problem finding marginal distribution for a PDF

Question

I need to find the marginal distribution function $f_y$ for $$f_{xy}(u,v)= \begin{cases} 1\over u, & \text{$u\ge 1, 0\le v \le {1 \over u}$} \\ 0, & \text{else} \\ \end{cases}$$

my problem is with the domain $1<u< \infty$ for there is no convergence for the integral.

Why do you use $u,v$ for the variables and not $x,y$. Just notation, nothing false. — Jimmy R., Feb 24 '14 at 16:13
that's from a past test, and that's how the prof. wrote the question... — user107761, Feb 24 '14 at 16:15
For any given $v$, $u$ goes from $1$ to $1/v$ not to $\infty$. — Jimmy R., Feb 24 '14 at 16:16
why? and then, when x goes to $\infty$ and when till the function's graph? — user107761, Feb 24 '14 at 16:18
No, x (or u) depends on y (or v). You cannot let it go to $\infty$. Wait, I will try to write an answer — Jimmy R., Feb 24 '14 at 16:19
Do, you see what I mean? You can always draw a picture of the support of the function (not of the function itself!). That will help you to see, the range in which $u$ can vary for every given $v$. — Jimmy R., Feb 24 '14 at 16:32

score 1 · Answer 1 · answered Feb 24 '14 at 16:25

As usual, including the indicator functions in the densities makes the problems disappear... Here the joint density is $$ f_{X,Y}(u,v)=\frac1u\cdot\mathbf 1_{0\lt v\lt1}\cdot\mathbf 1_{1\lt u\lt1/v}, $$ hence $$ f_Y(v)=\int_\mathbb Rf_{X,Y}(u,v)\,\mathrm du, $$ that is, $$ f_Y(v)=\mathbf 1_{0\lt v\lt1}\int_\mathbb R\frac1u\,\mathbf 1_{1\lt u\lt1/v}\,\mathrm du=\mathbf 1_{0\lt v\lt1}\int_1^{1/v}\frac1u\,\mathrm du=-\log v\cdot\mathbf 1_{0\lt v\lt1}. $$

problem finding marginal distribution for a PDF

1 Answers1