# Self finance conditions - proof check

Find expressions for the process $$\psi=(\psi(t),\ 0\leq t\leq T)$$ , so the portfolio $$(\phi,\ \psi)$$ is self-financing when:

(1) $$\phi(t)= \int_{0}^{t}S_{s}ds$$

(2) $$\phi(t)=S_{t}$$

where $$\phi(t)$$ is an Ito process.

Where is the error in my Ito product rule? I can't figure out what the cross-variation terms should be. Any help would be greatly appreciated.

# 1)

Given: $$V_t=A_t\psi_t + S_t\phi_t$$

We have the below condition: $$\phi_t= \int_{0}^{t}S_{s}ds$$

condition for a portfolio to be self financing: $$dV_t= \psi_t dA_t +\phi_t dS_t$$

Input our condition of: $$\phi_t= \int_{0}^{t}S_{s}ds$$

$$dV_t= \psi_t dA_t + (\int_{0}^{t}S_{s}ds)dS_t$$

# 2)

given: $$V_t=A_t\psi_t + S_t\phi_t$$

We have the below condition: $$\phi_t= S_t$$

So We now have: $$V_t=A_t\psi_t + S_t^{2}$$

See below for correct solution

• Please provide all the information needed to understand your question. For example, what is $A_t$ or $\psi_t$? – Cettt Mar 14 '18 at 7:43
• ϕ(t) = number of stocks. ψ(t) = number of bonds. A(t) = $e^{rt}$ – user531618 Mar 14 '18 at 8:24
• When you take an exercise from a textbook, it would be nice to provide the reference. This one is Exercise 5.1 from Etheridge's "A Course in Financial Calculus". Also, please try to make your question self-contained. – LocalVolatility Mar 14 '18 at 11:45
• I will do that in the future. Thanks for the tips. – user531618 Mar 14 '18 at 15:47

## 1 Answer

Let me define $B_t=A_t=e^{rt}$ $-$ to avoid confusing it with the geometric average $1/t\int S_u\text{d}u$. Your portfolio value is: $$V_t =\psi_tB_t+\phi_tS_t$$ To be self-financing we need to enforce one of the following equivalent conditions: \begin{align} & \text{[1]} \quad \text{d}V_t =\psi_t\text{d}B_t+\phi_t\text{d}S_t \\[3pt] & \text{[2]} \quad B_t\text{d}\psi_t+\text{d}\psi_t\text{d}B_t+S_t\text{d}\phi_t+\text{d}\phi_t\text{d}S_t=0 \end{align}

You haven't specified any dynamics for the asset $S_t$ but we will assume that the cross-term $\text{d}t\text{d}S_t$ is equal to $0$ which is true for all common models such as Black-Scholes or Heston. We will also assume the process $\psi_t$ is a diffusion: $$\text{d}\psi_t=a_{\psi}(t,S_t)\text{d}t+b_{\psi}(t,S_t)\text{d}W_t$$

Case 1: $\phi_t=\int_0^t S_u\text{d}u$

Let us first compute the differential of $\phi_t$. Note $\phi_t=\phi(t)$ is a function of time $t$ thus: \begin{align} \text{d}\phi_t=\text{d}\left(\int_0^tS_u\text{d}u\right)=S_t\text{d}t \end{align}

Hence: \begin{align} B_t\text{d}\psi_t+\text{d}\psi_t\text{d}B_t+S_t\text{d}\phi_t+\text{d}\phi_t\text{d}S_t & = B_t\text{d}\psi_t+\text{d}\psi_t\text{d}B_t+S_t^2\text{d}t+S_t\text{d}S_t\text{d}t \\[3pt] & = B_t\text{d}\psi_t+rB_t\text{d}\psi_t\text{d}t+S_t^2\text{d}t \end{align}

Because the process $\psi_t$ is a diffusion, the cross-term $\text{d}\psi_t\text{d}t$ should also be null, hence: \begin{align} B_t\text{d}\psi_t+rB_t\text{d}\psi_t\text{d}t+S_t^2\text{d}t & = B_t\text{d}\psi_t+S_t^2\text{d}t \end{align}

Therefore: $$\psi_t=\psi_0-\int_0^te^{-ru}S^2_u\text{d}u$$

Case 2: $\phi_t=S_t$

\begin{align} B_t\text{d}\psi_t+\text{d}\psi_t\text{d}B_t+S_t\text{d}\phi_t+\text{d}\phi_t\text{d}S_t & = B_t\text{d}\psi_t+\underbrace{\text{d}\psi_t\text{d}B_t}_{0}+S_t\text{d}S_t+(\text{d}S_t)^2 \\[3pt] & = B_t\text{d}\psi_t+S_t\text{d}S_t+(\text{d}S_t)^2 \end{align}

Therefore: $$\psi_t=\psi_0-\int_0^te^{-ru}S_u\text{d}S_u-\int_0^te^{-ru}(\text{d}S_u)^2$$

If we assume $S_t$ follows a diffusion of the form: $$\text{d}S_t = a_S(t,S_t)\text{d}t+b_S(t,S_t)\text{d}W_t$$

Then: $$\psi_t=\psi_0-\int_0^te^{-ru}\left(S_ua_S(u,S_u)+b^2_S(u,S_u)\right)\text{d}u-\int_0^te^{-ru}S_ub_S(u,S_u)\text{d}W_u$$

• You are welcome @user531618. Do not hesitate to accept the answer if you feel I have addressed the issues in your question. – Daneel Olivaw Mar 14 '18 at 15:50