Probability of profitable option position

Suppose I write an European put option, S=100, X=100, Volatility=10%, RF=2%

If this option has a maturity of 1 year, it will have a value of Y according to an option pricing model

If the option has a maturity of 20 years, it will have a value > Y according to an option pricing model

Most tools will tell you that the option is more likely to expiring the money for the 20 year option, this is because they are all closely linked to option valuation models which are assuming symmetrical volatility (sure those models never claim that they are trying to simulate the real world, the models work due to other mathematical reasons.

However if we were to talk real world probabilities, most assets have a positive drift, would not that mean that the longer term option has a lower probability of maturing in the money because of that drift? My VBA simulation shows as expected when you account for drift that probability of expiring in the money would decrease…

Thank you

An option can have the probability of expiring in-the-money decreases and its value increases when time to maturity increases. I guess you find it’s weird but just remember the distribution form of price of put option changes in function of time to maturity t.

Besides, the price of European put option could increase when time to maturity increases, but if time to maturity is sufficient large, the price will decrease.

For instance, under the condition r>1/2*sigma2 , d1=(r/sigma+1/2*sigma)*t0.5 and d2 = (r/sigma-1/2*sigma)*t0.5 increase when t increase

  • the probability of expiring in the money is N(-d2) decreases when t increases because -d2 decreases.
  • the price X.e-rt N(-d2) - X.N(-d1) increases. It’s not intuitive and quite complex, so just remark that N(-d2) (even e-rt N(-d2) ) decreases slower than N(-d1​​​​​​​) does when t increases and not enough large. But if t is sufficient large, e-rt tends to 0, N(-d2) and N(-d1​​​​​​​) are bounded, so the price of put option tends to 0.

“Besides, the price of European put option could increase when time to maturity increases, but if time to maturity is sufficient large, the price will decrease.”

Thanks, interesting point that I don’t recall but checks out as correct. If you had to explain that effect in simple terms, what is causing the price of the put to increase then eventually starting to decrease?

Also back to talking about real world probabilities of not having that option exercised against me at maturity (i am the seller of the put), what would be the best way to arrive at that probability? I know you can obtain that from whatever model you use, for example if you used a binomial we can look at the final nodes and count how many of them are below the strike price. But is that really giving me the probability of the option being in the money at maturity? Nowhere in model did we consider that the asset can be drifting upwards in its price thus making it unlikely that the option will end up in the money (for example compare two assets both with a vol of 20% but one with expected return of 1% and one with expected return of 10%)…

Thanks!

comments here are missing a key point… there are actually two probabilities to consider…

the probability of expiring ITM and a probability for it (going) ITM at any time before expiration - when you could close for a profit. the latter % is higher but more difficult to calculate

http://www.optionstrategist.com/products/probability-calculator

this calculator seems to use drift so it sort of confirms I am on to something, but I could still use a lot of help…

Thanks

Thanks, trying to focus on expiration to nail the concept first…

How would you go about calculating…

I don’t agree :slightly_smiling_face:. I think you missed a point in the question. We talked about European put option, not American put option.

The second probability you mentioned is for American put option. Yes, it’s higher than the probability of expiring ITM for European put option.

You mean the formula for probability of expiring in the money of European put option? It’s N(-d2) as I wrote above.

For American put option, no closed formula.

wrong… American options are exercisable prior to expiration but you can also trade (close position) European prior to expiration.

if stock is trading at $90 and you hold a $100 call… the probability of the stock moving above $100 (or your call showing an unrealized gain) at any time prior to expiration is HIGHER that the probability of it expiring ITM > $100.

think of a stock jumpinp on earnings release only to pullback sharply in the days/weeks after… closing the option (American/ European) immediately may be the correct decision to capture vol/theta premium

Sorry, what you said is not wrong. But I’m not wrong either :slightly_smiling_face:. Of course, I know “American options are exercisable prior to expiration but you can also trade (close position) European prior to expiration”. But what is its relevance here? And I have never said the hitting probability at any time was less than the hitting probability at maturity.

