Replicating a European Call in the Bachelier Model

# Bachelier model: number of stocks in replicating strategy


# Bachelier model: number of stocks in replicating strategy












Given:

Consider a two-asset, continuous time model (B,S) where \begin{equation} dB_t = B_t r dt, \quad dS_t = \mu dt + \sigma dW_t. \end{equation} The question is:

- Show that there exists a trading strategy which replicates the payout of a European call with strike K and maturity T.

- Show that the number of shares of stock in the replicating portfolio is always non-negative but never greater than one.

## Answer by user9403 (score 1)

https://quant.stackexchange.com/a/15675

Part 1:

Show that there exists a trading strategy which replicates a European Call.

Proof: I am actually going to prove a stronger statement: that there exists an admissible trading strategy which replicates any payoff in this market. By the First Fundamental Theorem of Asset Pricing, there is no arbitrage if there exists a change of measure such that, for all assets, the following holds: $$X_t=\mathbb{\tilde{E}}[e^{-r(T-t)} X_T | \mathcal{F}_t]$$ By Girsonav's theorem, the following change of measure can be made: $$d\tilde{W}_t=dW_t+\frac{\mu-rS}{\sigma}dt$$ Substituting this into the dynamics of $$dS_t=\mu dt+\sigma dW_t$$ yields $$dS_t=rS_tdt+\sigma d\tilde{W}_t $$ The expected value of this process is $$\mathbb{\tilde{E}}[S_T | \mathcal{F}_t]=S_t+\mathbb{\tilde{E}}\left[\int_t ^ T rS_u du |\mathcal{F}_t\right] $$ Taking the expectation inside the integral on the right hand side, $$\mathbb{\tilde{E}}[S_T | \mathcal{F}_t]=S_t+\int_t ^ T r\mathbb{\tilde{E}}\left[S_u |\mathcal{F}_t\right] du $$ Letting $$f(t, T)=\mathbb{\tilde{E}}[S_T | \mathcal{F}_t]$$ And taking the differential of both sides, $$df=rfdt \implies \frac{df}{dt}=rf$$ This is an ODE with initial condition $$f(t, t)=S_t $$ This ODE has the unique solution $$S_t e^{r(T-t)} $$ Thus $$\mathbb{\tilde{E}} [S_T e^{-r(T-t)}]=S_t e^{r(T-t)} e^{-r(T-t)}=S_t $$ Hence this model does not admit arbitrage. By the Second Fundamental Theorem of Asset Pricing, there exists a unique change of measure if and only if every payoff can be replicated. In this model with a single Brownian motion, the change of measure is one such that $$ dS=rS dt+\sigma(dW+\theta dt)=\mu dt+\sigma dW $$ Solving for theta, $$ rS+\theta \sigma= \mu $$

Clearly this has a single solution, namely $$\theta=\frac{\mu-rS}{\sigma} $$

This proves 1.

Part 2: Show that the Delta of the option is between zero and one.

By Feynman-Kac and Ito's lemma, $$e^{-rt} g(S_t, t, T)=e^{-rT}\mathbb{E}[h(S_T)|\mathcal{F}_t]$$ implies that g has the following dynamics: $$\frac{\partial g}{\partial t} dt+\frac{\partial g}{\partial S} dS_t+\frac{\partial^2 g}{2\partial S^2} \sigma^2 dt -rg dt $$ Comparing this with the dynamics of the self-financing replicating portfolio $$X_t=\Delta S_t+\Gamma B_t $$ $$dX_t=\Delta dS_t+\Gamma dB_t $$ By the First Fundamental Theorem of Asset pricing, $$X_t=g(S_t, t, T) $$ It is thus clear that $$\Delta=\frac{\partial g}{\partial S}$$

Writing the expectation of the payoff as an integral, $$X_t=e^{-r(T-t)}\int_K ^ \infty (S_T-K) d\mathbb{\tilde{P}} $$ $$=e^{-r(T-t)}\int_{K-S} ^ \infty (S e^{r(T-t)} +y-K) p(y) dy $$ Taking the derivative with respect to S, $$\frac{\partial g}{\partial S}= \int_{K-Se^{r(T-t)}} ^ \infty p(y) dy -e^{-r(T-t)}(K-Se^{r(T-t)}+Se^{r(T-t)}-K)p(y)=\mathbb{\tilde{P}}(S_T>K) $$ Thus the delta of the option can be written as a probability, which is always between zero and one.

This proves 2.

## Answer by emcor (score 0)

https://quant.stackexchange.com/a/15677

The stock weight $\pi_t\geq0$ is nonnegative as $C_T=\max(S_T-K,0)$ is increasing in $S_T$ (always long).

$\pi_t\leq 1$ cannot be greater than one, because one can receive at most 1 stock from the call at maturity, so you dont pay more than 1 stock price for it. Otherwise, one could buy the (cheaper) call and sell the replicating portfolio for riskfree profit at maturity.