Stochastic Differentials - Ito's formula for a self-financing portfolio

Suppose I have a portfolio of stocks $(S)$ and savings account ($\beta_t$) then, the value is

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

and for this portfolio to be self replicating, we need by Ito's lemma $$dV = a_t dS_t + b d \beta_t$$

Now let $$a_t = 2B_t, b_t = -t - B_t^2 - 20B_t, S_t = 10 + B_t, \beta_t = 1$$ With $$B_t = \text{Brownian Motion at time t}$$

How can I show if this portfolio is self-financing?

I can write

$$V = a_t S_t + b_t \beta_t = 2B_t(10+B_t) - (t + B_t^2)$$ $$= 20B_t + 2B_t^2 - t - B_t^2 = 20B_t + B_t^2$$

Since $$S_t = 10 + B_t \to dS_t = dB_t ?$$ And $$\beta_t = 1 \to d \beta_t = 0 ?$$

Now I am having difficulty in evaluating $dV$ in these terms. Can someone help?

$$dV = \{....?\}$$

The portfolio is self-financing. You simply forgot a term in $b$ and a $-t$ term in $V$: \begin{eqnarray} V_t &=& a_t S_t + b_t \beta_t = (2B_t ) (10+ B_t) + (- t - B_t^2 - 20B_t)1 \\ &=& 20B_t + 2B_t^2 - t - B_t^2 - 20B_t \\ &=& B_t^2 - t \end{eqnarray} Applying Ito's lemma \begin{eqnarray} dV_t &=& (2B_t dB_t + \frac{1}{2}2d\langle B,B\rangle_t) - dt \\ &=& 2B_t dB_t \\ &=& a_t dS_t + b_t d\beta_t \end{eqnarray} Since $dS_t = dB_t$ and $d\beta_t = 0$, we have \begin{eqnarray} dV_t &=& a_t dS_t + b_t d\beta_t \end{eqnarray} which is a characterization of a self-financing portfolio.
• Could I further ask, if I wanted to show whether this was an arbitrage strategy or not, I would just need to verifty: 1) $V_0 = 0$ 2) $P(V_T \geq 0) = 1$ 3) $P(V_T \geq 0) >0$ Correct? How would I handle the last two cases? I took the $\mathbb{E} V_T = 0$ thus concluded $P(V_T \geq 0) \neq 1)$ and thus the portfolio is not an arbitrage strategy. is this correct? – piman314 May 6 '15 at 2:35
• 3) Is false. The correct formulation is $P(V_T>0) > 0$ (else 2) would imply 3) and a strategy yielding 0 would be an arbitrage). To prove that this is not an arbitrage strategy, you can indeed use that $E[V_T] = 0$. For that you should prove that a non negative random variable with zero expectation is almost surely zero (not too difficult). – AFK May 6 '15 at 3:32
• Yes! this is exactly what I did, since $$E[V_T] = 0$$ then $$P[V_T < 0] > 0, P[V_T>0]>0$$ therefore its impossible for $P(V_T \geq 0 = 1)$ Thanks for the confirmation – piman314 May 7 '15 at 11:21