The OP ask about European put option and he gives an European put option at the money (S = K = 100).

So your example is not relevant here :wink:.

Yes, to study the behavior of European put option in function of t, you could write the mathematical formula of the price in excel, and test it. The intuition is as following

  • In short term, as I said above, e-rt N(-d2) decreases slower than N(-d1) does. So, X * (e-rt N(-d2) - N(-d1)) increases.
  • In long term, e-rt —> 0 , N(-d2) is bounded by 1, N(-d1)—> 0 (because d1 --> infinity) . So, X * (e-rt N(-d2) - N(-d1)) —> 0 . The price decreases to 0.

I’m not sure I understand your question. As seller of put (I think you work in sell side?), you use neutral probability and not real world probability.

PS: I just realize that for the probability of expiring ITM, we must use real world probability even we are in sell side (or buy side).

Suppose you want to use real world probability, you can definitely calculate the probability of the option being in the money at maturity (if my memory is still good, just replace the risk-free rate by the expected rate in N(-d2) ).

The right question should be: what is the meaningness of this probability? If you believe and suppose the world is reflected by your real world probability, this probability will be the “right” probability under this assumption. But this assumption is right or wrong, no one knows.

So to give you a background on how this started, we were discussing options strategies for our clients and someone suggested that clients sell long term put options as over a long time an asset with a positive expected return will very likely be above the level of the put strike.

Question A, ignoring the cost of the option, can we say the above is true in theory… Someone who sells very long term options in practice will rarely be exercised again (especially if it is on an index)

Question B, if the above is true how long is long enough. Clearly if an asset is at a 100 today and I sell an option at strike 90 that matures in 100 days it is more likely to be exercised against me than an option that matures in 10 day.

Question C, if A is correct, can this even be a strategy or the cost of the options will remove any arbitrage? Because remember the cost of the option only includes volatility but not expected return. So an asset with a very high expected return is very likely to finish above the strike level but this cost is not reflected in the price of the option.

If someone can point to a book that someone without a degree in math can read, I will happily read it.

Thanks!

This statement is not true, or we can say it is partially true depending on the value of r, sigma, S0 and K. In fact, the probability of European put option expiring in the money is p = N(-d2) (linkhere). If S0 <> K, the problem is more complicated so I suppose S0 = K, so d2 = (r/sigma-1/2*sigma)*t0.5 and the probability is equal to p = N(-(r/sigma-1/2*sigma)*t0.5)

  • If r/sigma-1/2*sigma >0 or r >1/2*sigma2 : we have p decreases when time to maturity t increases. In layman’s term, it means : clients sell long term put options as over a long time an asset with a positive expected return, the asset’s value S will more likely be above the level of the put strike, put option is less likely in the money.
  • If r/sigma-1/2*sigma <0 or r <1/2*sigma2 : we have p increases when time to maturity t increases. In layman’s term, it means : clients sell long term put options as over a long time an asset with a positive expected return, the asset’s value S will less likely be above the level of the put strike, put option is more likely in the money.

Remarks:

  • For European put option, if S0 <> K, you can expect the probability increases (or decreases) with t small and decreases (or increases) when t is long enough.
  • For American put option. The answer is completely different. The probability of expiring in the money is always increases in function of time to maturity.
  • I think I didn’t use much math here, and this is the maximum I can do without math. I’m happy if someone can explain this with less math :slightly_smiling_face:.

Question A: It depends on the value of r and sigma.

Question B: we must know r and sigma, and the probability is not monotone in function of time to maturity.

Question C: I think the question is not well posed. But perhaps I’m wrong.

thanks, gives me a lot to read up on…

For the question C, If you want to hedge European put option, just use delta neutral hedging method which removes all arbitrages. I don’t know why you bother about the probability of expiring ITM.

Maybe you want to study and hedge digital put option, not put option.