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I saw this interesting argument that you can use the value of a put spread to find the approximate market implied probability of underlying finishing below the mid point of the two strikes, assuming the distance of strikes and DTE are reasonably small.

Namely, for two put options on the same underlying with strikes $K_2>K_1$ and stock price at expiry $S_T$, we have

$$ \frac{P(S_t,K_2)-P(S_t,K_1)}{(K_2-K_1)-\big(P(S_t,K_2)-P(S_t,K_1)\big)}\approx P(S_T<\frac{K_2+K_1}{2}) $$

In other words, the odds you get for buying a put spread equals to the probability of stock finishing below mid of strikes.

My question is how to derive this result, and does it applies to call spreads as well.

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    $\begingroup$ To add to the excellent answer below, think about intuitively: draw the payoff of your strategy. It's clear that it only starts paying off when the underlying is under $K_2$ and ends up paying 1 if your below $K_1$. Now bring $K_1$ and $K_2$ closer together such that $K = K_1 = K_2$: you've created a strategy that pays of 1 if the underlying is under $K$. It's price up to some discount factor, should therefore be the probability that the underlying ends below $K$. You can also understand now why if that price is negative this is an arb opportunity (negative price for an a.s. positive payout) $\endgroup$ Commented Jul 30 at 20:45

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This is a standard result in quantitative finance for vanilla (i.e. European style) options. It can be applied to puts as well as calls (thru put-call-parity).

Expressing the put option price as a function of the risk neutral density $q(S_T)$, it is calculated as:

$$ P(S_t,K)=e^{-r(T-t)}\int_0^{\infty}(K-S_T)^+q(S_T)\mathrm{d}S_T $$

The first derivative of this expression with respect to $K$ is

$$ \begin{align} \frac{\partial P}{\partial K}&=e^{-r(T-T)}\int_{0}^{K}q(S_T)\mathrm{d}S_T\\ &=e^{-r(T-t)}\mathrm{P_Q}\left(S_T\leq K\right) \end{align} $$ with $\mathrm{P_Q}\left(S_T\leq K\right)$ the risk neutral probability the put being in the money. For call options, we get a like result

$$ \begin{align} C(S_t,K)&=e^{-r(T-t)}\int_0^{\infty}(S_T-K)^+q(S_T)\mathrm{d}S_T\\ \frac{\partial C}{\partial K}&=-e^{-r(T-T)}\int_{K}^{\infty}q(S_T)\mathrm{d}S_T\\ &=-e^{-r(T-t)}\left(1-\mathrm{P_Q}\left(S_T\leq K\right)\right)\\ &=-e^{-r(T-t)}\mathrm{P_Q}\left(S_T> K\right) \end{align} $$

with $\mathrm{P_Q}\left(S_T> K\right)$ the probability of the call option ending in the money.

You can also have a look at the standard Black-Scholes option pricing equation:

$$ \begin{align} C_{\mathrm{BS}}(S_t,K)&=S_t\mathrm{N}\left(d_1\right)-Ke^{-r(T-t)}\mathrm{N}\left(d_2\right)\\ \Rightarrow\frac{\partial C_{\mathrm{BS}}}{\partial K}&=-e^{-r(T-t)}\mathrm{N}\left(d_2\right) \end{align} $$ with $\mathrm{N}\left(d_2\right)$ the risk-neutral probability of ending in the money.

The equations iny our post are finite difference approximations to the first order derivatives above, driven by the fact that we can usually observe options prices only at discretely sampled strike prices.

You can, of course, then use second order finite difference approximations to directly approximate the probability density function thru the Breeden-Litzenberger formula.

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