5 ms·

If the virus spreads exponentially, you would expect most cases to be presymptomatic at any given time.

Traster 7y ago
Even an exponentially spreading virus is only in its exponential stage when the number of infected is significantly lower than the total population.
- graycat 7y ago
  Right, e.g., follows the (a) logistic curve.
oldgradstudent 7y ago
In an infinitely large population.
- bonzini 7y ago
  Also in a finite population, as long as the virus is spreading exponentially (which it doesn't do forever in a finite population).
bluenose69 7y ago
This is an astute point, elegantly stated.
- oldgradstudent 7y ago
  Vacous, but elegantly stated. In a small population like an aircraft carrier, the virus stops spreading exponentially very quickly.
  - jhayward 7y ago
    At roughly 10% infected the epidemic curve is in the steeply exponential portion of the curve, not the flattened end. And, as a request, perhaps you could glance at the 'principle of charity' and choose some other term than 'vacuous' to describe an argument that your response clearly did not give sufficient thought to.
    - oldgradstudent 7y ago
      > And, as a request, perhaps you could glance at the 'principle of charity' and choose some other term than 'vacuous' to describe an argument that your response clearly did not give sufficient thought to. Vacous is also a technical term[1]. The proposition If A then B is said to be vacously true if A does not hold.
watwut 7y ago
Not when the amount of people is limited. Exponential spread in isolated group of 3000 people necessary becomes non exponential at some point.
- graycat 7y ago
  > becomes non exponential at some point From some simple calculus, becomes the (a) logistic curve.
  - watwut 7y ago
    Regardless of calculus, it becomes constant at one point. Latest when everyone is infected.
    - graycat 7y ago
      Yes, the logistic curves are lazy S curves and approach a limit asymptotically from below. Here's the calculus: At FedEx, the BoD wanted some revenue projections. We knew (1) the current revenue and (2) the revenue for serving all the target customers. So, for time t in days, let y(t) be the revenue per day at time t, t = 0 corresponds to the present, y(0) is the current revenue. Let b be the revenue when serving all the target customers. Assume that the rate of growth in revenue, in dollars per day, that is, the calculus y'(t) = d/dt y(t) is directly proportional to (a) the current number of customers talking about the service and (b) the number of target customers not yet customers hearing about the service, that is proportional to y(t) and (b - y(t)). Then for some constant of proportionality k we have the first order ordinary differential equation initial value problem y'(t) = k y(t) (b - y(t)) Then in TeX source code, the solution is y(t) = { y(0) b e^{bkt} \over y(0) \big ( e^{bkt} - 1 \big ) + b} which is the famous logistic curve. So, this curve starts as an exponential and convex, rises, becomes concave, and approaches b asymptotically from below. [On a Friday, I did the calculus, and a Senior VP of Planning and I picked a reasonable value of k, and I submitted the graph. The next day at 8 AM was a BoD meeting. The graph was presented at the start, and two guys from BoD Member General Dynamics asked how the graph was constructed. At noon in my office I got a call from another SVP asking if I knew about the graph and could come to the meeting. When I arrived, everyone was unhappy; the two General Dynamics were standing in the hall with their bags packed and plane tickets back to Texas. FedEx was about to die. I calculated a few points on the graph; the General Dynamics guys stayed; and the rest is history. The problem was, from the beginning of the meeting at 8 AM until noon, no one at the meeting could explain how the graph was calculated. The General Dynamics guys lost patience with FedEx, went to their rooms, packed their bags, and as a last effort returned to the meeting.] Well the little calculus exercise is, say, an axiomatic derivation of viral growth, say, as for TV sets in US homes a few years around 1950. Well, a glance at the derivation shows that the solution should also be good for viral growth of viruses. Can get the same thing from a continuous time, discrete state space Markov process subordinated to a Poisson process as in the book on stochastic processes by E. Çinlar, long at Princeton in operations research, etc.

Even an exponentially spreading virus is only in its exponential stage when the number of infected is significantly lower than the total population.

graycat 7y ago
Right, e.g., follows the (a) logistic curve.

oldgradstudent 7y ago

In an infinitely large population.

bonzini 7y ago
Also in a finite population, as long as the virus is spreading exponentially (which it doesn't do forever in a finite population).

bluenose69 7y ago

This is an astute point, elegantly stated.

oldgradstudent 7y ago
Vacous, but elegantly stated. In a small population like an aircraft carrier, the virus stops spreading exponentially very quickly.
- jhayward 7y ago
  At roughly 10% infected the epidemic curve is in the steeply exponential portion of the curve, not the flattened end. And, as a request, perhaps you could glance at the 'principle of charity' and choose some other term than 'vacuous' to describe an argument that your response clearly did not give sufficient thought to.
  - oldgradstudent 7y ago
    > And, as a request, perhaps you could glance at the 'principle of charity' and choose some other term than 'vacuous' to describe an argument that your response clearly did not give sufficient thought to. Vacous is also a technical term[1]. The proposition If A then B is said to be vacously true if A does not hold.

watwut 7y ago

Not when the amount of people is limited. Exponential spread in isolated group of 3000 people necessary becomes non exponential at some point.

graycat 7y ago
> becomes non exponential at some point From some simple calculus, becomes the (a) logistic curve.
- watwut 7y ago
  Regardless of calculus, it becomes constant at one point. Latest when everyone is infected.
  - graycat 7y ago
    Yes, the logistic curves are lazy S curves and approach a limit asymptotically from below. Here's the calculus: At FedEx, the BoD wanted some revenue projections. We knew (1) the current revenue and (2) the revenue for serving all the target customers. So, for time t in days, let y(t) be the revenue per day at time t, t = 0 corresponds to the present, y(0) is the current revenue. Let b be the revenue when serving all the target customers. Assume that the rate of growth in revenue, in dollars per day, that is, the calculus y'(t) = d/dt y(t) is directly proportional to (a) the current number of customers talking about the service and (b) the number of target customers not yet customers hearing about the service, that is proportional to y(t) and (b - y(t)). Then for some constant of proportionality k we have the first order ordinary differential equation initial value problem y'(t) = k y(t) (b - y(t)) Then in TeX source code, the solution is y(t) = { y(0) b e^{bkt} \over y(0) \big ( e^{bkt} - 1 \big ) + b} which is the famous logistic curve. So, this curve starts as an exponential and convex, rises, becomes concave, and approaches b asymptotically from below. [On a Friday, I did the calculus, and a Senior VP of Planning and I picked a reasonable value of k, and I submitted the graph. The next day at 8 AM was a BoD meeting. The graph was presented at the start, and two guys from BoD Member General Dynamics asked how the graph was constructed. At noon in my office I got a call from another SVP asking if I knew about the graph and could come to the meeting. When I arrived, everyone was unhappy; the two General Dynamics were standing in the hall with their bags packed and plane tickets back to Texas. FedEx was about to die. I calculated a few points on the graph; the General Dynamics guys stayed; and the rest is history. The problem was, from the beginning of the meeting at 8 AM until noon, no one at the meeting could explain how the graph was calculated. The General Dynamics guys lost patience with FedEx, went to their rooms, packed their bags, and as a last effort returned to the meeting.] Well the little calculus exercise is, say, an axiomatic derivation of viral growth, say, as for TV sets in US homes a few years around 1950. Well, a glance at the derivation shows that the solution should also be good for viral growth of viruses. Can get the same thing from a continuous time, discrete state space Markov process subordinated to a Poisson process as in the book on stochastic processes by E. Çinlar, long at Princeton in operations research, etc.