X and Y are independent? $f(x,y) = frac{e^{-y}:e^{-frac{x}{y}}}{y}$.
$begingroup$
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
asked Dec 6 '18 at 7:57
Cuca BeludoCuca Beludo
226
$endgroup$
add a comment |
$begingroup$
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
asked Dec 6 '18 at 7:57
Cuca BeludoCuca Beludo
226
$endgroup$
$begingroup$
$f_X(x)$ 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
add a comment |
$begingroup$
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
asked Dec 6 '18 at 7:57
Cuca BeludoCuca Beludo
226
$endgroup$
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
asked Dec 6 '18 at 7:57
Cuca BeludoCuca Beludo
226
asked Dec 6 '18 at 7:57
Cuca BeludoCuca Beludo
226
asked Dec 6 '18 at 7:57
Cuca BeludoCuca Beludo
226
asked Dec 6 '18 at 7:57
Cuca BeludoCuca Beludo
226
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
add a comment |
$begingroup$
$f_X(x)$ 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
add a comment |
1 Answer
1
active
oldest
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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.
answered Dec 6 '18 at 8:09
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
$endgroup$
add a comment |
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1 Answer
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1 Answer
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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.
answered Dec 6 '18 at 8:09
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 6 '18 at 8:09
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 6 '18 at 8:09
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
$endgroup$
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.
answered Dec 6 '18 at 8:09
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
answered Dec 6 '18 at 8:09
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
answered Dec 6 '18 at 8:09
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
answered Dec 6 '18 at 8:09
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
65.1k42766
add a comment |
add a comment |
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$begingroup$
$f_X(x)$ 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