Solving a Two-Instrument Delta and Gamma Hedge

Summary

The document shows how to calculate positions in two derivatives that offset an option portfolio’s delta and gamma. It sets up a pair of linear equations: the hedge instruments’ weighted sensitivities must cancel the portfolio’s existing delta and gamma. Substituting the listed sensitivities into that system yields the same equations as writing the portfolio exposures together with the hedge positions and setting each total to zero.

The two formulations are equivalent, so different answers indicate an arithmetic or equation-solving error rather than a different hedging technique. The response identifies this equivalence but does not work through the corrected position values or discuss practical implementation. The method assumes the stated sensitivities apply to the positions being combined; it addresses local delta and gamma neutrality and gives no treatment of costs, changing market conditions, or other risk exposures.

Key ideas

• Delta and gamma neutrality can be formulated as two linear equations in the hedge positions.
• The hedge instruments’ weighted sensitivities must offset the portfolio’s existing exposures.
• Writing hedge exposures equal to the negative portfolio exposures is equivalent to setting total exposures to zero.
• Different results from these formulations point to an algebra or arithmetic mistake.
• The document does not address transaction costs or risks beyond delta and gamma.

Tags

# Making portfolio Delta and Gamma neutral using 2 derivatives


# Making portfolio Delta and Gamma neutral using 2 derivatives












We have an option portfolio with delta =2 and gamma 3 and we want to making this portfolio delta and gamma neutral using two derivatives D1 and D2:

```
------------------------
|        |Delta | Gamma|
------------------------
| Option | 2    | 3    |
------------------------
| D1     | -1   | 2    |
------------------------
| D2     | 5    | -2   |
------------------------
```

I have tried two ways for solving this and they both give different answers:

1)

$w_{D1}*\Delta_{D1} + w_{D2}*\Delta_{D2} = -2$ $w_{D1}*\Gamma_{D1} + w_{D2}*\Gamma_{D2} = -3$

With answers: $w_{D1}$ = -4/9 and $w_{D2}$ = -1/9

2)

$2 -1w_{D1} + 5w_{D2} = 0$;

$3 + 2w_{D1} + -2w_{D2} = 0$

With answers: $w_{D1}$ = -19/8 and $w_{D2}$ = -7/8

Can someone tell me where I do go wrong and give an interpretation of the results? Which technique should be used?

## Answer by Magic is in the chain (score 1, accepted)

https://quant.stackexchange.com/a/50671

The two formulations seem to be exactly the same. If I take the equations from the first method:

$w_{D1}*\Delta_{D1} + w_{D2}*\Delta_{D2} = -2$

$w_{D1}*\Gamma_{D1} + w_{D2}*\Gamma_{D2} = -3$

And substitute for delta and gamma of the two options:

$-w_{D1}+ 5 w_{D2}= -2$

$2w_{D1} -2w_{D2} = -3$

which after shifting the constants to the left becomes exactly the same set as in method 2:

$2-w_{D1}+ 5 w_{D2}= 0$

$3+2w_{D1} -2w_{D2} = 0$

Maybe there is a typo in the solution method you used when solving the first set of equations.

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