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I have two interrelated questions that have been bothering me for some time. I have read all the stuff online and it still doesn't make sense to me:

Let us assume:

  • 0% interest rate (both hedge funding and discounting rate)
  • 0% dividends

Hence, Forward=Spot.

  1. Why is the price of a ATM put equal to the price of a ATM call? What is happening to the log-normal distribution here, shouldn't the call be more expensive? (two 50% increases lead to a greater payoff(225% Spot) than 2 50% decreases (25% Spot).

  2. Very related.. why is then a 110% Call Option worth more than a 90% Put option (under the same conditions as above)?

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If I did understand your question correctly, it is off-topic (too basic). So please speak if it's not the case, or I'll close this otherwise. –  SRKX Apr 30 '13 at 12:41
No way this is a basic question. Why is an ATM Call equally expensive to an ATM Put (assume 0% rates)? What is the intuition, not the math. Why is a 110% Call MORE EXPENSIVE than a 90% but? (this should not be the case given the first point) –  Remus Stanescu Apr 30 '13 at 14:21
Calm down straightaway. This is basic, you forgot to mention something important the first time (the ATM point), and you're being aggressive. Please look at the answers, and show a bit of courtesy to people who try to help you. –  SRKX Apr 30 '13 at 15:21
I apologize, I was not being aggressive. I tried to emphasize the main point of my post, not to scream. I do not think this is basic since I haven't found the answer yet....please do share if you have it. –  Remus Stanescu Apr 30 '13 at 16:23
I think you're thinking too much, intuition is basic: a put = a call (a put can be made to have the same payoff as a call by using the underlying). It would be absurd for those two things that are the same to have different prices. With regard to your second question: don't compare 110% call with 90% put, compare the call with a call and a put with a put and you'll probably see it straight away given skew. –  frickskit Sep 11 '13 at 12:58
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5 Answers

A simple intuitive answer why the OTM Call is more expensive than the OTM Put is because of the skewness of the log-normal distribution. Think about it, what is the probability that the stock price is above 110 at expiration and what is the probability it is below 90? This should answer your question.

Written in probability terms:

  • The median of the distribution of future stock prices is $S(t) e^{r-\frac{1}{2}\sigma^2} \sim S(t)$ (for reasonable and realistic $\sigma^2$ )

  • $P(S(T)> 110) \sim 1/2 - P(100<S(T)<110)$ which is the probability that the call finishes in the money.

  • $P(S(T)<90) \sim 1/2-P(90<S(T)<100)$ which is the probability that the put finishes in the money

  • The distribution is skewed so that $P(90<S(T)<100) > P(100<S(T)<110)$ which follows from the fact that the probability density function at the median is downward sloping.

  • Thus, $P(S(T)>110) >> P(S(T)<90)$ -> call price > put price

This is not a rigorous proof but you asked for a qualitative answer.

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Could you use the math formatting please? It's easier to read... –  SRKX Apr 30 '13 at 15:30
SRKX, thanks a thousand for the edit and apologies –  Matt Wolf Apr 30 '13 at 15:38
It's fine I did it, you simply can use latex within \$ signs... –  SRKX Apr 30 '13 at 15:38
Put Call Parity is both intuitive and mathematical, two ATM options are not subject to skew nor distributional assumptions of the underlying. If that is not intuitive then I do not recommend considering a career in options –  Matt Wolf Apr 30 '13 at 16:55
Nice attitude towards those that tried to help you and that may be many years of professional work experience senior to you. –  Matt Wolf Apr 30 '13 at 17:06
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The put call parity is given as follows:

$$c_t-p_t = S_t - \frac{X}{e^{r(T-t)}}$$

If you assume $r=0$, you get

$$c_t-p_t = S_t - X$$

So, $c_t \neq p_t$.

The rationale behind it is much more financial than mathematical. You have to look at the payoff on both side of the equation, and you see that both portfolio will give the same payoff at time $T$ (the expiration of the options). So they have to have the same price today otherwise there is an obvious arbitrage opportunity. It holds for any model you assume for the process $S$.

