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

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

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

1. Long $N(x)$ quantity stocks

2. 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$, hence the question. •$x=\frac{\ln \frac{S}{K}+\left(R-\frac{1}{2}\sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}$depends on$S$. – Gordon May 28 '17 at 17:51 • Thankyou Gordon for pointing this out. I have taken this into account. Assume$h_1$denotes the quantity of stocks bought, then$h_1=N(x)$. Similarly$h_2$denotes the quantity of dollar bonds sold, and$h_2=KR^{-T}N(x-\sigma\sqrt{T})$. Now to prove that the portfolio can self finance when$S$changes, we must prove that the condition$dP=h_1dP_1+..+h_ndP_n$for self financing holds. The LHS turns out to be$N(x)$. And the RHS also turns out to be$N(x)$which means that for infinitesimal changes in$S$, the portfolio is self financing. I am unable to make a similar proof for changes in$T$. – swanar May 29 '17 at 16:43 • I have never heard the concept of "self financing for changes in time". What does it mean? – Alex C May 30 '17 at 2:14 • Hello Alex. Self financing in time means if all variables except time are kept constant, then the portfolio must fulfill the equation for self financing which is Eqn 1. In more simple terms, call options are assets which depreciate with time. If price of the underlying doesn't change and the option is OTM then it will decay to 0. For the portfolio claimed to replicate an option, in the situation where the underlying does not change in value, with OTM option, I cannot see how time decay is reducing the price of the portfolio while simultaneously satisfying the condition for self financing. – swanar May 31 '17 at 17:03 • When you hold a call option and the stock price ends up near where it started you lose money because of theta decay. When you hold a replicating portfolio you lose the same amount of money through a different mechanism: you are constantly buying shares after every small rise and selling after every small drop, resulting in negative trading profits. Keep in mind that the stock price is constantly "vibrating" up and down, it cannot stay constant since we are assuming a volatility$\sigma>0\$. (Gordon can show you the mathematical proof of what I just said). – Alex C May 31 '17 at 22:54