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












0












$begingroup$


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?






probability probability-theory statistics probability-distributions







share|cite|improve this question









$endgroup$












  • $begingroup$
    $f_X(x)$ cannot depend on $y$. In fact, the variables are not independent.
    $endgroup$
    – Kavi Rama Murthy
    Dec 6 '18 at 8:06










  • $begingroup$
    Note that your expression for $f_X(x)$ also depends on variable $y$ hence makes no sense at all.
    $endgroup$
    – drhab
    Dec 6 '18 at 8:21










  • $begingroup$
    @drhab - it may make some sort of sense as a conditional likelihood
    $endgroup$
    – Henry
    Dec 6 '18 at 8:33
















0












$begingroup$


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?






probability probability-theory statistics probability-distributions







share|cite|improve this question









$endgroup$












  • $begingroup$
    $f_X(x)$ cannot depend on $y$. In fact, the variables are not independent.
    $endgroup$
    – Kavi Rama Murthy
    Dec 6 '18 at 8:06










  • $begingroup$
    Note that your expression for $f_X(x)$ also depends on variable $y$ hence makes no sense at all.
    $endgroup$
    – drhab
    Dec 6 '18 at 8:21










  • $begingroup$
    @drhab - it may make some sort of sense as a conditional likelihood
    $endgroup$
    – Henry
    Dec 6 '18 at 8:33














0












0








0





$begingroup$


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?






probability probability-theory statistics probability-distributions







share|cite|improve this question









$endgroup$




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?






probability probability-theory statistics probability-distributions




probability probability-theory statistics probability-distributions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 6 '18 at 7:57









Cuca BeludoCuca Beludo

226




226












  • $begingroup$
    $f_X(x)$ cannot depend on $y$. In fact, the variables are not independent.
    $endgroup$
    – Kavi Rama Murthy
    Dec 6 '18 at 8:06










  • $begingroup$
    Note that your expression for $f_X(x)$ also depends on variable $y$ hence makes no sense at all.
    $endgroup$
    – drhab
    Dec 6 '18 at 8:21










  • $begingroup$
    @drhab - it may make some sort of sense as a conditional likelihood
    $endgroup$
    – Henry
    Dec 6 '18 at 8:33


















  • $begingroup$
    $f_X(x)$ cannot depend on $y$. In fact, the variables are not independent.
    $endgroup$
    – Kavi Rama Murthy
    Dec 6 '18 at 8:06










  • $begingroup$
    Note that your expression for $f_X(x)$ also depends on variable $y$ hence makes no sense at all.
    $endgroup$
    – drhab
    Dec 6 '18 at 8:21










  • $begingroup$
    @drhab - it may make some sort of sense as a conditional likelihood
    $endgroup$
    – Henry
    Dec 6 '18 at 8:33
















$begingroup$
$f_X(x)$ cannot depend on $y$. In fact, the variables are not independent.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 8:06




$begingroup$
$f_X(x)$ cannot depend on $y$. In fact, the variables are not independent.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 8:06












$begingroup$
Note that your expression for $f_X(x)$ also depends on variable $y$ hence makes no sense at all.
$endgroup$
– drhab
Dec 6 '18 at 8:21




$begingroup$
Note that your expression for $f_X(x)$ also depends on variable $y$ hence makes no sense at all.
$endgroup$
– drhab
Dec 6 '18 at 8:21












$begingroup$
@drhab - it may make some sort of sense as a conditional likelihood
$endgroup$
– Henry
Dec 6 '18 at 8:33




$begingroup$
@drhab - it may make some sort of sense as a conditional likelihood
$endgroup$
– Henry
Dec 6 '18 at 8:33










1 Answer
1






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$begingroup$

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.






share|cite|improve this answer









$endgroup$













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    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 8:09









        Kavi Rama MurthyKavi Rama Murthy

        65.1k42766




        65.1k42766






























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