1
$\begingroup$

We know the delta of a portfolio of options is simply the sum of deltas of the individual options. But are there any additional known properties about the total delta (or other greeks) of a portfolio of options?

More specifically, how does the total delta of a portfolio change the more options one adds? If there's a random collection of calls and puts on the same underlying but with varying strikes and maturities, shouldn't the likelihood of offsetting deltas increase? Is it possible to apply the central limit theorem here to derive some general rules how the greeks of such a portfolio behave as more options are being added to such a portfolio?

$\endgroup$
2
  • $\begingroup$ Yes, you are right. In a large option portfolio considerable cancellation of delta usually takes place. Something market makers know and experience every day. I am not sure what assumptions you would have to make to usefully quantify this however. Not much is known about the distribution of incoming option orders AFAIK. $\endgroup$
    – noob2
    Nov 6 '21 at 17:09
  • $\begingroup$ ok, many thanks you for your answer. $\endgroup$
    – asardon
    Nov 9 '21 at 8:19
3
$\begingroup$

A bit too long for a comment.

TLDR: It's hard because there is interaction between your strategy and market movements.

If you're executing a option strategy aiming for certain exposures to the Greeks I don't think a statistical approach makes sense. You should be getting the exposure you're aiming for.

If you're providing liquidity a statistical approach might make more sense: After all, you're putting out two sided quotes and getting fills. If you're doing a decent job you're getting fills on both sides and your positions cancel out.

I don't think this approach will work in practice though. When the market moves, fills on one side will dominate. This exposure is risky and not desired so you attempt to mitigate this creating some feedback in the process. The nature of the market and the feedback are dependent on the specific market, the time and your risk management measures, i.e. the feedback process is quite complex and ever changing. So, calculating these statistics a priori seems hard and probably not worthwhile.

Of course, if you have enough data you can analyse the time series of your Greeks and analyse that data, this analysis is only valid for your system. Hopefully, the future will not be dramatically different and your model can be used to forecast the near future as well.

$\endgroup$
1
  • $\begingroup$ ok, many thanks you for your answer. $\endgroup$
    – asardon
    Nov 9 '21 at 8:19
2
$\begingroup$

From a (quite) theoretical point of view, if all you are doing is buying puts and calls across all strikes $0\leq K \leq u$ up to some maximum strike $u$, you are aggregating the corresponding call deltas $N(d_1(K))$ and put deltas $N(d_1(K))-1$ across strikes. Per strike position, this results in a total delta of $2N(d_1(K))-1$.

Now, let us assume a dense strike range, $0\leq K \leq u$ and impose a Black Scholes world without dividends. Let me state two facts:

  1. For a positive random variable $X$ , the integral over its cumulative density function (cdf) over some range $[0,u]$ ($u$ sufficiently large; possibly infinite) equals its expected value $$E_F(X)=\int\limits_0^u1-F(x)dx$$
  2. The option delta $N(d_1(K))$ equals the expected probability of $S_T\geq K$ under the stock measure $\hat{\mathbb{Q}}$, hence $$N(d_1(K))=1-E_\hat{\mathbb{Q}}(\mathbb{1}_{S_T\leq K})=1-F_\hat{\mathbb{Q}}(S_T)$$

1 + 2 together imply $$I\equiv\int_{K=0}^uN(d_1(K))\mathrm{d}K=E_{\hat{\mathbb{Q}}}(S_T)=S_0e^{(r+\sigma^2)\tau}$$ with $r,\sigma,\tau$ the risk free rate, implied volatility and the time to maturity, and thus, your total position delta would equal

$$ \begin{align} \Delta&=\int_{K=0}^u(2N(d_1(K))-1)\mathrm{d}K\\ &=2I-u\\ &=2S_0e^{(r+\sigma^2)\tau}-u \end{align} $$ When setting the upper limit $u$, you could set it such that $N(d_1(u))< 10^{-10}$.

$\endgroup$
1
  • $\begingroup$ wow, cool, didn't think of this, thank you so much for sharing your thoughts, this is neat. $\endgroup$
    – asardon
    Nov 9 '21 at 8:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.