Black–Scholes Call Replication and Self-Financing Rebalancing

# Is the replication porfolio for a European Call, self financing for changes in time?


# Is the replication porfolio for a European Call, self financing for changes in time?












I was reading slide 29 here: http://people.hss.caltech.edu/~jlr/courses/BEM103/Readings/JWCh11.pdf (mirror)

Sub-heading: "An interpretation of the Black-Scholes formula"

It is saying that the below is a replication strategy of a call option.

- Long $N(x)$ quantity stocks

- Sell $K R^{-T}N(x-\sigma \sqrt{T})$ quantity one dollar bonds

As far as I understand, any replication strategy (portfolio) must be a self financing portfolio $P$. As per definition of a self-financing portfolio : A portfolio $P$ consisting of $n$ instruments, denoted $P_i$ , with quantities $h_i$ respectively, is self financing iff $dP=h_1 dP_1+h_2d P_2+...h_ndP_n$.

Now, I can very clearly see that the portfolio which the slide claims to be a replicating portfolio is indeed self financing if we differentiate with respect to $S$ but I cannot see how the portfolio is self financing when we consider the effect of $T$ i.e. time remaining until expiry. What I am seeing is that the above portfolio will be able to self finance for infinitesimal changes in the underlying stock price but I'm not able to see how it is able to self-finance the changes when the portfolio is changing due to the effect of $T$. Can someone please tell me if the above replication strategy can also replicate the effect of $T$?

Appendix:

How I concluded that the portfolio is self financing for infinitesimal changes in $S$:

The replicating portfolio is given by $P = N(x)S-K R^{-T}N(x-\sigma \sqrt{T}) $

Lets try to prove that for infinitesimal changes in $S$, $P$ will be able to self-finance itself.

To prove self financing property for changes in $S$, we have to prove that:

$\partial_S P=h_1 \partial_S P_1+h_2 \partial_S P_2+...h_n\partial_S P_n$. [Eqn 1]

Here $h_i$ is quantity of individual portfolio elements and $P_i$ is the price of that element.

For our case:

$h_1=N(x)$ = quantity of stock $S$

$h_2=K R^{-T}N(x-\sigma \sqrt{T})$ = quantity of 1 dollar bonds sold

$P_1=S$ [The stock]

$P_2=1$ [$1 Bonds]

All other $h_i$ and $P_i$ are zero.

Begin Proof:

First lets compute LHS of Eqn 1: Its an established result: LHS = $N(x)$

Now lets compute RHS of Eqn 1: RHS = $N(x)$

LHS=RHS. Hence Eqn 1 Proved. Hence portfolio will self finance for small changes in $S$.

I couldn't outline a similar proof for infinitesimal changes in $T$ and hence the question.

Intuitive explanation of the proof:

Let there be an infinitesimal change in $S$ given by $dS$; in this case the replicating portfolio will change by $N(x)dS$. Now if we want to dynamically hedge the portfolio, the additional change in $P$ would be given by the multiplication of the change $dS$ and the result of differentiation of $P$ w.r.t. $S$, which turns out to be $N(x)dS$. We see that the change in the replicating portfolio w.r.t. $S$ is equal to the change in the dynamically hedged portfolio. Hence $P$ is self financing for changes in $S$. Note that I was not able to produce the same results when I considered infinitesimal changes in $T$, hence the question.