Estimating Derivative P&L by Integrating Delta Across Price Levels

# Gamma and delta P&L example question


# Gamma and delta P&L example question












I'm trying to get a basic understanding of this example delta ladder

```
Price  Delta
80     43
90     31
100    25
110    11
120    -5
130    -20
140    -12
150    10
160    15
170    30
```

with spot at 110

So it's long gamma on the upside above 130 but short initially, and short gamma downside.

I want to know in a easy to understand way, what is the delta and gamma P&L if the price were to end up at each of the above price points.

My understanding for say 130 is

```
Delta = 11*(130-110) = 220
Gamma = ((-20-11) * (130-110)) /2  = -310
Total P&L = delta + gamma = -90
```

Thus if the price were to go from 110 to 170 then it's

```
P&L from 110-130 = -90
P&L from 130-170 =
   Delta = -20 * (170-130) = -800
   Gamma = ((30-20) * (170-130))/2 = 200
   Total = -600
Total P&L = -90-600 = -690
```

If this is correct, then is it not the same as the area under the curve ? ie, I can just do a discrete integral?

## Answer by David Durrleman (score 2, accepted)

https://quant.stackexchange.com/a/16145

First, I think you made a mistake in your computations above. Where you wrote $(30-20)$, I think you really meant $(30-(-20))$ i.e. $30+20$, yielding a gamma P&L of $1000$ instead of $200$. Your total P&L over $[90,170]$ would then be $110$ instead of $-690$. It doesn't matter for my answer either way, just thought I'd point it out for confused readers.

By definition, $\Delta = \frac{\partial V}{\partial S}$, where $V$ is the price of a financial derivative and $S$ is the price of its underlying.

So if $S$ experiences a move from $S_0$ to $S_1$, it follows logically that

$$\Delta V = V(S=S_1) - V(S=S_0) = \int_{S_0}^{S_1}{\Delta \mathrm{d}S}$$

So, your P&L is indeed the area under the curve. Now, unfortunately, what you are computing here is not that. Indeed, by writing your P&L as the sum of these simplistic delta and gamma terms, what you really are saying, mathematically, is:

$$\Delta V = \int_{S_0}^{S_1}{\Delta \mathrm{d}S} = \Delta(S=S_0)\cdot(S_1-S_0) + \frac{\Delta(S=S_1)-\Delta(S=S_0)}{2}\cdot(S_1-S_0)$$ i.e. $$\Delta V = \frac{\Delta(S=S_0)+\Delta(S=S_1)}{2}\cdot(S_1-S_0)$$

This would hold if for example $\Delta$ is assumed linear over $[S_0,S_1]$, but unfortunately, isn't true in the general case. For example, over $[110, 130]$, your computed $\Delta$ is slightly concave (it would have to be equal to $4.5$ at $120$ to be linear), so your estimation of the P&L is slightly off. Over $[130,170]$, as the interval is larger and the shape more complex, the error is obviously worse.

A better estimation when you know some values of $\Delta$ over a discrete interval would be to assume that it is piecewise linear in between observations. It would be equivalent to using your method, but over the smallest possible intervals, which is indeed as you suggest the same as doing a discrete integral. In this case

$$\int_{S_i}^{S_j}{\Delta \mathrm{d}S} = \sum_{k=i}^{k=j-1}\left(\frac{\Delta(S=S_k)+\Delta(S=S_{k+1})}{2}\cdot(S_{k+1}-S_k)\right)$$

In your specific case, the computation would yield

$$\Delta V = \left(\frac{11-5}{2}+\frac{-20-5}2+\frac{-12-20}2+\frac{-12+10}2+\frac{15+10}2+\frac{30+15}2\right)\cdot10=85$$

So as you can see, it gives a quite different result ;-)