Risk-Neutral Pricing and Delta Hedging in a Binomial Model

# European Markovian option


# European Markovian option












Background information: Consider a European contingent claim with payoff $V(S_T)$, where $V: \mathbb{R}_+\rightarrow \mathbb{R}$ is a function which assigns a value to the payoff based on the price of the asset at terminal time $T$. Such European contingent claims are called Markovian claims. The claim is that the price $V_0$ of the European contingent claim $V(S_T)$ at time 0 is given by $$V_) = \frac{1}{(1+R)^T}\hat{\mathbb{E}}[V(S_T)]$$ provided that there is no arbitrage. Here $\hat{\mathbb{E}}$ is the expectation with respect to the (unique) risk neutral probability. As a free bonus we construct a replicating portfolio through a backward recursive scheme and show that at state $i$ at time $t$, the price $V_t^{i}$ of the European contingent claim is a function of the price of the underlying asset $S_t(i)$.

First notice that the price of the contingent claim at maturity time $T$ is equal $V(S_T(i))$ if the state $i$ occurs. Then, assume that the price of contingent claim $V_{t+1}^j$ is known at time $t+1$ for all states $j = 0,\ldots,t + 1 $ and is a function $V_{t+1}$ of $S_{t+1}(j)$, i.e. $V_{t+1}(S_{t+1}(j))$. Conditional on the state $i$ at time $t$, the state of the market at time $t+1$ is either $i+1$ or $i$. Therefore, we can use the result of the single period binomial tree to conclude that $$V_t^i = \frac{1}{1+R}(V_{t+1}^{i+1}\hat{\pi}_u + V_{t+1}^{i}\hat{\pi}_l) = \frac{1}{1+R}\hat{\mathbb{E}}[V_{t+1}|S_t = S_t(i)]$$ Here we used the fact the no arbitrage implies the existence of risk neutral probabilities with conditional probabilities $\hat{\pi}_u = \frac{1+R-l}{u-l}$ and $\hat{\pi}_l = \frac{u-1-R}{u-l}$ given state $i$ at time $t$. Notice that the above expression also suggests that $V_t^j$ is a function of $S_t(j)$, $$V_t(S);= \frac{1}{1+R}\hat{\mathbb{E}}[V_{t+1}|S_t = S] \ \ \ \ \ (1.6)$$ To replicate the payoff at time $t+1$, one needs to keep $$\frac{V_{t+1}^{i+1} - V_{t+1}^{i}}{S_{t+1}(i+1) - S_{t+1}(i)} = \frac{V_{t+1}(S_t(i)u) - V_{t+1}(S_t(i)l)}{S_t(i)(u-l)}$$ units of risky asset and $$\frac{uV_{t+1}^i - lV_{t+1}^{i+1}}{(u-l)(1+R)} = \frac{uV_{t+1}(S_t(i)l) - lV_{t+1}(S_t(i)l)}{(u-l)(1+R)}$$ risk free bonds in the replicating portfolio. In other words, the replicating portfolio is a self-financing portfolio by (1.1) with $w_0 = V_0$ and portfolio strategy given by $\Delta_0(S_0),\ldots,\Delta_{T-1}(S_{T-1})$ $$\Delta_t(S):= \frac{V_{t+1}(Su) - V_{t+1}(Sl)}{S(u-l)} \ \ \ \ \ \ (1.7)$$ The number of units of risky asset in replicating portfolio, given by (1.7), is called the Delta of the contingent claim at time $t$.

In the replicating portfolio for a European Markovian option with payoff $V(S_T)$, let $Y_t := S_T\Delta_t(S_t)$ be the amount of investment on the underlying asset. Then, use (1.6) and (1.7) to show $$Y_t = \frac{1}{1 + R}\hat{\mathbb{E}}[Y_{t+1}|S_t], \ \ \text{for} \ t = 0,\ldots, T-1$$

Attempted solution: Assume we have a replicating portfolio for a European Markovian option with pay off $V(S_T)$, let $$Y_t := S_t\Delta_t(S_t)$$ Now we know that $$\Delta_t(S_t) = \frac{V_{t+1}(S_tu) - V_{t+1}(S_tl)}{S_t(u-l)}$$ which represents the quantity of units of the risky asset we need to replicate. Also, we need $$\frac{uV_{t+1} - lV_{t+1}}{(u-l)(1+R)} = \frac{uV_{t+1}(S_tl) - lV_{t+1}(S_tl)}{(u-l)(1+R)}$$ risk free bonds in the replicating portfolio as well. These two quantities represent the replication of the pay off the investment defined by 1.6 in the book $$V_t(S) = \frac{1}{1+R}\hat{\mathbb{E}}[V_{t+1}|S_t = S]$$ Then given the fact the $Y_t = S_t\Delta_t$ then clearly $$Y_t = \frac{1}{1 + R}\hat{\mathbb{E}}[Y_{t+1}|S_t], \ \ \text{for} \ t = 0,\ldots, T-1$$

I am not sure if this is correct in any way,shape, or form. Any suggestions is greatly appreciated.

## Answer by Gordon (score 1, accepted)

https://quant.stackexchange.com/a/23265

We show that \begin{align*} Y_t^i = \frac{1}{1+R}E\big( Y_{t+1} \mid S_t = S_t(i)\big).\tag{1} \end{align*} Note that \begin{align*} Y_{t+1} &= S_{t+1} \Delta_{t+1}(S_{t+1})\\ &=\frac{V_{t+2}(S_{t+1}u) - V_{t+2}(S_{t+1}l)}{u-l}. \end{align*} Then \begin{align*} \frac{1}{1+R}E\big( Y_{t+1} \mid S_t = S_t(i)\big) &= \frac{1}{1+R}E\left(\frac{V_{t+2}(S_{t+1}u) - V_{t+2}(S_{t+1}l)}{u-l} \mid S_t = S_t(i)\right)\\ &= \frac{1}{1+R}E\bigg(\pi_u\frac{V_{t+2}(S_{t+1}(i+1)u) - V_{t+2}(S_{t+1}(i+1)l)}{u-l}\\ &\qquad\qquad +\pi_l\frac{V_{t+2}(S_{t+1}(i)u) - V_{t+2}(S_{t+1}(i)l)}{u-l} \mid S_t = S_t(i)\bigg)\\ &= \frac{1}{1+R}E\bigg(\pi_u\frac{V_{t+2}(S_{t+2}(i+2)) - V_{t+2}(S_{t+2}(i+1))}{u-l}\\ &\qquad\qquad +\pi_l\frac{V_{t+2}(S_{t+2}(i+1)) - V_{t+2}(S_{t+2}(i))}{u-l} \mid S_t = S_t(i)\bigg)\\ &=E\left(\frac{V_{t+1}^{i+1}-V_{t+1}^{i}}{u-l}\mid S_t = S_t(i) \right)\\ &=E\left(Y_t^i\mid S_t = S_t(i) \right)\\ &=Y_t^i. \end{align*} The identity (1) is thus proved.