Finding the Mode of a Lognormal Underlying and Call Payoff
Summary
The document asks how to identify the most likely value associated with a call option when the underlying price is lognormally distributed. It starts from the discounted expected call payoff and distinguishes the mode of the underlying price distribution from its mean. The response gives the lognormal mode in terms of the parameters of the normally distributed logarithm, then considers the call payoff as the underlying price above strike, floored at zero.
Under the simplifying assumption of a zero interest rate, the response says that when the strike lies below the underlying distribution’s mode, the most likely positive payoff is the underlying mode minus the strike. When the strike is at or above that mode, it identifies zero as the most likely payoff. This is an informal explanation rather than a full derivation, and the response itself questions the practical value of the mode compared with expected payoff. It does not analyze positive rates or fully discuss the point mass at zero created by the payoff floor, so the result should be read with those limits in mind.
Key ideas
- The mode of a lognormal variable differs from its mean and median.
- A call payoff is zero below the strike and equals the excess of the underlying price above the strike otherwise.
- For a zero-rate simplification, the response locates the most likely payoff by comparing the strike with the underlying mode.
- The analysis is informal and does not address positive rates or fully examine the payoff distribution’s point mass at zero.
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Full text
# How to deduce the mode associated with a call option value?
# How to deduce the mode associated with a call option value?
Option value being expressed:
$$OV=e^{-rT}E[max(V-K,0)] \tag{1}$$
Where $V$ is the price of the underlying security, $K$ is the strike price, and $r,T$ are discount rate and time to exercise date, as usual.
In the "Key Result" on pg. 352 of this edition of Hull's book, Hull shows how the Black Scholes Merton formula for call option pricing results from $(1)$ if $V$ is lognormally distributed.
My question is: How can I deduce the mode (i.e., the most likely value, as opposed to the expected value) of the distribution implicit in $(1)$?
## Answer by Alex C (score 1)
https://quant.stackexchange.com/a/33107
If the question is about the underlying, Wikipedia has the answer. If the log of a variable is normal $N(\mu,\sigma)$ then the variable itself is lognormal with mode at $e^{\mu-\sigma^2}$ and mean at $e^{\mu+\frac{1}{2}\sigma^2}$. The Median (or 50% point) is at $e^\mu$. Wikipedia link that I find very useful in working with option maths.
If the question is about the call option, we can analyze it in terms of different cases depending how high the strike price is. I'll assume $r=0$ for simplicity.
If $K < e^{\mu-\sigma^2}$ then the cutoff is at a point on the curve where the density is still increasing (i.e. we are to the left of the mode). In this case the most likely value for the option is the excess of the mode over the strike: $e^{\mu-\sigma^2}-K$
In the opposite case, the density for the stock declines monotonically to the right of K. So the most likely value for the stock is K, and after subtracting K we are left with a most likely option value of 0.
But TBH I am not sure why anyone cares what the most likely case is. It makes more sense to analyze the situation based on the expectation.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.