On independence of R.V












0












$begingroup$


We know that if $X$ and $Y$ have joint density $f(x,y)$ and if the ranges of $X$ and $Y$ are not dependent then X,Y are independent iff $f(x,y)= f(x) g(y)$ for some f,g .



Now, can we get same result, if we instead have



$$ P(X leq x, Y > y) = f(x) g(y) $$



That is, if we have the above, does this imply that the random variables are independent?






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you mean $P(Xleq x,Y>y)=P(Xleq x)P(Y>y)$? If so then yes you will have same result.
    $endgroup$
    – John_Wick
    Dec 16 '18 at 23:06










  • $begingroup$
    i meant for some function f and g
    $endgroup$
    – Jimmy Sabater
    Dec 16 '18 at 23:09


















0












$begingroup$


We know that if $X$ and $Y$ have joint density $f(x,y)$ and if the ranges of $X$ and $Y$ are not dependent then X,Y are independent iff $f(x,y)= f(x) g(y)$ for some f,g .



Now, can we get same result, if we instead have



$$ P(X leq x, Y > y) = f(x) g(y) $$



That is, if we have the above, does this imply that the random variables are independent?






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you mean $P(Xleq x,Y>y)=P(Xleq x)P(Y>y)$? If so then yes you will have same result.
    $endgroup$
    – John_Wick
    Dec 16 '18 at 23:06










  • $begingroup$
    i meant for some function f and g
    $endgroup$
    – Jimmy Sabater
    Dec 16 '18 at 23:09
















0












0








0





$begingroup$


We know that if $X$ and $Y$ have joint density $f(x,y)$ and if the ranges of $X$ and $Y$ are not dependent then X,Y are independent iff $f(x,y)= f(x) g(y)$ for some f,g .



Now, can we get same result, if we instead have



$$ P(X leq x, Y > y) = f(x) g(y) $$



That is, if we have the above, does this imply that the random variables are independent?






probability







share|cite|improve this question











$endgroup$




We know that if $X$ and $Y$ have joint density $f(x,y)$ and if the ranges of $X$ and $Y$ are not dependent then X,Y are independent iff $f(x,y)= f(x) g(y)$ for some f,g .



Now, can we get same result, if we instead have



$$ P(X leq x, Y > y) = f(x) g(y) $$



That is, if we have the above, does this imply that the random variables are independent?






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 23:56









Bernard

123k741116




123k741116










asked Dec 16 '18 at 22:56









Jimmy SabaterJimmy Sabater

3,023325




3,023325












  • $begingroup$
    Did you mean $P(Xleq x,Y>y)=P(Xleq x)P(Y>y)$? If so then yes you will have same result.
    $endgroup$
    – John_Wick
    Dec 16 '18 at 23:06










  • $begingroup$
    i meant for some function f and g
    $endgroup$
    – Jimmy Sabater
    Dec 16 '18 at 23:09




















  • $begingroup$
    Did you mean $P(Xleq x,Y>y)=P(Xleq x)P(Y>y)$? If so then yes you will have same result.
    $endgroup$
    – John_Wick
    Dec 16 '18 at 23:06










  • $begingroup$
    i meant for some function f and g
    $endgroup$
    – Jimmy Sabater
    Dec 16 '18 at 23:09


















$begingroup$
Did you mean $P(Xleq x,Y>y)=P(Xleq x)P(Y>y)$? If so then yes you will have same result.
$endgroup$
– John_Wick
Dec 16 '18 at 23:06




$begingroup$
Did you mean $P(Xleq x,Y>y)=P(Xleq x)P(Y>y)$? If so then yes you will have same result.
$endgroup$
– John_Wick
Dec 16 '18 at 23:06












$begingroup$
i meant for some function f and g
$endgroup$
– Jimmy Sabater
Dec 16 '18 at 23:09






$begingroup$
i meant for some function f and g
$endgroup$
– Jimmy Sabater
Dec 16 '18 at 23:09












1 Answer
1






active

oldest

votes


















2












$begingroup$

$P(Xleq x, Y>y)=f(x)g(y)$ implies $P(Xleq x)=lim_{yrightarrow -infty}P(Xleq x, Y>y)=f(x)g(-infty).$ Similarly taking $xrightarrow infty$ we get $P(Y>y)=f(infty)g(y).$ Also taking both the limits simultaneously we get $f(infty)g(-infty)=lim_{yrightarrow -infty}lim_{xrightarrow infty}P(Xleq x, Y>y)=1.$



So we get $P(Xleq x)P(Y>y)=f(x)g(-infty)f(infty)g(y)=f(x)g(y)=P(Xleq x, Y>y).$ From there just taking derivatives w.r.t. both $x$ and $y$ we get $f_{X,Y}(x,y)=f_{X}(x)f_Y(y).$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Passing by infinities is not needed, simply use $$P(Xleqslant x,Y>y)=P(Xleqslant x)-P(Xleqslant x,Yleqslant y)$$
    $endgroup$
    – Did
    Dec 17 '18 at 0:36











