If $F_X(x), F_Y(y)$ are two CDF's for r.v.'s $X$ and $Y$, will $F_{X,Y}(x,y) < F_X(x) cdot F_Y(y)$ if $X$...
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If $F_X(x), F_Y(y)$ are two CDF's for r.v.'s $X$ and $Y$, will $F_{X,Y}(x,y)
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)For random variables $X$ and $Y$, $F_{X,Y}(x,y)=F_X(x)F_Y(y)$ if and only if $f_{X,Y}(x,y)=f_X(x)f_Y(y)$If the PDF satisfies $f(x_1,y_1)f(x_2,y_2)leq f(x_1,y_2)f(x_2,y_1)$ then the CDF's satisfy $F(a,b)leq F_X(a)F_Y(b)$.What is CDF $F_X(x) $ and $F_Y(y)$?Prove $X$ and $Y$ are not independentHow to interpret the the fact that sum of two independent uniform r.v. forms a triangular distribution?Is $int_{-infty}^{infty}f_X(z-y)f_Y(y)dy$ a PDF where $f_x,f_Y$ are both PDFSProve that if $X,Y$ are independent random variables with densities $f_X, f_Y$ then,Showing that two random variables are independentHow to check if two random variables are independent without calculating their density functions?Clarifying the statement $f(x,y) = f_X(x) f_Y(y)$ for all $x, y$ (when $X$ and $Y$ are independent)
$begingroup$
If $F_X(x), F_Y(y)$ are two CDF's for r.v.'s $X$ and $Y$, if $X$ and $Y$ are independent, then:
$$
F_{X,Y}(x,y) = F_X(x) cdot F_Y(y)
$$
However, I am wondering more generally if
$$
F_{X,Y}(x,y) < F_X(x) cdot F_Y(y)
$$
if $X$ and $Y$ are not independent or if it depends how the joint distribution is constructed?
probability
asked Mar 23 at 23:33
user321627user321627
1,000515
$endgroup$
add a comment |
$begingroup$
If $F_X(x), F_Y(y)$ are two CDF's for r.v.'s $X$ and $Y$, if $X$ and $Y$ are independent, then:
$$
F_{X,Y}(x,y) = F_X(x) cdot F_Y(y)
$$
However, I am wondering more generally if
$$
F_{X,Y}(x,y) < F_X(x) cdot F_Y(y)
$$
if $X$ and $Y$ are not independent or if it depends how the joint distribution is constructed?
probability
asked Mar 23 at 23:33
user321627user321627
1,000515
$endgroup$
add a comment |
$begingroup$
If $F_X(x), F_Y(y)$ are two CDF's for r.v.'s $X$ and $Y$, if $X$ and $Y$ are independent, then:
$$
F_{X,Y}(x,y) = F_X(x) cdot F_Y(y)
$$
However, I am wondering more generally if
$$
F_{X,Y}(x,y) < F_X(x) cdot F_Y(y)
$$
if $X$ and $Y$ are not independent or if it depends how the joint distribution is constructed?
probability
asked Mar 23 at 23:33
user321627user321627
1,000515
$endgroup$
If $F_X(x), F_Y(y)$ are two CDF's for r.v.'s $X$ and $Y$, if $X$ and $Y$ are independent, then:
$$
F_{X,Y}(x,y) = F_X(x) cdot F_Y(y)
$$
However, I am wondering more generally if
$$
F_{X,Y}(x,y) < F_X(x) cdot F_Y(y)
$$
if $X$ and $Y$ are not independent or if it depends how the joint distribution is constructed?
probability
probability
asked Mar 23 at 23:33
user321627user321627
1,000515
asked Mar 23 at 23:33
user321627user321627
1,000515
asked Mar 23 at 23:33
user321627user321627
1,000515
asked Mar 23 at 23:33
user321627user321627
1,000515
asked Mar 23 at 23:33
user321627user321627
1,000515
1,000515
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $X=Y$ then $F_X(x)F_Y(x)=(F_X(x))^{2} <F_X(x)=F_{X,Y}(x,x)$ provided $F_X(x)<1$.
answered Mar 23 at 23:41
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
For cases not perfectly dependent, is there a general rule of thumb?
$endgroup$
– user321627
Mar 24 at 2:42
$begingroup$
No. there is no reason to expect an inequalilty of the type you have come up with.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 4:55
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $X=Y$ then $F_X(x)F_Y(x)=(F_X(x))^{2} <F_X(x)=F_{X,Y}(x,x)$ provided $F_X(x)<1$.
answered Mar 23 at 23:41
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
For cases not perfectly dependent, is there a general rule of thumb?
$endgroup$
– user321627
Mar 24 at 2:42
$begingroup$
No. there is no reason to expect an inequalilty of the type you have come up with.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 4:55
add a comment |
$begingroup$
If $X=Y$ then $F_X(x)F_Y(x)=(F_X(x))^{2} <F_X(x)=F_{X,Y}(x,x)$ provided $F_X(x)<1$.
answered Mar 23 at 23:41
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
For cases not perfectly dependent, is there a general rule of thumb?
$endgroup$
– user321627
Mar 24 at 2:42
$begingroup$
No. there is no reason to expect an inequalilty of the type you have come up with.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 4:55
add a comment |
$begingroup$
If $X=Y$ then $F_X(x)F_Y(x)=(F_X(x))^{2} <F_X(x)=F_{X,Y}(x,x)$ provided $F_X(x)<1$.
answered Mar 23 at 23:41
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
If $X=Y$ then $F_X(x)F_Y(x)=(F_X(x))^{2} <F_X(x)=F_{X,Y}(x,x)$ provided $F_X(x)<1$.
answered Mar 23 at 23:41
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Mar 23 at 23:41
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Mar 23 at 23:41
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Mar 23 at 23:41
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
74.9k53270
$begingroup$
For cases not perfectly dependent, is there a general rule of thumb?
$endgroup$
– user321627
Mar 24 at 2:42
$begingroup$
No. there is no reason to expect an inequalilty of the type you have come up with.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 4:55
add a comment |
$begingroup$
For cases not perfectly dependent, is there a general rule of thumb?
$endgroup$
– user321627
Mar 24 at 2:42
$begingroup$
No. there is no reason to expect an inequalilty of the type you have come up with.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 4:55
$begingroup$
For cases not perfectly dependent, is there a general rule of thumb?
$endgroup$
– user321627
Mar 24 at 2:42
$begingroup$
For cases not perfectly dependent, is there a general rule of thumb?
$endgroup$
– user321627
Mar 24 at 2:42
$begingroup$
No. there is no reason to expect an inequalilty of the type you have come up with.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 4:55
$begingroup$
No. there is no reason to expect an inequalilty of the type you have come up with.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 4:55
add a comment |
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