Note that because it is independent of the model, the fact that for ATM option $c_t = p_t$ if $r=0$ does not tell you anything about the price of the options. They could be worth thousands or 0, they just have to be equal.

Assume for a minute they're not equal: $c_t=2$ and $p_t=1$ and $S_t=100$. Then I can sell the call, buy the put, buy the stock by borrowing 100 (with no cost). So I know that I got 2 from the call, paid for the put so I have the stock plus 1.

At maturity, if the stock went up, the put I have is worth 0, the call I sold will force me to give away the stock for 100 which I will give back as I borrowed them... But I made a profit of 1 over the operation. If the stock does not move, both option are worth 0 at time $T$, so I can sell my stock for 100 and pay back the loan, and made a profit of 1. If the stock went down, the call payoff is 0, I can sell the stock at 100 because I own the put, and use the 100 to repay the loan, making a profit of 1. So in any case, I make profit of 1, it's an arbitrage opportunity.

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Given that I specified that the two options are ATM, they will have the same price (S=X). That should not be the case given the log-normal distribution of prices. The second point says this. Spot =100, Forward = 100, Call Strike = 110, Put Strike = 90. Why is the call more expensive than the put? –  Remus Stanescu Apr 30 '13 at 12:49
@RemusStanescu actually, I see no ATM precision in your question... Could you please edit it to make it a bit more clear? –  SRKX Apr 30 '13 at 13:31
yes, sorry forgot to specify. There is nothing wrong in my statement. Why is point 2 unclear? –  Remus Stanescu Apr 30 '13 at 14:24
@RemusStanescu now that you corrected it by specifying it's at ATM, yes it's correct. I'll edit the answer later... –  SRKX Apr 30 '13 at 14:39
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As SRKX mentioned the price relation between the Put and Call (with equal strikes) follows from the Put-Call parity. The whole point behind Put-Call parity is that it does not depend on the underlying distribution which describes your stock price. Put-Call parity is consequence of no arbitrage. It makes no reference to whatever model you are using to describe your stocks.

If the prices of the Put and Call do not satisfy Put-Call parity then you automatically have an arbitrage opportunity.

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Correct. So then, why is a 110% more expensive than a 90% Put? Please don't say put call parity again... –  Remus Stanescu Apr 30 '13 at 14:46
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@RemusStanescu Question 2) was answered quite intuitively but incorrectly by Freddy (he'd be right if he focused on conditional expectations rather than probabilities: indeed, P(s>110) < P(s<90) assuming lognormal dynamics for the underlying stock.) This follows from its negative skewness, which is key to your question. First note that call and put prices are nothing more than conditional expectations of the stock price against the respective payoffs (this follows from their being contingent claims.) The negative skewness of the lognormal distribution, which has mean greater than its median, is such that in probabilistic terms, since the average outcome of the stock price at expiry has a less than 1/2 chance of being realized, the only way for calls to be worth more than puts for equally out of the money strikes is for the conditional expectations to weight higher values that offset the loss from probability. And that is just what occurs with the Lognormal. Thus one can say that conditional on being in the money, calls are always worth more than puts. In visual terms consider the longer right tail of the lognormal distribution, which goes off to infinity (whereas put payoffs are bounded by 0) so that higher values despite being less likely in symmetric terms, compensate the final expectation to the loss in probability mass. Remember, pricing is usually nothing more than integrating a payoff over a probability distribution of some dimension. see here:http://en.wikipedia.org/wiki/File:Comparison_mean_median_mode.svg

SRKX does enough to answer your ATM question. see here for more: http://www.ederman.com/new/docs/qf-Illusions-dynamic.pdf In terms of the above, ATM puts = ATM calls when S=K (r=q=0) because there is a correction of 1/2*{sigma^2} to the drift of the stock dynamics that offsets the probability that P(S>K)<.5 when S is lognormally distributed. But this is a model dependent reason: just read the paper in the link above for the stronger, model independent reason.

