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?
Best Answer
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 $1/y_1$.