When constructing a replicating portfolio for a short position in a call option under Black Scholes, I am not able to pinpoint the source of gains from theta decay. When theta decay materializes, I don't understand how the replicating portfolio generates the gains necessary to finance the new, higher value (less negative), position.

To be concrete, I will try illustrate the question with a numerical example. Suppose we have the following parameters: $K=10$, $S(0)=10$, $\sigma=0.1$, $r=0$, $T=10, q=0$.

The call price formula from Wikipedia is



$$d_1 = \frac{\log(S/K)+ (\sigma^2/2)(T-t)}{\sigma \sqrt{T-t}} \\ d_2 = d_1 - \sigma \sqrt{T-t} $$

At $t=0$, we have $C(10,0)=1.256$. To replicate a short position in the option, we use that the initial delta is $C_S(10,0)=\Phi(d_1)=0.5628$.

We begin by taking a short position of delta in the underlying and invest the remaining proceeds in the money market. We therefore have a short position worth $-10*C_S(10,0)=-5.628$ in the risky asset and a money market position of $-C_S(10,0)+10*C(10,0)=4.372$. We can check that $4.372-5.628=-1.256=-C(10,0)$.

Now suppose we are at $t=1$ and the price of the underlying doesn't change, i.e. $S(1)=10$. The value of the call is now $C(10,1)=1.192$, which means that a short position in the call should net 0.064 in profit. However, before adjusting the hedge, the value of our replicating portfolio seems to be the same: We have a short position of -5.628 and a money market position of 4.372, giving $4.372-5.628=-1.256 \leq -C(10,1)=-1.192$.

To continue the replication, we would need a short position in the underlying equal to $-10*C_S(10,1)=-5.596$ and a money market position of $C_S(10,1)+10*C(10,1)=4.404$. But we need an additional 0.064 which we don't seem to have: where do we get the 0.0032 units of the risky asset to decrease the short position and the 0.032 units of numeraire to increase the money market position? It would seem like we are missing 0.064 and can't continue self-financing replication.

Perhaps this has something to do with using a discrete time increment in the example?

  • $\begingroup$ Update: The issue does not seem to be associated with the time step, as spreading the change over smaller time increments still leads to the same aggregate error. This is extremely perplexing to me. $\endgroup$
    – gill
    Commented Mar 11, 2021 at 16:49
  • $\begingroup$ the closest question I was able to find on here was this thread: quant.stackexchange.com/questions/34427/… where the last comment by describes an intuition related to the volatility of the stock, but remains imprecise. $\endgroup$
    – gill
    Commented Mar 11, 2021 at 17:11
  • $\begingroup$ I haven't checked your numbers, but in general to reduce your short position you need to buy stock, and to buy stock you take money out of the money market account. Furthermore if there is not enough in the MMA you just do an "overdraft" i.e. the balance in the MMA goes negative and you henceforth have to pay interest rather than receive it. HTH. $\endgroup$
    – nbbo2
    Commented Mar 11, 2021 at 17:57
  • $\begingroup$ Sure and thanks, though here continuing the replication seems to require buying stock and increasing the money market position and we obviously do not have the funds to do both. So it seems that we at an impasse and cannot continue the replication according to BS. Hence the question: where do the effects of theta show up when replicating an option? $\endgroup$
    – gill
    Commented Mar 11, 2021 at 19:54
  • 1
    $\begingroup$ First am I right in saying since you sold the call option you should be buying the stock not selling it. $\endgroup$
    – dm63
    Commented Mar 12, 2021 at 10:33

1 Answer 1


I haven't checked your numbers, but the delta hedging principle is that if the realized stock volatility is equal to the pricing/hedging volatility, then theta and gamma compensate each other.

Since you have assumed $r = q = 0$ the pricing PDE is $$ \frac{\partial C}{\partial t} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 C}{\partial S^2} = 0 $$ which you can rewrite using the option greeks notation as $$ \Theta + \frac{1}{2}\sigma^2 S^2 \Gamma=0 $$ Now assume you are delta hedging your option on a small time step $\delta t$. Your PnL is $$ \delta \text{pnl}=C(S+\delta S, t + \delta t)-C(S,t)-\Delta \delta S \approx \frac{1}{2}\Gamma \delta S^2+\Theta \delta t $$ Combining with the pricing PDE you obtain $$ \delta \text{pnl}\approx \frac{1}{2}\sigma^2 S^2 \Gamma\left(\left(\frac{\delta S}{\sigma S}\right)^2 - \delta t\right) $$

So if the market does not move ($\delta S/S=0$) and you are long gamma, you lose money. Conversely if the market move (up or down) is $|\delta S/S| > \sigma \sqrt{\delta t}$ you make money on a long gamma position.

  • $\begingroup$ That is completely correct. The intuition is also clear when one holds the actual option. However, the stated theta from the BS PDE does not appear to show up when dynamically replicating the option using the underlying stock and the MM asset. I understand it might be more tedious, but I invite you to check the computation in the example. I find that it clarifies the exact problem I'm struggling with. $\endgroup$
    – gill
    Commented Mar 12, 2021 at 14:32
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    $\begingroup$ I think you are confused because you want the money market account to be equal to $C - \Delta S$ at all times. This is only true at $t=0$ (by construction of setting up the initial hedging portfolio). At $t>0$ the amount in the money market account is simply the initial amount plus the sum of all delta adjustments. It will be equal to $C - \Delta S$ only if realized volatility is equal to the model volatility (and hedging was done in continuous time), in which cas the realized PnL is zero. But if the spot does not move, there is a realized PnL. $\endgroup$ Commented Mar 12, 2021 at 15:09
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    $\begingroup$ you will carry on hedging by borrowing/lending whatever amount is required to delta hedge the option. You will find that the hedging portfolio actually replicates the option only if the theta cancels out the gamma times the stock realized variance, that is if the realized volatility is same as the model volatility, otherwise you will incur a PnL leakeage. $\endgroup$ Commented Mar 12, 2021 at 15:21

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