A more pragmatic model

In a previous article, we explored a method to get the simplest model that describes the growth. However, this simple model is not good enough when we want to fit it into observed data.

Imagine a crowd of infected people arrived at an island some unknown time before a pandemic broke out. After a few weeks of the virus infecting people without interference, we find that we can accurately determine how many people are infected on two specific dates. Can this minimal observation help us to analyse the growth of infections?

The problem explore here is:
Find a model that describes the number of people infected with a virus such that we can fit the model with two observations.

The weakness of the previous attempt #

What is the problem with our previous model? This is what we found:
$$\begin{array}{r}y(t)=e^{\rho t}\end{array}$$
Where $y(t)$ is the total number of infected after day $t$ and $\rho$ is constant.

To get to this answer, we had to assume that $\rho$ is a constant value and that we know the day when the first person started to infect others.

What we consider now is a situation in which we do not know the first day, or even how many people were infected at the start of that day.

What we do have at our disposal is observations about two other days. Let's call one of the observed days, day $t_i$ and the other we call day $t_j$.

What we need is a model that allows us to use these observations to calculate the model parameters.

Deriving the new model #

The process to derive the model is very much the same as previously. This time we go through it with a little less fanfare.

We start with for differential equation:
$$\begin{array}{r}{y}^{\prime }=ϵ+\rho y\end{array}\text{\hspace{1em}(1)}$$
Where $\rho$ is some unknown growth factor and $\epsilon$ is the number of people that are infected every day by those that were part of the initial crowd.

Note that we assume that $\epsilon$ and $\rho$ are constants.

We still have the initial value:
$$\begin{array}{r}y(0)=z_0\end{array}\text{\hspace{1em}(2)}$$
Where $z_0$ is the constant that describes the number of people that were infected on day zero - the number of people in the crowd.

Solving this ODE, gives:

$$\begin{array}{r}y\left(t\right)=\frac{-ϵ+\left(ϵ+\rho z_0\right)e^{\rho t}}{\rho }\end{array}\text{\hspace{1em}(3)}$$

This new model has three parameters and is, therefore, more versatile than the previous model.

Using the data points #

We have two data points. We can write these out as known values:
$$\begin{array}{r}y(t_i)=y_i\\ y(t_j)=y_j\end{array}$$

And now we have two equations to solve the two unknowns in Eq. 3. First, put the $t_i$ data into Eq 3 and solve for $\epsilon$.
$$\begin{array}{r}ϵ=\frac{\rho \left(y_i-z_0{e}^{\rho t_i}\right)}{e^{\rho t_i}-1}\end{array}\text{\hspace{1em}(4)}$$

And then, do the same for $t_j$:
$$\begin{array}{r}ϵ=\frac{\rho \left(y_j-z_0{e}^{\rho t_i}\right)}{e^{\rho t_j}-1}\end{array}\text{\hspace{1em}(5)}$$

The two expressions are equal, and we combine them to solve for $z_0$:

$$\begin{array}{r}{z}_0=\frac{-y_j{e}^{\rho t_i}+y_j+y_i{e}^{\rho t_j}-y_i}{e^{\rho t_j}-e^{\rho t_2}}\end{array}\text{\hspace{1em}(6)}$$

Finally, we combine Eq. 6 and Eq. 4 to get

$$\begin{array}{r}ϵ=\frac{\rho \left(y_j{e}^{\rho t_i}-y_i{e}^{\rho t_j}\right)}{e^{\rho t_j}-e^{\rho t_i}}\end{array}\text{\hspace{1em}(7)}$$

Conclusion #

A more expressive model has been defined in Eq. 3, and we can calculate the parameters of this model using only two observations. This model is more useful if you want to fit the model to actual observations. But essentially, it is based very much on the same principles as before.

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