On edit (to place all comments in one place:) I did not purge anything. Looks like the moderator did, and threw away all of the formulas i used to prove that if we remain in the risk neutral world, one cannot make the approximation *(1)* $S(t) \exp^{r-\frac{\sigma^2}{2}} \sim S(t)$ without introducing arbitrage. All the probabilities in this question require the risk neutral measure which allows for the correct integration of $SN\left(d_1\right)-Ke^{-r(T-t)}N\left(d_2\right)$ where

$d_1=\frac{\ln(S/K)+\left(r+\frac{1}{2}\sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}$,

$d_2=d_1-\sigma \sqrt{T-t}$ and $N(x)=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^x e^{-\frac{1}{2}s^2}ds$ (assuming no divs and the formula for the call price, with no loss in generality.) Notice $d_1, d_2$ are limits of integration. If one makes the approximation *(1)*, the integral evaluates so as to introduce arbitrage into the BS formula, in particular, PC parity is violated. Again, in simplest terms, just look at $N(d_2)$, which is the risk neutral probability that S>k at expiry. Plug in the numbers and you'll see proof positive that in the BS framework, it is not the unconditional probability of being in the money that makes calls more expensive versus puts for symmetrically OTM strikes (in a world where there is no implied volatility skew, i.e., each strike gets the same vol), it is the expectation conditional on the terminal stock price being in the money that makes it so. Of course i've lost my breath trying to make you see this, to actually prove it to yourself, to no avail. You simply aren't thinking closely about what I'm saying. And for the love of 'i have a life', I ain't copying anything from anywhere. Why would I care to do that? I am here because I noticed your reference to the unconditional probability of being in the money is wrong in the risk neutral world, and simply found it interesting to flesh this out. In case you're wondering, my intuition was guided simply by checking the claim with the good ole, $N(d_2)$.

I actually pasted one of my purged comments in a notepad that is still open, so let me offer this up for further proof:The right comparison is between: $P(S(T)> 110) \sim 1/2 - P(median<S(T)<110)$ & $P(S(T)<90) \sim 1/2-P(90<S(T)<median)$ As soon as vol>0, the median is less than the mean which means there is some substantial probability mass getting wedged between 100, the mean, and the median. If one then accounts for this, the following becomes true: $P(median<S(T)<110)$ > $P(90<S(T)<median)$, and thus $P(S(T)<90)$ > $P(S(T)>110)$,
and therefore we are back to step 1, trying to understand why the call is worth more than the put. It is then the reasoning based on expectation that provides the answer.

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Comments have been purged. –  chrisaycock May 2 '13 at 15:26
I downvoted for the following reasons: You did not answer either question, instead you claimed other answers are incorrect without the slightest reasoning why. Then you claimed the answer can only be found through conditional probabilities. Not only did you not show at all how your conditional probability solution looks like but it is factually wrong that the question can only be answered through conditional probabilities. Your links did not support your answer other than tossing in wikis and academic papers. What can I say, I simply think most of your answer is taken from other posts. Sorry. –  Matt Wolf May 2 '13 at 15:27
Sorry I just commented above while you purged. I wanted to explain my rational for the downvote. –  Matt Wolf May 2 '13 at 15:27
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As already answered, your first question is call-put parity and this is an arbitrage relation independent of model assumption. Your second question (under zero rates and dividends, in the Black-Scholes model) relates to call-put symmetry : $$Call(spot=S_0,strike=K)=Put(spot=S_0,strike= \frac{S_0^2}{K})\times \frac{K}{S_0}$$ It can be easily derived from the B&S formula. A good reference is Peter Carr, "PUT-CALL SYMMETRY: EXTENSIONS AND APPLICATIONS" in Mathematical Finance.

From this symmetry formula, you get that a 110% call is approximatively 10% more expensive than a 90% put.

The intuition behind this for 110% call and 90% put : Consider a favorable upward move for the call: $S_T=110\% \times (1+x)$ with $x>0$ and the symmetric equiprobable downward move $S_T=90\% \times (1-x)$ favorable for the put. For those equiprobable moves, the payout are respectively $110\% \times x$ for the call and $90\% \times x$ for the put. You see that the call payout is $20\%$ higher. As the probability of a favorable move is close to $1\over 2$, you can expect the call price to be $10\%$ above the put price.

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