$a_t S_t$ = 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?


2 Answers 2


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.

  • $\begingroup$ 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 $\endgroup$
    – elbarto
    Commented May 19, 2015 at 9:27
  • $\begingroup$ 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 $dV_t$. $\endgroup$
    – torbonde
    Commented May 19, 2015 at 14:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.