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Q:My investment portfolio has one share of one call and one put, what would be the delta of my portfolio ?

delta of call:0.45

delta of put: -0.14

My thought process:

To begin with since im dealing with a single share this automatically means that the delta of this share is 1 [ according to what i found in the book of hull] therefore would my portfolio delta be : share number x call delta x put delta = portfolio delta

1 x 0.45 x -0.14 = 0.31

It feels as if im missing something very important and ive been over the same chapter for hours now.

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I am not sure what you mean with share of one call and put. Based on the later formula I assume you have on underlying unit, plus a call and put which also has one share as notional value. Spot has delta 1 Call 0.45 Put -0.14 Overall is sum of all deltas, here 1.31.

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  • $\begingroup$ One share as if you buy contracts they are usually bought in 100's but what if the case my portfolio contains only 1? $\endgroup$ – Pedro May 8 at 19:40
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    $\begingroup$ The way your Greeks are written (fraction of 1) means they are "raw" greeks in any case (# of shares does no matter). If you add more shares, it just scales up, and 1.31 becomes 2.62 and so forth. If you have 100, delta would be 131. In terms of actual notional delta, assume it is IBM for a spot price of 145. If you had 100, your delta of 131 will translate into 18995 delta notional. $\endgroup$ – AKdemy May 8 at 21:02
  • $\begingroup$ What I did not understand: does your portfolio include 1 share of the stock or not? Is it just 2 options, one put and one call? If so you just add 0.45 and (-0.14) to get a delta of 0.31. $\endgroup$ – noob2 May 8 at 21:37
  • $\begingroup$ I am not 100% sure either (as mentioned above). In case there is no position in the underlying, my examples are off by 1. Meaning it would only be 0.31, 0.62, 31 and 4495 in actual USD. If the 0.31 is the answer (in the book); it means you only have two options (and no position in the underlying). $\endgroup$ – AKdemy May 8 at 21:42
  • $\begingroup$ Thank you both, i believe the answer is with no position in the underlying but ill give it a further read :D $\endgroup$ – Pedro May 9 at 6:25

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