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I am stuck in this exercise from my textbook:

Consider a one-period market model with $N+1$ assets: a bond, a stock and $N-1$ call options. The prices of the bond are $B_0=1$ and $B_1 = 1+r$, where $r$ is a constant. The prices of the stock are given by a constant $S_0$ and a random variable $S_1$ taking values in $\{0, 1, \ldots, N-1, N \}$ for a given integer $N \geq 4$. Finally, let the time-0 price of the call option with strike $K \in \{ 1, \ldots, N-1 \}$ be denoted by $C(K)$. Now we introduce a contingent claim with time-1 payout $\xi_1 = g(S_1)$, where $g$ is the function $$g(M) = \mathbf{1}_{ \{M = K_0 \} }, \quad 0 \leq K_0 \leq N. $$ Assuming that the market has no arbitrage, we want to find the time-0 price $\xi_0$ in the following cases: $$ 2 \leq K_0 \leq N-2 \, ; \quad K_0 = N-1 \, ; \quad K_0 = 0 .$$

Let $Y$ be the state price density of the market such that $Y_0 =1$. We know that $$ \mathbb{E} [ YS_1 ] = S_0, \quad \mathbb{E} [ Y ( S_1 - K)^{+} ] = C(K), \text{ for } K \in \{1, \ldots, N-1 \}. $$

But how can we compute $$\xi_0 = \mathbb{E}[ Y \mathbf{1}_{ \{ S_1 = K_0 \} }] \quad ?$$

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The claim payoff you describe, $g(M)$, looks to me like a tight butterfly spread that pays off only in one state of the world. Can't you just replicate that by short two calls with strike $K_0$ and long two calls, with strikes one either side at $K_0\pm 1$? Then the price of your option would be $C(K_0+1)+C(K_0-1)-2\cdot C(K_0)$.

This is effectively the Breeden-Litzenberger formula that expresses the risk neutral distribution (here for discrete states). You can see that the price of the butterfly is the finite difference estimate of the second derivative of the call price with respect to the price.

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