# Replicating Strategy For An European Put Option $$dX_0(t)=\rho\,X_0(t)\,dt$$ thus $$X_0(t)=e^{\rho t}$$ To replicate the derivative $V=F(t,X_1(t))$ we form a self-financing portfolio with the stochastic process $X_1$ and deterministic process $X_0$ in the right proportion.

## One Period Binomial Option Pricing: Portfolio Replication Approach

Hence we need to use the replicating $(\theta_0(t),\theta_1(t))$. The self-financing assumption means that $$V=\theta_1(t)X_1(t)+\theta_0(t)X_0(t)\tag 0$$ so

$$dV=\theta_1(t)dX_1(t)+\theta_0(t)dX_0(t)\tag 1$$ by application of Ito's lemma, we have $$dV=\left(\frac{\partial V}{\partial t}+\alpha X_1(t)\frac{\partial V}{\partial X_1}+\frac{1}{2}\beta^2X_1^2(t)\frac{\partial^2 V}{\partial X_1^2}\right)dt+\beta X_1(t)\frac{\partial V}{\partial X_1}dB_t\tag 2$$ We assume the portfolios are self-financing, which implies that changes in portfolio value are due to changes in the value of the three instruments, and nothing else.

Under this setup, any of the instruments can be replicated by forming a replicating portfolio of the other two instruments, using the correct weights.

$(1)$ and $(2)$ $$\left(\frac{\partial V}{\partial t}+\alpha X_1(t)\frac{\partial V}{\partial X_1}+\frac{1}{2}\beta^2X_1^2(t)\frac{\partial^2 V}{\partial X_1^2}\right)dt+\beta X_1(t)\frac{\partial V}{\partial X_1}dB_t=\\(\alpha\theta_1(t) X_1(t)+\rho\theta_0(t) X_0(t))dt+\beta\theta_1(t) X_1(t)dB_t\tag 3$$ so $$\theta_1(t)=\frac{\partial V}{\partial X_1}\tag 4$$ Substituting in Equation $(4)$, we have $$\left(\frac{\partial V}{\partial t}+\frac{1}{2}\beta^2X_1^2(t)\frac{\partial^2 V}{\partial X_1^2}\right)=\rho\theta_0(t) X_0(t)\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad=\rho\theta_0(t)\left(\frac{V-\theta_1(t)X_1(t)}{\theta_0(t)}\right)\\ \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad\,\,\,\,=\rho V-\rho\theta_1(t)X_1(t)\\ \\ \\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad=\rho V-\rho\frac{\partial V}{\partial X_1}X_1(t)$$ in other words $$\frac{\partial V}{\partial t}+\rho X_1(t)\frac{\partial V}{\partial X_1}+\frac{1}{2}\beta^2X_1^2(t)\frac{\partial^2 V}{\partial X_1^2}-\rho V=0\tag 5$$ Indeed

$$\theta_0(t)=\frac{V-\theta_1(t)X_1(t)}{X_0(t)}\\ \\ \theta_1(t)=\frac{\partial V}{\partial X_1}$$

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