3 ms·
To stretch the analogy a little further, to take an observation of the market, you have to buy or sell which in turn affects the price. It's very much like the
by Junk_Collector 7y ago
To stretch the analogy a little further, to take an observation of the market, you have to buy or sell which in turn affects the price. It's very much like the effect of measuring a quantum particle, the impact of a photon is enough to change to observation so you have an inherent uncertainty to everything. Because of this, I don't think you can ever distinguish between noise and meaningful trading with respect to a Nyquist limit in the market. If you tried to for instance, look at the price of all stocks over time and find the 2-D frequency function required to represent that, it is going to creep up based on your measurement rate. Another way to describe it might be that your observability and controllability vectors are not orthogonal.
This also skates around the issue of defining what a Nyquist rate of the market even means in real terms. But, if you want to use control theory to model the market, it's important to know that faster does not inherently mean more accurate. In the simple analogy, increasing trading frequency will improve measurement results, up to a point, after which it will actually likely result in decreasing accuracy.
I've also ignored the whole conflation of frequency and group delay in these analogies to keep things simpler as well.
- bumby 7y agoI don't want to oversimplify your stance, I'm just trying to distill it to something I understand. In the analogy above, would you think this tipping point where the signal is no longer improved is related to the volume of the buyer compared to the overall volume? In other words, Warren Buffet buying a huge portion of a penny stock will drive past that frequency tipping point easily while lil ol' me buying a few shares of an index fund will have effectively little to no impact?