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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?

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

1 Answer 1

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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 $1/y_1$.

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  • $\begingroup$ Thank you, this answer was extremely helpful for me :) $\endgroup$
    – Kevin
    Commented May 1, 2020 at 14:05

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