# finding the maximum likelihood and probability distribution function to fit to data

There are data on a pathogen population $$N(t)$$ measured over time (t). I am using the ODE of, $${dN\over dt}=-\lambda N$$ to model its decline over time.

So, $$N(t)=N_0 e^{- \lambda t}$$.
However, I want to obtain a maximum likelihood function related to $$N(t)$$. For that I have to get the probability distribution function to represent $$N(t)$$.

To convert $$N(t)$$ to a PDF, I used a normalizing factor $$c$$,
and $$c={\lambda\over N_0}$$.

So, can I represent $$N(t)$$ as a exponential distribution? However, as $$N(t)$$ is discrete I don't know what the distribution should be.

The final aim is to using maximum likelihood estimation (MLE), to estimate the parameter $$\lambda$$.
The data is as follows:

$$t=[0;1;2;3;4;5;6];$$

$$N(t)=[350000,210000,80000,20000,100,100,100];$$

Although I can simply fit the $$N(t)$$ data using a method such as least squares, I want to know how to do it using MLE and how to find the PDF

$$N(t)$$ is a deterministic population model, telling you population at time $$t$$. It decays with parameter $$\lambda$$.

You wish to estimate the parameter $$\lambda$$.

You have the observations $$(t, N(t))$$ for various $$t$$.

What is the likelihood of observing $$(t, N(t))$$ for a specific $$t$$?

You would need to come up with a sensible model (with parameter $$\lambda$$) for this. At present, your model is deterministic. If $$N(t) \sim Exp( \lambda t)$$, that means that you'd expect $$N(t)$$ to take large values with less probability as $$t$$ increases, which seems fairly reasonable.

At present, your $$N_0 e^{-\lambda t}$$ isn't a probability density, not sure where you got that idea.

Define $$f_t(x;\lambda)$$ to be the density function at population size $$x$$ with parameter $$\lambda$$ at time $$t$$.

Then what you'd do is take the likelihood of your data.

Likelihood $$= \prod_{t=1}^T f_{t}(N(t); \lambda)$$.

Take the natural logarithm of your likelihood.

To maximise it, you differentiate the logarithm you just found with respect to $$\lambda$$, then set it equal to zero.

To be honest what I'd do here is do a least squares fit of $$\sum_{t=1}^{T} (N(t)-N_0 e^{-\lambda t})^2$$ and minimise this under $$\lambda$$.

• Thank you for the answer. Actually, the part that I am struggling with is defining $f_t(x;\lambda)$. Can you please give any help on defining that. I know $N_0 e^{-\lambda t}$ is not a probability density and that's why a normalizing factor of $c={\lambda\over N_0}$ was used as the PDF should integrate to 1. – sam_rox Apr 5 at 7:35
• Yeah I can see the problem. I was thinking about $f_t(x;\lambda) = c e^{-\lambda t x}$. Then the chance of getting a high value for $x$ goes down as $t$ increases. – George Dewhirst Apr 5 at 7:38
• If it is$f_t(N(t);\lambda) = c e^{-\lambda t N(t)}$ does it integrate to 1? Shouldn't it be $f_t(N(t);\lambda) = c e^{-\lambda t \over N(t)}$ – sam_rox Apr 5 at 7:57
• Yeah it integrates to $1$ by changing what $c$ is – George Dewhirst Apr 5 at 8:04
• @ George Dewhirst Thank you very much for helping me out so far. I also calculated the cumulative distribution function (CDF) and the expected population size of time $t$ of the PDF $f_t(N(t);\lambda) = c e^{-\lambda t N(t)}$. Can you please let me know if these are correct. CDF= $1-{N_0e^{-\lambda N(t)t}\over N(t)}$ and Expected value= $1 \over \lambda N_0$ – sam_rox Apr 5 at 11:23