Delta Hedging a European Put in the Black-Scholes Market

# Finding the replicating portfolio a European T-claim (put)


# Finding the replicating portfolio a European T-claim (put)












I have

$$dX_0(t) = ρX_0(t)dt ; \qquad X_0(0) = 1\\ dX_1(t) = αX_1(t)dt + βX_1(t)dB(t) ; \qquad X_1(0) = x_1 > 0$$

as the classical Black-Scholes market. I a trying to look for the replicating portfolio $\theta(t) = (\theta_0(t),\theta_1(t))$ for the following European T-claim:

$F(\omega) = (K -X_1(T,\omega))^+$ $\qquad$ (the European put)

Can someone please help me with this?

## Answer by user16651 (score 1, accepted)

https://quant.stackexchange.com/a/29488

$$dX_0(t)=\rho\,X_0(t)\,dt$$ 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.

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

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