If we are in a Black Scholes setup and a I have a Call option and hedged it by shorting delta amount of its underlying.

What does the second derivative of the call with respect to Price of the underlying tells me? Let's look at the graph. What does the terms "break Even", "decay" and "locally hedged mean in my situation?

This graph is a professors handwritten notes that I have copied.

I do this understand why the curvature look the way it does but I can't crack the code of what the break even points and the term "decay" means. And what does this curve tells us about PnL for my delta position. The overtall PnL should be zero, right?

Please provide an explanation of this graph. I do know all the equations so im interested in putting Words to them

enter image description here


2 Answers 2


We work in a Black-Scholes world. Consider the following delta-hedged portfolio:

$$ \Pi_t=V_t-\frac{\partial V}{\partial S}S_t$$

We assume the portfolio is self-financing$^{\text{(a)}}$, therefore: $$ \begin{align} \text{d}\Pi_t &= \text{d}V_t-\frac{\partial V}{\partial S}\text{d}S_t \\[3pt] \tag{1} & = \left(\frac{\partial V}{\partial t}\text{d}t+\frac{\partial V}{\partial S}\text{d}S_t+\frac{1}{2}\frac{\partial^2 V}{\partial S^2}(\text{d}S_t)^2\right)-\frac{\partial V}{\partial S}\text{d}S_t \\[3pt] & = \frac{\partial V}{\partial t}\text{d}t+\frac{1}{2}\frac{\partial^2 V}{\partial S^2}(\text{d}S_t)^2 \\[3pt] & = \Theta_t\text{d}t+\frac{\Gamma_t}{2}(\text{d}S_t)^2 \end{align}$$

In Black-Scholes world, theta is negative whereas gamma is positive, thus: $$\begin{align} \Theta_t\text{d}t & \leq 0 \\[6pt] \frac{\Gamma_t}{2}(\text{d}S_t)^2 & \geq 0 \end{align}$$

How does this relate to your graph? Taking discrete steps $\Delta t$ and $\Delta S_t$ (i.e. as in real-life trading), like so:

  • If $\Delta S_t=0$ then $\Delta\Pi_t=\Theta_t\Delta t \leq 0$ which corresponds to the decay in your graph: the option loses value as time passes.
  • However, because gamma $\Gamma_t$ is positive and the variation of the stock price is squared, you conclude that any stock price move, whether it is downwards or upwards, increases the value of the position. Why? For the sake of clarity let us assume the risk-free rate is null; then your delta-hedged portfolio, whose value grows at the risk-free rate, remains constant and equal to $\pi$. Now:
    • If the stock price rises in the interval $\Delta t$, the value of your call option $V_t$ rises. To keep the value of the portfolio $\Pi_t$ equal to $\pi$, you need to sell more stocks $S_t$: you sell when price is high.
    • If the stock price falls in the interval $\Delta t$, the value of your call option $V_t$ falls. To keep the value of the portfolio $\Pi_t$ equal to $\pi$, you need to buy back stocks $S_t$: you buy when price is low.
  • Breakeven is the point where your time decay is offset by your gain on stock rebalancing: $$ \frac{\Gamma_t}{2}(\Delta S_t)^2=-\Theta_t\Delta t$$
  • The term locally hedged simply means that the expansion $\text{(1)}$ is only valid locally.

Thus what the graph means is that a delta-hedged portfolio where we are long the call benefits from stock volatility because you buy low and sell high. You can also check the hedging profit/loss formula in the following question: "derivation of the hedging error in a black scholes setup".

$\text{(a): }$ this portfolio is not strictly speaking self-financing but this does not have any impact for the matter in question, see my answer to "Dynamic Delta Hedging And a Self Financing Portfolio" for more details.

  • $\begingroup$ I interperate the break-even points like this; The LOSS in (long) Call position equals the PROFIT in the (short) stock Price. Am I right? $\endgroup$
    – Lisa
    Commented Mar 22, 2018 at 18:04
  • $\begingroup$ ... and the other way around of course $\endgroup$
    – Lisa
    Commented Mar 22, 2018 at 18:29
  • $\begingroup$ @Lisa that's correct. $\endgroup$ Commented Mar 22, 2018 at 18:33
  • $\begingroup$ Yes, that is right. And what happens when $\Delta S$ is zero? The (short) stock position breaks even (zero profit), the (long) call position loses value due to time decay $\frac{\partial C}{\partial t}$. That why it says "decay" there. Decay is how much the price of the call drops in 1 day if the stock price stays the same. $\endgroup$
    – Alex C
    Commented Mar 22, 2018 at 21:35
  • $\begingroup$ No, the breakeven here is the point at which the loss from theta (time decay) is offset by the gamma pnl which is the excess pnl from the call over the delta stock pnl. On an up move the call makes more than the short stock loses, on a down move the call loses less than the short stock makes. That excess pnl is offset by the time decay of the option. The breakeven points are the points at which the excess pnl is exactly equal to the time decay. $\endgroup$
    – Ivan
    Commented Mar 24, 2018 at 19:34

There are serious issues with how this graph is drawn, which impede understanding. The y axis is unlabeled and should be labeled "profits" or $\pi$ on a hedged position. The other axis should be labeled "stock price change" or $\Delta S$, not stock price, and the middle point of the axis (where the parabola has a minimum) should be labeled zero. Points to the left of the midpoint represent negative $\Delta S$ and points to the right represent positive $\Delta S$ i.e. an increase in the stock price. The words 'Locally hedged' could be replaced with 'discretely hedged', meaning you are long the option and short the stock and you adjust the delta every $\Delta t$ rather than continuously. You have asked about the word 'decay': the tendency for an option to lose money over time is sometimes called "time decay", but it would be just as good to write "losses" there instead of "decay". In this part of the curve the ordinate is negative, meaning that we are making losses.

Now with the graph correctly drawn and labeled, what does it say?

As you can see: for small (near zero) changes in price, the profits are negative (i.e. the profit curve is below the axis). In the other 2 parts of the curve (for big positive changes in stock price and for big negative changes) profits are positive. Of course on average, if the option is priced and hedge correctly profits on a hedged position are zero. But on any day whether you make or lose money depends on whether $\Delta S$ is big or small in absolute value. And this property is what "having gamma" is all about.

  • $\begingroup$ I kind of prefer this answer as it talks about discrete time, which I think is what is needed for this graph to make sense. $\endgroup$
    – Yian Pap
    Commented Mar 23, 2018 at 1:34
  • $\begingroup$ @Yian Pap not necessarily, profits can come from 1) discrete rebalancing, but also 2) realized volatility being different from implied volatility. $\endgroup$ Commented Mar 23, 2018 at 10:04
  • 1
    $\begingroup$ What I mean is that your (correct) relationship (1) explains/describes the given graph if dt (and thus dS) are not infinitesimal. $\endgroup$
    – Yian Pap
    Commented Mar 23, 2018 at 11:59

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