Consider a derivative of digital type which pays this kind of payoff at time $T$: \begin{align*} g(S_T,k) &= \begin{cases} P_0,~S_T>k \\ S_T, ~S_T\leq k \end{cases} \end{align*}

with $S_T$ being the current price of the underlying at maturity time $T$, $P_0$ the price of the underlying at the issue time 0 and $k$ - kind of the strike price with barrier feature.

Apparently, function $g$ is discontinuous at $S_T=k$ and has a jump there. The idea is to approximate it with a set options, call $c(S_T,k_1)$ and put $p(S_T,k_2)$ that have strikes: $k_1 < k < k_2$. Then, to construct a linear piece-wise function that will look as following: $$ \hat g(S_T,k_1,k_2)=a_0+a_1 S_T+a_2 c(S_T,k_1) + a_3 p(S_T,k_2). $$

The question is how to get the coefficients. Which complementary equations may be used?

  • 3
    $\begingroup$ Note that \begin{align*} g &= P_0 \pmb{1}_{S_T > K} + S_T \pmb{1}_{S_T\le K}\\ &= P_0 \pmb{1}_{S_T > K} + S_T \left(1-\pmb{1}_{S_T> K}\right)\\ &=(P_0-K)\pmb{1}_{S_T > K} + S_T - (S_T-K)^+. \end{align*} $\endgroup$
    – Gordon
    May 17, 2019 at 15:44
  • $\begingroup$ Moreover, $$\pmb{1}_{S>K} \approx \frac{(S_T-(K-\varepsilon))^+-(S_T-K)^+}{\varepsilon},$$ for a sufficiently small $\varepsilon$ (e.g., $10^{-4}K)$. $\endgroup$
    – Gordon
    May 18, 2019 at 11:36

1 Answer 1


We should be able to replicate the payoff exactly in each of the two regions $S_{T}\leq k_{1}$ and $S_{T}\geq k_{2}$. From the first, $$a_{0}+a_{1}S_{T}+a_{3}(k_{2}-S_{T}) =S_{T}$$ so, matching coefficients, $a_{0}+a_{3}k_{2}=0$ and $a_{1}-a_{3}=1$. From the second, $$a_{0}+a_{1}S_{T}+a_{2}(S_{T}-k_{1})=P_{0}$$ so, matching coefficients, $a_{0}-a_{2}k_{1}=P_{0}$ and $a_{1}+a_{2}=0$.

Sorry I haven't time to check this works. Hope it helps.

  • $\begingroup$ Solved by $a_{0}=(P_{0}-k_{1}) k_{2}/(k_{2}-k_{1})$, $a_{1}=(k_{2}-P_{0})/(k_{2}-k_{1})$, $a_{2}=(P_{0}-k_{2})/(k_{2}-k_{1})$ and $a_{3}=(k_{1}-P_{0})/(k_{2}-k_{1})$. It would probably come out a bit nicer if you swapped the put for a call at the same strike. $\endgroup$
    – Ali
    May 18, 2019 at 12:36
  • $\begingroup$ Thank you! This does work! $\endgroup$
    – harvey
    May 19, 2019 at 0:02

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