@eladar saidWhat I do is least squares fit on the actual data. As Deppthought pointed out it is very ambitious to try to make a model from very few data points.
Ponderable,
How did you calculate your margin of error for your 95 percent confidence interval? I do not think it is the same technique as stats.
In fact I think to make a logistic model with just the onset of the growth is impossible. I did for fun a handful of logistic models wit the expected (by the US president) number of death to be 100000 to 250000. I get of course all kinds of curves (depending on the date I set for reaching the number. Even if I apply the condition taht the curves are symmetrical their ascend don't deviate too much from ym original curve which holds unexpecetdly well.
And just to make this point: I just did an exponential curve on very few points, putting down a number which is reasonable for an unchecked exponential growth. I expect still that it will end with a reasonable hit at the given date (day 42, April 11th). The growth won't be impeded by sparsity of infectable people up to that number. But we will see what the measures will have as effect (I fear however that mortality will go up with sparsity of equipment).
@Ponderable
If I am not mistaken, after you find the sum of the squared residuals and take that square root, you are finding the standard deviation of the sample, so now you divide that by the square root of the number of samples to estimate the true sd, then multiply by 1.96.
@ponderable saidI've spent a couple of hours today playing around with my spreadsheet. The results, including total number of deaths and when the point of inflection is depends rather sensitively on how much of the data one includes trying to estimate slope and intercept of the log of the number of deaths. The early data is the noisiest and the later data is already away from pure exponential growth, as you can see just by clicking "logarithmic" on the graphs on worldometer.com. In fact the UK graph has this kink around March 13th. I'll try looking at deaths per day, and see if I can get at the parameters from;
What I do is least squares fit on the actual data. As Deppthought pointed out it is very ambitious to try to make a model from very few data points.
In fact I think to make a logistic model with just the onset of the growth is impossible. I did for fun a handful of logistic models wit the expected (by the US president) number of death to be 100000 to 250000. I get of c ...[text shortened]... measures will have as effect (I fear however that mortality will go up with sparsity of equipment).
f'(t) ~ (f(t + 1) - f(t))/1
f'(t) = (m/a)f(t) (a - f(t)) = m f(t) - (m/a)f(t)²
@Ponderable
For your total deaths exponential function, what day number would be April 4ths total deaths?
I think I might be off by a day.
I did a new regression from the earlier data, but started with March 14 being day 14 instead of day 1. New equation
Daily deaths = .26829×1.2873^t
Total deaths= .19584×1.37327^t
Using e, that would be about...
.26829*e^(.25259*t)
.19584*e^(.31720*t)
Good news is for the last couple of days these models have been overprojecting. The residuals are negative.
I've plotted daily deaths against cumulative deaths and it is looking like an approximation to a quadratic curve. So if g(t) is the number of deaths on day t, and
f(t) = Σₛ₌₀...ₜ g(s) then we have:
g(t) ~ f'(t) = (m/a)f(t)(a - f(t)) = m f(t) - (m/a)f(t)² = m f(t) (1 - f(t)/a)
Which means we can try to extract the important parameters using quadratic regression. Irritatingly OpenOffice doesn't have that as a built in, but I think it shouldn't be too hard to implement (failing that I can just write a c program, but it feels like using a press-hammer to crack a nut).
It dawned on me how to do this. Let g(t) be daily deaths and f(t) be cumulative deaths, then:
g(t)/f(t) = m f(t) - (m/a)
and we can do linear regression. Using the UK death figures from the 14/3 I get:
m = 0.2595 (0.0239)
(m/a) = 2.705E-5 (1.51E-5)
for a total number of deaths of 9,600. But the uncertainty in m/a is huge. A guesstimate of the 95% confidence interval gives: [0, 20295].
For the US figures so far I get 72,000 deaths, m = 0.581 (0.021) (m/a) = -8.061E-6 (1.17E-5), so a hack's guess at the 95% confidence interval is: [0, 282000] for the US.
Some more playing around later. At the expense of using rather few data points but having the criterion that the number of total deaths should be more than 1,000 which means statistical fluctuations die down I have:
UK data
m = 0.2235 (0.0317)
m/a = 1.209E-5 (1.079E-5)
a = 15,870; 95% CI = 0 to 40,100
US data
m = 0.5617 (0.0173)
(m/a) = 2.966 E-6 (5.03E-6)
a = 189,350; 95% CI = 0 to 830,000
Next I'll try to get a handle on when the inflection point is and see if I can reproduce the mortality curves - which I couldn't with the exponential estimation approach.
@deepthought saidI'm getting behind on the models! I'm having trouble seeing this algebra ( currently not sure if I'm loosing it, or you are again doing some algebraic manipulation above my head):
It dawned on me how to do this. Let g(t) be daily deaths and f(t) be cumulative deaths, then:
g(t)/f(t) = m f(t) - (m/a)
and we can do linear regression. Using the UK death figures from the 14/3 I get:
m = 0.2595 (0.0239)
(m/a) = 2.705E-5 (1.51E-5)
for a total number of deaths of 9,600. But the uncertainty in m/a is huge. A guesstimate of the 95% confiden ...[text shortened]... = -8.061E-6 (1.17E-5), so a hack's guess at the 95% confidence interval is: [0, 282000] for the US.
If
g(t) = m f(t) ( 1 - f(t)/a )
Then
g(t)/f(t) = m ( 1 - f(t)/a )
g(t)/f(t) = m - (m/a)*f(t)
How are you getting?
g(t)/f(t) = m*f(t) - m/a
https://covid19.healthdata.org/
2644 is the max predicted death rate in 11 days from now.
My quadratic model predicts 1644. It is underestimating at the moment but by the time we reach the max it should be a good estimate.
With 1644 as the upper bound, the max total death is bounded from above by 53596.
The middle value with this new upper bound would be 50309.
Of course this assumes their projected max death day is correct.