Arrow-Debreu State Prices in the Derman-Kani Implied Tree

# Arrow-Debreu Price in "The Volatility Smile and its implied Tree"


# Arrow-Debreu Price in "The Volatility Smile and its implied Tree"












I was reading the old, but still interesting paper "The volatility smile and its implied tree" by Derman and Kani. I have a two questions about the derivation of the $2n+1$ equations, both of them regarding Arrow-Debreu price. On page 6, is shown in figure 4 how the model is set up. On the next page the authors writes down:

$$C(K,t_{n+1})=\exp{(-r\Delta t)}\sum_{j=1}^n\{\lambda_jp_j+\lambda_{j+1}(1-p_{j+1})\}\max{\{S_{j+1}-K,0\}}$$

where $C(K,t_{n+1}$ denotes the price of a call option with strike $K$ and expiry and expiry $t_{n+1}$. The first term in the sum are the probabilities. However, since $p_j$ is already the risk neutral probability why do the authors multiply them by $\lambda_j$? $\lambda_j$ is the known Arrow-Debreu price at node $(n,i)$. I've never heard of an Arrow-Debreu price. After checking the web it is still unclear to me, what the reason is for this equation.

Moreover using the above equation, there should be a $\lambda_{n+1}$, which is not the case! So is it just set to $0$?

## Answer by Mr. Rodriguez (score 3, accepted)

https://quant.stackexchange.com/a/15842

We use Derman and Kani's notations.

### Arrow-Debreu prices

The Arrow-Debreu price $\lambda_i$ is the price of the security $\Lambda_i$ paying \$1 in node $(n, i)$, and \$0 in all other states $(n, j)$, for $j \neq i$.

Let $\mathbb{P}_{n,j}$ be the risk-neutral probability of getting to state $(n,j)$, from state $(1,1)$.

The price of $\Lambda_i$ is the risk-neutral expectation of its discounted payoff, which is simply $$\lambda_i = e^{-rt_n}\mathbb{P}_{n,i}$$

### Call price

The call price is the risk-neutral expectation of its discounted payoff: \begin{eqnarray} \tag{1} C(K, t_{n+1}) & =& e^{-rt_{n+1}}\mathbb{E}(S_{t_{n+1}} - K)^+ \\ &=& e^{-rt_{n+1}} \sum\limits_{j=1}^{n} \mathbb{P}_{n+1,j+1} (S_{j+1} - K)^+ \end{eqnarray}

Now in the binomial tree, there are two ways to get to state $(n+1, j+1)$:

- either you were in state $(n,j)$ (probability $\mathbb{P}_{n,j} $) and you went up (probability $p_{j}$),

- or you were in state $(n,j+1)$ (probability $\mathbb{P}_{n,j+1}$) and you went down (probability $1-p_{j+1}$)

Therefore: \begin{eqnarray} \mathbb{P_{n+1,j+1}} &=& p_j\mathbb{P_{n,j}} + (1-p_{j+1})\mathbb{P}_{n,j+1} \\&=& e^{rt_n}\left[p_j \lambda_j + (1-p_{j+1})\lambda_{j+1} \right] \tag{2} \end{eqnarray}

The only case where (2) doesn't hold is when you are in an extremal node, you can resolve this issue by setting $\lambda_{n+1}=0$.

You get Derman and Kani's formula by plugging (2) back into (1).

The intuition here is that the Arrow-Debreu prices capture all the information you need from $t_0$ to $t_n$, so that all you have to worry about is what happens between $t_n$ and $t_{n+1}$