Justify a backward differential equation

Question

Regards of 4.5.1, how we get 4.5.5?

Answer 1

Note that for the replicating portfolio to be self-financing it suffices that ⁽¹⁾: $$\lambda_t=\frac{V_t-h_tS_t}{B_t}$$ where I have changed the notation by designating by $B_t$ the money market account: $$B_t=B_0e^{rt}$$ Hence, because the portfolio is self-financing, its dynamics are: $$\begin{align} dV_t&=\left(\frac{V_t-h_tS_t}{B_t}\right)dB_t+h_tdS_t \\[5pt] &=r(V_t-h_tS_t)dt+(\mu h_tS_tdt+\sigma h_tS_tdW_t) \end{align}$$ Now, you can either be in a position to lend ($V_t>h_tS_t$) or borrow money. If the rates to lend $r_1>0$ and borrow $r_2<0$ are different then the equation above changes to: $$dV_t=r_1\max(V_t-h_tS_t,0)dt-r_2\min(V_t-h_tS_t,0)dt+h_tdS_t$$ You lend at $r_1$ if you have excess cash in your hedging account, namely the value of your hedge $h_tS_t$ is lower than the value of the derivative $V_t$, otherwise you borrow at $r_2$. Note that: $$\begin{align} V_t>h_tS_t \Leftrightarrow &\ r_1\max(V_t-h_tS_t,0)-r_2\min(V_t-h_tS_t,0) \\ &= r_1(V_t-h_tS_t)>0 \\[3pt] V_t<h_tS_t \Leftrightarrow &\ r_1\max(V_t-h_tS_t,0)-r_2\min(V_t-h_tS_t,0) \\ &= -r_2(V_t-h_tS_t)<0 \end{align}$$ Namely, the $\max$ and $\min$ functions allow to separate the excess/deficit of cash cases.

_{(1) Proving this is extremely cumbersome. Basically you need to prove that: $$B_td\lambda_t+d\lambda_tdB_t+S_tdh_t+dS_tdh_t=0$$ If I find the time (and energy) I'll try to post a derivation, otherwise you can check this answer to see how to do this.}