Why Vanilla Option Payoffs Can Have a Point Mass at Zero
Summary
The document explains how a call option’s payoff distribution follows from the distribution of its underlying asset. For a European call with strike K, the payoff is zero whenever the underlying finishes at or below K; above the strike, the payoff is the underlying’s terminal value minus K. Thus the payoff distribution has an accumulation of probability at zero, plus a positive portion inherited from the underlying’s distribution, shifted by the strike.
This structure means a call payoff is generally not normally distributed, since it cannot be negative, while a normal variable can. If the call is out of the money, the zero-payoff probability can combine with the positive payoff density to produce a bimodal-looking histogram. The document also cautions against assuming option prices themselves are log-normal: models typically specify a stochastic process for the underlying and derive the option payoff from it. These observations concern payoff distributions; they do not determine a universal distribution for exotic options, whose payoff rules and dependence on the underlying path can vary.
Key ideas
- A call payoff is zero when the underlying finishes at or below the strike.
- Above the strike, the call payoff distribution is inherited from the underlying distribution after subtracting the strike.
- The payoff distribution has a probability mass at zero, so it is generally not normal.
- An out-of-the-money call can show a bimodal payoff histogram because of the zero mass and positive payoff values.
- Option valuation usually models the underlying process and derives the payoff rather than assuming a distribution for option prices.
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Full text
# Distribution of pay-off of an exotic option
# Distribution of pay-off of an exotic option
Can any assumptions be made about the pay-off of an exotic option? For example, might we say the distribution of the pay-off a vanilla option would be Normal?
I have built a valuation tool that estimates the price of a replicating delta-hedging strategy through Monte Carlo methods by trading the structure of a forward curve. It seems that a histogram of the pay-offs have two relative maximums. Can anyone explain this?
Are options (/ real options) prices logNormally distributed, or does the standard assumption not hold given convexity?
Thanks
## Answer by Gordon (score 1)
https://quant.stackexchange.com/a/25356
In general, an option payoff cannot be normal, as the payoff is generally positive, while a normal variable can be negative.
For a standard call option, the distribution function can be computed from the distribution of the underlying stock. Specifically, consider the vanilla European option payoff $X=(S_T-K)^+$. Then, for $x < 0$, \begin{align*} P(X \le x) = 0, \end{align*} while for $x>0$, \begin{align*} P(X \le x) &= P(S_T \le K+x), \end{align*} which can be computed, if the distribution of $S_T$ is given.
## Answer by dm63 (score 0)
https://quant.stackexchange.com/a/25353
Let's take a call option on a stock with exercise price $K$. What is the risk-neutral probability of a payoff $x$? $$P(x=0) = P(\text{stock} \le K)$$ Also we have $P(\text{payoff} = x > 0) = P(\text{stock}=x+K)$. Hence the required density function $f$ has two parts, (a) an accumulation point at zero representing the probability of being out of the money and (b) a truncated part of the stock distribution.
This will indeed give a bimodal distribution if the call is out of the money.
## Answer by SmallChess (score 0)
https://quant.stackexchange.com/a/25364
I think you are confused with what's exactly log-normally distributed. The distribution of option prices can't be normal or log-normal because the prices can't be negative.
In general, we don't model option prices, we model the underlying stochastic processes (i.e: geometric brownian motion, mean-reverting etc). We then use the distribution of those processes to derive the payoff.
Exotic option is really not much different to vanilla option. More complicated payoff structure, but similar idea.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.