# Unsure if the way I calculated the Maximum Likelihood estimator is correct.

We take a distribution $$D = {x_1,...,x_n}$$ from a diamond shaped area with length of $$\theta$$ and diagonals of $$2\theta$$.

I have the following density function : $$f_{\theta }\left( x\right) = \dfrac{1}{2\theta^2}$$ if $$||x||_1 = |x_1| + |x_2| \leq \theta$$ ($$x_1$$ and $$x_2$$ are the axis of the diagram containing the diamond shape) and $$0$$ otherwise.

The density function means that if the norm of a point x is inferior to $$\theta$$ then our point x is within the diamond shape. I want to find which value of the parameter $$\theta$$ makes it more likely that this is the case for all the points in my distribution D.

We want to find \theta_{MLE} = \begin{aligned} argmax \\ \theta \in \mathbb{R} \end{aligned} f_{\theta }\left( x_1,x_2,x_3,...,x_n\right) (We assume independence).

To make it easier to calculate I use the log to get sums instead of products :

\theta_{MLE} = \begin{aligned} argmax \\ \theta \in \mathbb{R} \end{aligned} \sum log (f_{\theta }\left( x_i\right)) (We assume independence).

I let $$L(\theta) = log(f_{\theta }\left( x_i\right))$$

Then I do the following partial derivative: $$\dfrac{\partial L(\theta)}{\partial \theta} = -\dfrac{1}{\theta^3}$$

Now, I let $$f'_{\theta }\left( x\right) = -\dfrac{1}{\theta^3}$$ if some condition regarding and $$0$$ otherwise.

Can I assume that

\theta_{MLE} = \begin{aligned} argmax \\ \theta \in \mathbb{R} \end{aligned} \sum log (f'_{\theta }\left( x_i\right)) is the maximum likelihood estimation for the parameter $$\theta$$? If not, what went wrong in my approach?

The reason I'm using $$argmax$$ is because we try to find the maximum likelihood of $$\theta$$ for each of the points in D. The greatest value of the maximum likelihood overall for the distribution D is the greatest value of the maximum likelihood of a point in D that is greater than all the other values likelihood estimation we have for other points in D. That's why we keep it by doing argmax.

• Why are you maximizing the log of the partial derivative in the last line? Sep 29, 2020 at 18:41
• The support of the distribution is the single most important thing here. So as written, your density is meaningless. Sep 29, 2020 at 18:45
• I made a mistake the support for the distribution is that $||x||_1 = |x_1| + |x_2| \leq \theta$ or the value of the function is 0. Sep 29, 2020 at 19:02
• I have edited the question details. Sep 29, 2020 at 19:03
• So I think your density is$f_\theta(x,y)=\frac1{2\theta^2}$ if $|x|+|y|\le \theta$ and $f_\theta(x,y)=0$ otherwise, where $\theta>0$. And you are trying to find MLE of $\theta$ based on a sample of $n$ observations $(X_1,Y_1),\ldots,(X_n,Y_n)$. Is that right? Sep 29, 2020 at 19:12

For $$\theta>0$$, suppose the random vector $$(X,Y)$$ has density

$$f_\theta(x,y)=\begin{cases}\frac1{2\theta^2}&,\text{ if }|x|+|y|\le \theta \\ 0 &, \text{ otherwise} \end{cases}$$

This describes a uniform distribution on the region $$\left\{(x,y)\in \mathbb R^2:|x|+|y|\le \theta\right\}$$, which is what you call the diamond shaped area.

Suppose you are want to find maximum likelihood estimator (MLE) of $$\theta$$ based on a random sample of $$n$$ paired observations $$(X_1,Y_1),\ldots,(X_n,Y_n)$$ drawn from the distribution $$f_{\theta}$$. This means the vectors $$(X_1,Y_1),\ldots,(X_n,Y_n)$$ are all independently distributed with common density $$f_\theta$$.

The likelihood function given this sample is then the joint density of $$(X_1,Y_1),\ldots,(X_n,Y_n)$$:

\begin{align} L(\theta)&=\prod_{i=1}^n f_\theta(x_i,y_i) \\&=\begin{cases}\frac1{(2\theta^2)^n} &,\text{ if }|x_1|+|y_1|\le \theta,\ldots,|x_n|+|y_n|\le \theta \\ 0 &, \text{ otherwise}\end{cases} \\&=\begin{cases}\frac1{(2\theta^2)^n} &,\text{ if }\max\limits_{1\le i\le n}(|x_i|+|y_i|)\le \theta \\ 0 &, \text{ otherwise}\end{cases} \end{align}

This is a decreasing function of $$\theta$$, so $$L(\theta)$$ reaches its maximum for the minimum possible value of $$\theta$$ and that value is your MLE. Differentiation is not required for solving this optimization problem; in fact the likelihood is not differentiable at $$\theta=\max\limits_{1\le i\le n}(|x_i|+|y_i|)$$.

The support of the parent distribution $$f_\theta$$ depends on the unknown parameter $$\theta$$, so you simply cannot ignore it. The joint support of $$(X_1,Y_1),\ldots,(X_n,Y_n)$$ practically determines the parameter space having observed the sample.