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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?

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

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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.

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  • $\begingroup$ Thanks AFK, I had written it down correctly, however transcribed it wrong when typing it up. My main issue was on how to apply Ito's lemma, but thanks for clearing that up for me. :) $\endgroup$ – piman314 May 6 '15 at 2:33
  • $\begingroup$ 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? $\endgroup$ – piman314 May 6 '15 at 2:35
  • $\begingroup$ 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). $\endgroup$ – AFK May 6 '15 at 3:32
  • $\begingroup$ 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 $\endgroup$ – piman314 May 7 '15 at 11:21

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