# How can I make this portfolio self-financing?

$a_t S_t$ = number of shares ($S_t$ is stock price at $t$), $S_0 = 1$

$b_t \beta _t$ = saving account value , $d \beta_t = r \beta_t dt$, $r=$ interest rate

So the value of the portfolio:

$$V_t = a_t S_t + b_t \beta_t$$

Is self-financing if

$$dV_t = a_t dS_t + b_t d \beta_t$$

If $a_t = 1-t$, how can I choose $b_t$ such that my portfolio is self-financing?

$$V_t = (1-t)S_t + b_t \beta_t$$

How do I formulate $dV_t$ now? Don't I require more information, in particular, what is $S_t$?

Is there a need to use the stochastic product rule?

We have $$V_t = a_t S_t + b_t \beta_t.$$

By Ito's product rule, \begin{align*} dV_t & = d(a_t S_t) + d(b_t \beta_t) \\ & = a_t dS_t + S_t da_t + da_t dS_t + b_t d\beta_t + \beta_t db_t + db_td\beta_t. \end{align*}

Since $da_t$ and $db_t$ have no $dW_t$ term, the cross terms are both zero and we have \begin{align*} dV_t & = a_t dS_t + S_t da_t + b_t d\beta_t + \beta_t db_t. \end{align*}

Now just plug in your value for $da_t$ and solve one equation in the unknown $b_t$: \begin{align*} dV_t & = a_tdS_t - S_t dt + b_t d\beta_t + \beta_t db_t \triangleq a_t dS_t + b_t d\beta_t \\ & \iff \beta_t db_t = S_t dt \\ & \iff b_t = b_0 + \int_0^t S_u/\beta_u du. \end{align*}

Now you may be able to solve for $b_t$ explicitly depending on your model for $S_t$.

In the Black-Scholes model, you would have $d S_t = \mu\, d t + \sigma\, d W_t$ where $W$ is a Brownian motion. So if $V_t = a_t S_t + b_t \beta_t$, then $$dV_t = a_t\, d S_t + S_t\, d a_t + da_t\,dS_t + b_t\,d\beta_t + \beta_t\,d b_t + db_t\, d\beta_t$$ by the product rule. In your case, when $a_t = 1-t$ you will have $$dV_t = (1-t) \, dS_t - S_t\, dt + b_t\,d\beta_t + \beta_t\,db_t$$ since $da$ and $d\beta$ have no $dW$-term. Hence, you will need to pick $b$ such that $\beta\,db_t = S_t\,dt$, for the portfolio to be self-financing.

• Can you elaborate on how you used the product rule here: $$V_t = a_t S_t + b_t \beta_t$$ $$dV_t = a_t dS_t + S_t da_t + da_t dS_t + b_t d\beta_t + \beta _t db_t + db_t d \beta_t$$ ? when you take $dV_t$ does that mean on the RHS you get: $$d(a_t dS_t) + d(b_t \beta _ t)$$ or am I misinterpreting what you've done May 19, 2015 at 9:27
• Yes, so first off, I used that $dV_t = d(a_tS_t) + d(b_t\beta_t)$. Secondly, the product rule says that $d(a_tS_t) = a_t\, dS_t + S_t\, da_t + dS_t da_t$, and the same thing for $b$ and $\beta$ instead of $a$ and $S$. This gives my first expression for $dV_t$. May 19, 2015 at 14:31