of 0 has no value. – Charles Fox Sep 10 '19 at 18:38
• @BobBaerker I'm not sure what you would like for me to verify. I am still a beginner. – Mardymar Sep 10 '19 at 19:19

There are several factors which cause what you have observed for same series options. Here's a simplified explanation.

Puts and calls are related through an arbitrage process called conversions and reversals. The formula for a conversion is:

Stock + Put - Call - Div + Strike - Carry Cost = 0

(For a reversal, it would be the same equation but with all of the + and - signs reversed)

To make this simple, assume that the stock is at the strike price. The factored equation becomes:

Put + Carry Cost = Call + Dividend

1) If there is no dividend, a put's value will be less than the call's value by the amount of carry cost on the stock. call.

2) If there is a dividend, it increases the price of the put, relative to the call, making the put look more valuable to a seller when indeed, it's not.

3) Calls have greater potential profit (infinity?) than puts (zero) so over the full range, it's not a normal distribution. But for the options you've mentioned, this effect is insignificant (nanoman can better address this component).

EDIT: Here's an option pricing calculator. Input all of the variables with no dividend and an interest rate of zero. ATM options will be priced identically. Move the strikes away from ATM and you'll see the call's price increase slightly more than the equidistant put due to the non normal distribution. Add in a the carry cost and then a dividend and you'll see the effect of each on the respective premiums.

https://www.optionseducation.org/toolsoptionquotes/optionscalculator

• I don't see how this answers my question. I was asking about a value that could be calculated on the difference between puts and call prices. This seems like some academic bs. – Mardymar Sep 11 '19 at 0:42
• What is a carry cost – Mardymar Sep 11 '19 at 2:11
• You don't see how this answers your question because as stated by you, you are a beginner and therefore, you lack an understanding of option pricing. This "academic BS" is the basis for option pricing - prices aren't pulled willy nilly out of thin air by market makers. Though I suspect that this won't be clear to you, the following article explains it: optionsuniversity.com/blog/… – Bob Baerker Sep 11 '19 at 2:23

The term "long" is generally used to refer to something that is moving in the same general direction as the underlying, so all calls are long, and all puts are long. I'll assume that by "long", you mean "out of money".

You've likely heard of the mean and variance of a distribution. Those are first and second order parameters, respectively. The mean tells you the expected value, which for a stock should be (apart from risk-discount issues) the stock price. The variance tells you how risky the stock is. The value of a call is basically the expected value of all prices above the strike price, while a put is worth the expected value of the prices below the strike price (this is a simplification, but I think it's close enough for this question).

There's the further parameter of skew, which is a third-order parameter. Basically, this tells you how far from being symmetrical the distribution is. For a symmetrical distribution, the options will of course be equally valuable; the distribution will look the same x above the mean as x below. But if one side has a longer tail, then that can cause them to have different values.

In statistics, skewness is defined as being the expected value of the cube of the \$z\$-score, E[((x-mu)/sigma)^2]. Larger skewness tends to mean more differences between options with the same difference from market price, but the exact difference isn't completely determined by the skewness, but depends on the distribution.

As an example of a distribution with significant skewness, suppose you roll twelve dice, and get a dollar for each six that comes up. The mean of this is 2, so the "market price" of this game is \$2. Now suppose we have a call option with a strike price of \$3, and a put of \$1. The put is worth money only if you roll no sixes, which happens 11% of the time. So it's worth \$0.11. The call, however, is worth money as long as there are more than 3 sixes, and the more sixes there are, the more it is worth. Each individual roll is worth less: the probability of getting 4 sixes is only 8%, 3% for 5, etc. But there are more numbers greater than 3, and each of them is worth more. Overall, the call is worth \$0.17.

This is a positive, or right, skew distribution. In a case like this, where a "stock" has really unlikely possibilities, but the payoff for those possibilities is really high, that tends to make calls worth more than puts. Actual stocks with highly positive skewness can often, as in this example, be modeled as being the sum of low probability possibilities. This is a highly simplified model, where each die has the same probabilities and payoffs, and each is independent of the other. While the real world isn't that simple, this gives a general idea of the principles involved.

• You did not answer my question.You say there is naturally a skew between puts and calls. But how is that skew calculated? – Mardymar Sep 11 '19 at 2:52
• I would be very careful when applying statistics that require independent variables to stocks and stock prices... It can be done, but carefully. – Stian Yttervik Sep 11 '19 at 6:24
• All puts and calls are not long. Long means that you own the option not that the option is out-of-the-money". – Bob Baerker Dec 19 '19 at 5:48
• @BobBaerker en.wikipedia.org/wiki/Long_(finance) Long means that you are bullish. I never said that it means "out-of-the-money". I said that it looks like the OP is using it to mean that. – Acccumulation Dec 19 '19 at 5:55
• "Long means that you are bullish"? Read the info in your Wikipedia link more carefully. It states: "...a long position in an option does not necessarily mean that the holder will profit if the price of the underlying instrument goes up." Being long a put is BEARISH. You've wandered far afield from the OP's question about being LONG a put and LONG a call with the same strike price. With securities, long and short refer to the direction of the bet. With options, it can be either way. Out-of-the-money refers to the distance of the bet. Direction and distance are mutually exclusive definitions. – Bob Baerker Dec 19 '19 at 15:11