Flatten the Curve

@eladar said
@joe-shmo

I disagree about the slowing. The general timeline for this thing was guessed. If you can slow down its growth by about a week, then you max out deaths with one fewer week of exponential growth.

But yeah it also slows growth at the moment buying time to get medical supplies and ventilator production going, something that should have been done about 3 months ago.

If CEO's knew enough to quit their job months ago, the government knew too.

@joe-shmo said
"I disagree about the slowing. The general timeline for this thing was guessed. If you can slow down its growth by about a week, then you max out deaths with one fewer week of exponential growth."

What makes you think by slowing the spread by one week, you have one less week of growth? The thing grows until mass immunity is established, without the intervention of treatm ...[text shortened]... t change the equilibrium point, it simply draws it out over time, flattening and widening the curve.

@eladar said
I still think it grows until the ultraviolet light and heat drives it off. There is a reason why viruses are very rare during the summer months.

@joe-shmo said
Well its going to come back before we have a vaccine for it in 18 months time.

@Eladar

I would only add the following: The common cold is not a novel virus for which we have no immunities. The seasonal change might not be as large an effect when its something completely new to us. Perhaps there is a seasonal element, and I hope there is, but perhaps not... just sayin.

Also, to go back to your original question about verifying the mitigation strategy.

I had to step away from thinking( time for outdoor fresh air and play with the kids ). The total number of dead on a given day according to their unmitigated model could be calculated by:

P/10^5 * Integral [0 to t] { f (t) } dt

Where:
P = US Population
t = time in days
f(t) = the sited death rate per day per 10^5 people.

I was initially concerned about accounting for a decreasing population and was getting all wonkers with units, but even if 2.2 million perish its only a 0.7% change. So I don't think its massively influential. none the less:

If we start with the fact that the rate of change of US population should be proportional to the death rate from COVID:

where k = 1/10^5

dP/dt = -k * P * f(t)

We can say that the population of the US on any given day can be found from solving that differential eq.

P(t) = P_o * e^[ -k Int{ f(t) } dt

Then the number of deaths up to any given day is

D(t) = P_o - P(t) = P_o * [ 1 - e^(-k* Int { f(t) } dt ) ]

We don't have an explicitly function for f(t), but we do have the graph. So we could do some approximation again with regressions on f(t). So that should show you on any given day how good we are doing compared to unmitigated strategy. However, right now the graph is virtually indistinguishable from 0. As is computing the death rate using a secant line approximation with the actual data.

(320 * 10^6)/10^5 * F(t) = 1027( Deaths on 25th) - 780 (Deaths on 24th) = 247/3200

The empirical result for f(t) is roughly 0.07 today, or basically in agreement with the unmitigated graph.

@joe-shmo said
@Eladar

I would only add the following: The common cold is not a novel virus for which we have no immunities. The seasonal change might not be as large an effect when its something completely new to us. Perhaps there is a seasonal element, and I hope there is, but... just sayin.

@Eladar

Well..if what your saying is accurate, then dragging it out has some benefits. And I hope it shows that.

@joe-shmo

If you are worried about having so many pulls you change the probability, the general rule is 10 percent.

As long as you do not change the population by 10 percent, your outcome will not be greatly affected.

@eladar said
@joe-shmo

If you are worried about having so many pulls you change the probability, the general rule is 10 percent.

As long as you do not change the population by 10 percent, your outcome will not be greatly affected.

@joe-shmo

I think you are missing a 1/t in your equation.

Average Value of a function is ...

1/(b-a) Integral from a to b of f (t) dt


Although technically I think you are suppose to have different letters in your integral than in your function.

So if you go from 0 to t, then it should be the integral of f (x) dx

@eladar said
@joe-shmo

I think you are missing a 1/t in your equation.

Average Value of a function is ...

1/(b-a) Integral from a to b of f (t) dt


Although technically I think you are suppose to have different letters in your integral than in your function.

So if you go from 0 to t, then it should be the integral of f (x) dx

@joe-shmo

Ok, I thought your f (t) was your death rate function.

Separating your variables helped me to see what you are doing, thanks.

D(t) = Int { f(t) dt } 

Without knowing the functional form of f(t) that is as far as we get symbolically. 



D (t)=F (t)-F (0)

@eladar said
@joe-shmo

Ok, I thought your f (t) was your death rate function.

Separating your variables helped me to see what you are doing, thanks.

D(t) = Int { f(t) dt } 

Without knowing the functional form of f(t) that is as far as we get symbolically. 



D (t)=F (t)-F (0)

@joe-shmo said
"Without knowing the functional form of f(t) that is as far as we get symbolically."



D (t)=F (t)-F (0)

What is your point with this? From my perspective, it seems as though you are either trying to show me that you understand, or ( what I'm leaning towards) you are trying to "put me in my place" by adding a "further" symbolic step that is effectivel ...[text shortened]... ively) the death rate function...just divided by a constant. So, I'm not sure if you are following?

@eladar said
I was just taking it one step further using the fundamental theorem of calculus.

With the 1/t I was just taking myself in a circle.