# How to find MLE from a cumulative distribution function?

I'm new to probability. Given the cumulative distribution function $$f_Y(y)=\theta e^{-y\theta}$$ defined from 0 to infinity, I would like to find the parameter $$\theta$$ such that it maximizes the likelihood function. I first thought that since PDF and CDF are strictly correlated between each other, I tried finding the first derivative of the CDF with respect to $$\theta$$: $$\frac{d}{d\theta}(\theta e^{-y\theta})=0 \implies \theta=\frac{1}{y}$$

Then I tried solving the PDF form the CDF: $$\frac{d}{dy}(\theta e^{-y\theta})=-\theta^2e^{-y\theta}$$ Which gives me the likelihood function for the continuous distribution. Naturally, I calculated the derivative with respect to $$\theta$$ of the likelihood function:

$$\frac{d}{d\theta}(-\theta^2e^{-y\theta})=0 \implies \theta=0 \vee \frac{2}{y}$$

My question is: why do I get two different values for $$\theta$$ with the two different approaches?

The textbook also suggests that for the sample $$Y_1, ..., Y_n$$, the MLE is $$1/{\bar{Y_n}}$$, which still is different from the two results I found. Can someone help me make some clarifications?

• $f_Y$ is a PDF for $y>0$, not CDF. Commented Apr 30, 2020 at 15:34
• Thanks, how can we deduce that it is the former and not the latter? Commented Apr 30, 2020 at 15:38
• In general: First you have to set up the likelihood function (!!!) which is based on a sample with a sample size of $n$ Commented Apr 30, 2020 at 15:43
• @Kevin By properties of PDF/CDF. Commented Apr 30, 2020 at 15:48
• PDF integrates to 1 over the domain it is define. So that's a quick check whether a function is a pdf or not.
– Mdoc
Commented Apr 30, 2020 at 18:02

The MLE estimator is the value of parameter, in your case of $$\theta$$, that maximizes the likelihood of observing a SAMPLE of observations, $$\{Y_1,...,Y_N\}$$. To compute MLE estimator you then need to set up a likelihood function. If the sample observations are i.i.d. then the likelihood function is given by the product of densities of each observation conditional on $$\theta$$.

In your case, the likelihood function is $$L = \prod_{i=1}^N \theta e^{-\theta y_i}$$

Maximizing this function w.r.t $$\theta$$ yields solution

$$\theta = \frac{N}{\sum_i^N y_i} = \frac{1}{\overline{y}_n},$$

where $$\overline{y}_n = \frac{1}{N}\sum_i^N y_i$$.

Notice that this solution agrees with your solution (proposed at the beginning of your post) when you have only one observation, namely $$N=1$$. In that case, MLE estimate is simply $$1/y_1$$.

• Thank you, this answer was extremely helpful for me :) Commented May 1, 2020 at 14:05