X and Y are independent? $f(x,y) = \frac{e^{-y}\:e^{-\frac{x}{y}}}{y}$.

Question

Let

$$ f_{X,Y}(x,y) = \frac{e^{-y}\:e^{-\frac{x}{y}}}{y}$$

It's okay to do this:

$$f_Y(y) = e^{-y}, y > 0 $$

This was achieved by solving:

$$f_Y(y) =\int_{0}^{+\infty} \frac{e^{-y}\:e^{-\frac{x}{y}}}{y} dx$$

and

$$f_X(x) = \frac{e^{-\frac{x}{y}}}{y}, x>0 $$

Obtained by:

$$ f_{X,Y}(x,y) = e^{-y} \frac{e^{-\frac{x}{y}}}{y}$$

$$f_X(x) = \int_{0}^{+\infty} \frac{e^{-\frac{x}{y}}}{y} dx = 1 $$

Therefore is density.

This guarantees independence?

Answer 1

Hint: if the variables are independent then we must have $f_{X,Y} (x,y)=g(x)h(y)$ for some $g$ and $h$. By first putting $y=x$ and writing $g$ in terms of $h$ try to get a contradiction.