What a Finite-Width Option Butterfly Reveals About Range Probabilities

# From Butterfly Price to Probability of $S_T$ Falling within a Range


# From Butterfly Price to Probability of $S_T$ Falling within a Range












If a butterfly in the limit represents a probability (by the Breeden-Litzenberger result), what can be said about the relative likelihood of a random variable $S_0$ from the price of a vanilla-option constructed butterfly with practical strikes (i.e. not infinitesimally close)?

For example, with the price of the butterfly $$B = C(S,K-\delta,t-T,\sigma, r) - 2*C(S,K,t-T,\sigma, r) + C(S,K+\delta,t-T,\sigma, r) $$

where $\delta = \text{{width of Butterfly}}/2$, can we say via some relationship of $B$ and $\delta$-or-$K$ the likelihood of being within a range at expiration?

I'm interested in what we can say about the likelihood with only the data from the butterfly spread. An approximation, like the one for the Black-Scholes eq'n: $C(S,t)\approx 0.4Se^{-r(T-t)}\sigma\sqrt{T-t}$.

Here's a first attempt:

Let's take $S=10$, $K=10$, $\delta=1$, $t-T=0.25$, $\sigma=0.25$, and $r=0.01$ with no dividends or cashflows until expiration.

$$B = $1.15 - 2\cdot $0.51 + $0.174 = $0.304$$

So we have $B=\$0.34$ and $\delta=\$1$. We assume the bet is fair - losses are equally as likely as profits. So the market thinks this "trade" has equal chances of making money as losing money. The butterfly has a breakeven profit in the range $[K-\delta +B, K+\delta-B]=[\$9.34,\$10.66]$. If we had a uniform distribution function, we can say the market has a 50-50 chance of being within or outside of this range. We have already assumed a normal distribution using the BSM prices (can this be relaxed?), so this is just an approximation. Can we do better?

What assumptions do we need? For example, I would assume we cannot ignore skew when $\delta$ is not approaching zero.

Edit - Note, the probability of expiring within a range and the probability of the trade being profitable are different.

## Answer by Brian B (score 1, accepted)

https://quant.stackexchange.com/a/41493

You are not going to have an upper bound on the probability without further conditions on the terminal distribution of your price process, no matter how small the market price $V_B$ of our butterfly might be.

Consider a terminal distribution containing a Dirac delta (point mass) at some $K-\delta < S < K$, with no other mass in the butterfly payoff region, and ignore discount factors. Let the Dirac mass be $M$. The value of our option is then simply calculated as

$$ V = M \cdot (S-(K-\delta)) $$

or, for brevity, $V = M \cdot \eta$.

Given the market price $V_B$ of our butterfly, we see that a terminal distribution with point mass corresponding to $\eta = V_B / M$ agrees with the market butterfly price, and can explain a probability of $M$.

(If $\eta > \delta$ then we can simply put some mass at a different point)

Since the Dirac delta terminal distribution can be found as a limit of continuous Ornstein Uhlenbeck processes, the same is true for continuous processes.

Now, a lower limit on probabilities can be found. The value of the option is a lower bound for the option price $V_S$ of the step function with payout $\delta$ in the $K-\delta < S < K+\delta $ region and zero otherwise.

We also have $$ V_S = \delta \cdot \mathrm{Prob}( K-\delta < S_T < K+\delta ) $$

from which we conclude

$$ \mathrm{Prob}( K-\delta < S_T < K+\delta ) = V_S / \delta > V_B /\delta $$

## Answer by Magic is in the chain (score 2)

https://quant.stackexchange.com/a/41453

Are you looking for something like this:

Then you can find this on pages 113-114 of Stephen Blyth’s Introduction to quantitative finance book.

Hope this helps!

## Answer by Ivan (score 2)

https://quant.stackexchange.com/a/41492

Here is the information you can infer from the butterfly: the (undiscounted) price of the butterfly (adjusted to pay \$1) is approximately equal to half the probability that $S_T$ will be in the range $[K-\delta; K+\delta]$.

$\frac{B_T(K-\delta;K+\delta)}{\delta.DF_T}\approx\frac{p(S_T\in[K-\delta; K+\delta])}{2}$

This is because the area covered by the butterfly payoff is half of that of a hypothetical product paying \$1 uniformly in the range $[K-\delta; K+\delta]$.

It is only approximate because the density of probability itself is not uniform within the range, so the half area that's shaved off may or may not correspond to one half of the "probability-weighted area".