I would like to have proven to me the above formula, mostly because I don't quite understand it. The formula is an approximation of the profit from gamma trading/gamma hedging, $$0.5 \Gamma (\Delta S)^2$$ So, my questions are, how to prove that, and secondly, what does it mean exactly by "profit"?


Today, an ATM 1-year 25 % volatility call is bought for 10, and we short $\Delta = 0.5$ in the underlying, which is worth 100. So working that out, we get portfolio value $\Pi = 10 - 50 = -40$, our portfolio value.

Some time later, the spot goes up to 105. The call goes up in value, from 10 to 13.

Currently we have shorted $0.5$ of the underlying, so we owe $0.5 \cdot 105 = 52.5$, so we have $\Pi = -39.5$.

So profit is 0.5.

Then we perform our re-hedge: if delta moved from 0.5 to 0.6, then we need to short 0.1 of the underlying. So, we add $-10.5$ to $\Pi$, i.e, $\Pi = -50$.

Where does the formula from above come into the picture here?


1 Answer 1


Assume you buy a plain vanilla call option at the price $V$ and the spot $S$. You immediately delta hedge buy selling $\partial V / \partial S$ units of the underlying asset.

The underlying asset now instantaneously jumps form $S$ to $S' = S + \Delta S$. The new value of the call option is $V'$. Your total p&l is

\begin{equation} \text{P&L} = V' - V - \frac{\partial V}{\partial S} \Delta S. \end{equation}

You can expand the change in the option price to the second order as

\begin{equation} V' = V + \frac{\partial V}{\partial S} \Delta S + \frac{1}{2} \frac{\partial^2 V}{\partial S^2} (\Delta S)^2 + \mathcal{O} \left( (\Delta S)^3 \right). \end{equation}

Substituting back yields

\begin{equation} \text{P&L} = \frac{1}{2} \frac{\partial^2 V}{\partial S^2} (\Delta S)^2 + \mathcal{O} \left( (\Delta S)^3 \right). \end{equation}

This is visualized in the below plots. They are based on $T = 1 / 12$, $K = 100$, $S = 100$, $r = 0\%$, $\sigma = 20\%$. The blue (green) line is the p&l of holding a long (short) position in the call option (underlying asset). The red line is the actual net portfolio p&l and the yellow one is the second order approximation of the latter using the gamma.

delta hedging p&l

  • $\begingroup$ Thanks. So the formula gives us the change in profit value due to an instant change in the spot price. If the change wasn't instant, I presume the formula doesn't hold since it doesn't include $\Theta$? ................................. Anyways. When the spot changes (not instantly, but after a day), and we reap some P&L, and then re-hedge our position so that we are still delta neutral, and then repeat that .... is this what you call "gamma trading"? Or is it just regular "delta hedging" done often? $\endgroup$
    – Jaood
    Commented Mar 12, 2017 at 16:42
  • $\begingroup$ Yes - if the change wasn't instantaneous, you'd (generally) pay theta as the holder of the long position. This makes sense by looking at the right plot. The p&l is positive for any instantaneous spot move. We'd have a free lottery if at the same time the option wouldn't lose time value. This argument is a bit simplistic as it ignores financing cost. That is what I would call delta-hedging the position. Gamma trading usually involves a more proactive view that the theta is small (large) relative to the expected gain from holding a gamma long (short) position. $\endgroup$ Commented Mar 12, 2017 at 21:27
  • $\begingroup$ I do not think this is the right argument. You err by simply Taylor expanding or expanding with Ito's lemma $V$ but dropping the term involving time. Once you do include the time partial derivative, you find that the so called profit is on average eaten by Theta. In essence, this mean no profit argument is the martingale derivation, closely related to the Feynman-Kac formular, of the option pricing. You should correct the answer by addressing this issue. $\endgroup$
    – Hans
    Commented Jul 15, 2017 at 22:25
  • 1
    $\begingroup$ @Hans: I agree with most of what you say. See also my comment immediately preceding yours which makes a very similar point. However, the OP was asking where that particular formula that he states comes from and this is exactly what my answer works out for him. I also clearly state that this only holds for instantaneous moves. $\endgroup$ Commented Jul 15, 2017 at 22:46
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    $\begingroup$ I see your comment earlier comment now. However, I think it is best if you explicitly write down the correct mathematical formula instead of dropping terms. It is wrong as it stands and is very misleading. You can always comment afterwards that one particular term is what the OP is after. Also, the instantaneous change of underlying would entail a jump. If you want to include jumps, then you have to change the option formula to compensate the mean drift from the jumps to again render the mean profit nil. You can not have the cake and eat it, or free lunch. $\endgroup$
    – Hans
    Commented Jul 15, 2017 at 23:08

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