MLE of Bivariate Vector Random Variable: Proof & Explanation

[squenshl]

Homework Statement

Consider the bivariate vector random variable $(X,Y)^T$ which has the probability density function $$f_{X,Y}(x,y) = \theta xe^{-x(y+\theta)}, \quad x\geq 0, y\geq 0 \; \; \text{and} \; \; \theta > 0.$$
I have shown that the marginal distribution of $X$ is $f_X(x|\theta) = \theta e^{-\theta x}, \quad x\geq 0 \; \; \text{and} \; \; \theta > 0.$

My question is why do these two distributions have the same maximum likelihood estimator $\hat{\theta} = \frac{1}{\bar{x}}??$

Homework Equations

The Attempt at a Solution

 

[Ray Vickson]

Homework Statement

Consider the bivariate vector random variable $(X,Y)^T$ which has the probability density function $$f_{X,Y}(x,y) = \theta xe^{-x(y+\theta)}, \quad x\geq 0, y\geq 0 \; \; \text{and} \; \; \theta > 0.$$
I have shown that the marginal distribution of $X$ is $f_X(x|\theta) = \theta e^{-\theta x}, \quad x\geq 0 \; \; \text{and} \; \; \theta > 0.$

My question is why do these two distributions have the same maximum likelihood estimator $\hat{\theta} = \frac{1}{\bar{x}}??$

Homework Equations

The Attempt at a Solution

In one case you observe $(X,Y)=(x,y)$ and estimate $\theta$ from $f_{XY}(x,y)$, which requires finding the maximum of a function of the form $a \theta e^{-x \theta}$, where $a = x e^{-xy}$ is a number. In the other case you observe $X=x$ and estimate $\theta$ from $f_X(x)$, which requires finding the maximum of a function of the form $a \theta e^{-x \theta}$, where $a = 1$. The "scale factor" $a$ does not affect the optimization.

Put it another way: the MLS estimator of $\theta$, based on $f_{XY}(x,y)$ is independent of $y$. Think of $f_X(x)$ as a sum over $y$ of curves $f_{XY}(x,y)$, where the variable is $\theta$ and $x, y$ are just an input parameters. Each of these constituent curves has a maximum at the same point $\theta = 1/x$, so the sum over $y$ is maximized at the same point.