Deriving Black–Scholes Equations for Merton Debt and Equity

# Merton model riskless self-financing derivation


# Merton model riskless self-financing derivation












Suppose $dA_t = A_t[\mu dt+\sigma dW_t]$ (assets' value) under the physical measure, plus the other assumptions of the Merton model.

Suppose further that debt and equity are tradeable assets that satisfy $A_t = D_t+E_t$ and follow processes $D_t = D(t,A_t)$, $E_t = E(t,A_t)$ for differentiable functions.

By considering a locally risk-free self-financing portfolio of bonds and equity(which by necessity will earn the risk-free rate of return), prove directly that both $D$, $E$ satisfy the Black-Scholes equation: $$\partial_t f+\frac{1}{2}\sigma^2 A^2 \partial_A^2 f+r A \partial_A f-r f = 0$$

## Answer by Gordon (score 4)

https://quant.stackexchange.com/a/26298

We construct a locally risk-free self-financing portfolio $X_t$, at time $t$, with $\Delta_t^1$ share of debt and $\Delta_t^2$ share of equity. That is, \begin{align*} X_t = \Delta_t^1 D_t + \Delta_t^2 E_t. \end{align*} Then, \begin{align*} dX_t &=\Delta_t^1 dD_t + \Delta_t^2 dE_t\\ &=\Delta_t^1\bigg[\Big(\frac{\partial D_t}{\partial t} + \mu A_t\frac{\partial D_t}{\partial A_t} + \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 D_t }{\partial A_t^2}\Big)dt + \sigma A_t \frac{\partial D_t}{\partial A_t} dW_t \bigg]\\ &\quad + \Delta_t^2\bigg[\Big(\frac{\partial E_t}{\partial t} + \mu A_t \frac{\partial E_t}{\partial A_t} + \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 E_t }{\partial A_t^2}\Big)dt + \sigma A_t \frac{\partial E_t}{\partial A_t} dW_t \bigg]. \end{align*} Since $X_t$ is locally risk-free, \begin{align*} \Delta_t^1\frac{\partial D_t}{\partial A_t} + \Delta_t^2\frac{\partial E_t}{\partial A_t}=0. \tag{1} \end{align*} Moreover, since $X_t$ earn the risk-free rate $r$, \begin{align*} dX_t = rX_t dt. \end{align*} From $(1)$, \begin{align*} r\Delta_t^1 D_t dt + r\Delta_t^2 E_t dt &= \Delta_t^1\Big(\frac{\partial D_t}{\partial t} + \mu A_t \frac{\partial D_t}{\partial A_t} + \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 D_t }{\partial A_t^2}\Big)dt \\ &\quad + \Delta_t^2\Big(\frac{\partial E_t}{\partial t} + \mu A_t \frac{\partial E_t}{\partial A_t} + \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 E_t }{\partial A_t^2}\Big)dt. \end{align*} That is, \begin{align*} &\Delta_t^1\Big(\frac{\partial D_t}{\partial t} + \mu A_t \frac{\partial D_t}{\partial A_t}+\frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 D_t }{\partial A_t^2} - rD_t \Big)dt \\ &+ \Delta_t^2\Big(\frac{\partial E_t}{\partial t} +\mu A_t \frac{\partial E_t}{\partial A_t}+ \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 E_t }{\partial A_t^2}-rE_t\Big)dt =0. \tag{2} \end{align*} Since $A_t = E_t + D_t$, \begin{align*} \frac{\partial D_t}{\partial t} &= -\frac{\partial E_t}{\partial t},\tag{3}\\ \frac{\partial D_t}{\partial A_t} &= 1- \frac{\partial E_t}{\partial A_t},\tag{4}\\ \frac{\partial^2 D_t }{\partial A_t^2} &= - \frac{\partial^2 E_t }{\partial A_t^2}.\tag{5} \end{align*} From $(1)$ and $(2)$, \begin{align*} &-\Delta_t^2\frac{\frac{\partial E_t}{\partial A_t}}{\frac{\partial D_t}{\partial A_t}}\Big(\frac{\partial D_t}{\partial t} + \mu A_t \frac{\partial D_t}{\partial A_t}+\frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 D_t }{\partial A_t^2} - rD_t \Big) \\ & \quad + \Delta_t^2\Big(\frac{\partial E_t}{\partial t} +\mu A_t \frac{\partial E_t}{\partial A_t}+ \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 E_t }{\partial A_t^2}-rE_t\Big) =0. \end{align*} Then, from $(3)$-$(5)$, and multiplying $\frac{\partial D_t}{\partial A_t}=1-\frac{\partial E_t}{\partial A_t}$ on both sides, \begin{align*} &-\frac{\partial E_t}{\partial A_t}\bigg[-\frac{\partial E_t}{\partial t} + \mu A_t \Big(1-\frac{\partial E_t}{\partial A_t}\Big)-\frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 E_t }{\partial A_t^2} - r(A_t-E_t) \bigg] \\ & \quad + \Big(1- \frac{\partial E_t}{\partial A_t}\Big)\Big(\frac{\partial E_t}{\partial t} +\mu A_t \frac{\partial E_t}{\partial A_t}+ \sigma^2 \frac{1}{2} A_t^2 \frac{\partial^2 E_t }{\partial A_t^2}-rE_t\Big) \\ =&\ \frac{\partial E_t}{\partial A_t}\frac{\partial E_t}{\partial t} - \mu A_t \frac{\partial E_t}{\partial A_t} + \mu A_t \left(\frac{\partial E_t}{\partial A_t}\right)^2 + \frac{1}{2} \sigma^2 A_t^2\frac{\partial E_t}{\partial A_t}\frac{\partial^2 E_t }{\partial A_t^2} +r(A_t-E_t)\frac{\partial E_t}{\partial A_t} \\ &\quad - \frac{\partial E_t}{\partial A_t}\frac{\partial E_t}{\partial t} -\mu A_t \left(\frac{\partial E_t}{\partial A_t}\right)^2 - \frac{1}{2} \sigma^2 A_t^2\frac{\partial E_t}{\partial A_t}\frac{\partial^2 E_t }{\partial A_t^2}+rE_t\frac{\partial E_t}{\partial A_t} \\ &\quad + \frac{\partial E_t}{\partial t} +\mu A_t \frac{\partial E_t}{\partial A_t}+ \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 E_t }{\partial A_t^2}-rE_t \\ =& \ \frac{\partial E_t}{\partial t} +r A_t \frac{\partial E_t}{\partial A_t}+ \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 E_t }{\partial A_t^2}-rE_t =0.\tag{6} \end{align*} Since $A_t$ also satisfies $(6)$, then so does $D_t$. The derivation is this complete.

## Answer by user20840 (score 0)

https://quant.stackexchange.com/a/26181

You need the derivation of BS eq.

https://en.wikipedia.org/wiki/Black%E2%80%93Scholes_equation#Derivation

where in your setup you need S = V and case 1: V = E, case 2: V = D.

## Answer by Jeppe Andersen (score 0)

https://quant.stackexchange.com/a/26285

- Use Ito's lemma to get dD

- Use dE = dV - dD

- Form a weighted portfolio of D and E, lets call it V, where the weights sum to 1

- Use the self-financing to get the dynamics of V

- Find the weights that makes V risk-free

- Set the drift of V equal to r