# what is the meaning of the differential of an arbitrary adapted random process?

I was working on the definition of the self-financing portfolio.

Say $V=\phi_tS_t+\psi_t A_t$ where $S_t$ and $A_t$ are the stock price and the money market price at time $t$, resp, and $\phi_t$ and $\psi_t$ are the shares that are invested in stock and money market at time $t$, resp.

Then I came across the formula like this: $dV_t$=$\phi_t dS_t$+$\psi_t dA_t$+ $S_t d\phi_t$+$A_t d\psi_t$+$dS_t d\phi_t$+$dA_t d\psi_t$.

I have no idea what the last four terms mean. Precisely speaking, I don't know what does $d\phi_t$ or $d\psi_t$ mean. They are just some adapted random process and nothing more I know about them.

Moreover, most of reference book or lecture notes (like this, page 16), claim the above formula is based on the Ito lemma. Doesn't Ito Lemma/formula only apply to Ito process?

Many Thanks,

I should clarify my confusion a little bit. If I understand correctly, the differential notations don't have mathematical meaning. However, for $dS_t$ and $dA_t$, their integrals do. So I understand $dS_t$ and $dA_t$. But then $d\phi_t$ or $d\psi_t$ makes no sense to me. Because neither themselves or the integrals are defined.

Moreover, the approach of @Neeraj in the equation $3$ is more like a discrete approach, which is great and I can follow the logic. On the other hand, the note says the equation follows from Ito's lemma. In fact, I even saw some notes say $dA_t d\psi_t=0$ because of Ito's lemma. Those remarks really puzzle me a lot.

I hope I illustrate my confusion a little bit better. Thanks,

You have given the value of portfolio at time $t$ is \begin{equation} V_t=\phi_t S_t + \psi_t A_t \quad \cdots \cdots (1) \end{equation} where $\phi_t$and $\psi_t$ denote the number of units of the security and cash account respectively that is held in a portfolio at time t.

So, the value of portfolio at time $t+dt$ would be $$V_{t+dt}=(\phi_t + d\phi_t)(S_t + dS_t) + (\psi_t + d\psi_t)(A_t+dA_t)\quad \cdots \cdots (2)$$ where $d\phi_t$ and $d\psi_t$ are additional units bought or sold. Solved equation (2) and subtract equation (1) from equation (2), you will get $$V_{t+dt}-V_t = dVt = \phi_t dS_t+\psi_t dA_t+ S_t d\phi_t+A_t d\psi_t+dS_t d\phi_t+dA_t d\psi_t \quad\cdots (3)$$ Remember, in self-financing portfolio, there is no exogenous infusion or withdrawal of money; the purchase of a new asset must be financed by the sale of an old one. So, $d\phi_t$ and $d\psi_t$ must be zero. So last four term in equation will vanish and it lead to $$dV_t= \phi_t dS_t+\psi_t dA_t$$

This is exactly what given in your lecture notes.

• Intuitively it is the change in $\phi_t$ betwwen $t$ and $t+dt$ Jan 23, 2016 at 21:39
• Thank you a lot. But what confuses me is what exactly $d\phi_t$ means. If I understand correctly, the differential notations don't have mathematical meaning. However, for $dS_t$ and $dA_t$, their integrals do. So I understand $dS_t$ and $dA_t$. But I really don't get $d\phi_t$ or $\psi_t$. Moreover, your equation $3$ is more like a discrete approach, which is great. The notes on the other hand says it follows Ito's lemma. That puzzles me. In fact, I even saw some notes says $dA_t d\psi_t=0$ because of Ito's lemma. Does this clarify my confusion a little bit? Thanks a lot in advance. Jan 23, 2016 at 22:10
• You say that $\phi_t$ "an arbitrary adapted random process". If I wrote it as $\phi(t,S)$ to make it clear that it is a function of time and the other random process in the problem, namely $S_t$ does that help? It is not arbitrary, it is yet to be defined more precisely but it will be a function of known things. Jan 24, 2016 at 14:38
• @AlexC do you mean that $\phi_t$ must be a function of $S_t$? (if so, I　now understand it.) Many thanks! Jan 25, 2016 at 5:39
• @FlowingCloud The differentials must have a meaning, maybe it's a trivial result, but still. You could divide by dV to get 1 on the left, and relative rates of change on the right...
– BAR
Mar 24, 2016 at 9:17

Let $[X,Y]_t$ be the quadratic covariation of two processes $X$ and $Y$ at time $t$. The actual precise meaning of $dXdY$ is $$dXdY= d[X,Y]_t$$ For example, the quadratic covariation of a Brownian motion $B_t$ with itself is $t$, so $$dB_t dB_t = d[B_t,B_t]_t=dt$$.

• This sounds convincing. Will think about it, thanks! Apr 20, 2017 at 4:16