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$begingroup$

$P(Xleq x, Y>y)=f(x)g(y)$ implies $P(Xleq x)=lim_{yrightarrow -infty}P(Xleq x, Y>y)=f(x)g(-infty).$ Similarly taking $xrightarrow infty$ we get $P(Y>y)=f(infty)g(y).$ Also taking both the limits simultaneously we get $f(infty)g(-infty)=lim_{yrightarrow -infty}lim_{xrightarrow infty}P(Xleq x, Y>y)=1.$



So we get $P(Xleq x)P(Y>y)=f(x)g(-infty)f(infty)g(y)=f(x)g(y)=P(Xleq x, Y>y).$ From there just taking derivatives w.r.t. both $x$ and $y$ we get $f_{X,Y}(x,y)=f_{X}(x)f_Y(y).$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Passing by infinities is not needed, simply use $$P(Xleqslant x,Y>y)=P(Xleqslant x)-P(Xleqslant x,Yleqslant y)$$
    $endgroup$
    – Did
    Dec 17 '18 at 0:36
















2












$begingroup$

$P(Xleq x, Y>y)=f(x)g(y)$ implies $P(Xleq x)=lim_{yrightarrow -infty}P(Xleq x, Y>y)=f(x)g(-infty).$ Similarly taking $xrightarrow infty$ we get $P(Y>y)=f(infty)g(y).$ Also taking both the limits simultaneously we get $f(infty)g(-infty)=lim_{yrightarrow -infty}lim_{xrightarrow infty}P(Xleq x, Y>y)=1.$



So we get $P(Xleq x)P(Y>y)=f(x)g(-infty)f(infty)g(y)=f(x)g(y)=P(Xleq x, Y>y).$ From there just taking derivatives w.r.t. both $x$ and $y$ we get $f_{X,Y}(x,y)=f_{X}(x)f_Y(y).$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Passing by infinities is not needed, simply use $$P(Xleqslant x,Y>y)=P(Xleqslant x)-P(Xleqslant x,Yleqslant y)$$
    $endgroup$
    – Did
    Dec 17 '18 at 0:36














2












2








2





$begingroup$

$P(Xleq x, Y>y)=f(x)g(y)$ implies $P(Xleq x)=lim_{yrightarrow -infty}P(Xleq x, Y>y)=f(x)g(-infty).$ Similarly taking $xrightarrow infty$ we get $P(Y>y)=f(infty)g(y).$ Also taking both the limits simultaneously we get $f(infty)g(-infty)=lim_{yrightarrow -infty}lim_{xrightarrow infty}P(Xleq x, Y>y)=1.$



So we get $P(Xleq x)P(Y>y)=f(x)g(-infty)f(infty)g(y)=f(x)g(y)=P(Xleq x, Y>y).$ From there just taking derivatives w.r.t. both $x$ and $y$ we get $f_{X,Y}(x,y)=f_{X}(x)f_Y(y).$






share|cite|improve this answer









$endgroup$



$P(Xleq x, Y>y)=f(x)g(y)$ implies $P(Xleq x)=lim_{yrightarrow -infty}P(Xleq x, Y>y)=f(x)g(-infty).$ Similarly taking $xrightarrow infty$ we get $P(Y>y)=f(infty)g(y).$ Also taking both the limits simultaneously we get $f(infty)g(-infty)=lim_{yrightarrow -infty}lim_{xrightarrow infty}P(Xleq x, Y>y)=1.$



So we get $P(Xleq x)P(Y>y)=f(x)g(-infty)f(infty)g(y)=f(x)g(y)=P(Xleq x, Y>y).$ From there just taking derivatives w.r.t. both $x$ and $y$ we get $f_{X,Y}(x,y)=f_{X}(x)f_Y(y).$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 23:19









John_WickJohn_Wick

1,616111




1,616111












  • $begingroup$
    Passing by infinities is not needed, simply use $$P(Xleqslant x,Y>y)=P(Xleqslant x)-P(Xleqslant x,Yleqslant y)$$
    $endgroup$
    – Did
    Dec 17 '18 at 0:36


















  • $begingroup$
    Passing by infinities is not needed, simply use $$P(Xleqslant x,Y>y)=P(Xleqslant x)-P(Xleqslant x,Yleqslant y)$$
    $endgroup$
    – Did
    Dec 17 '18 at 0:36
















$begingroup$
Passing by infinities is not needed, simply use $$P(Xleqslant x,Y>y)=P(Xleqslant x)-P(Xleqslant x,Yleqslant y)$$
$endgroup$
– Did
Dec 17 '18 at 0:36




$begingroup$
Passing by infinities is not needed, simply use $$P(Xleqslant x,Y>y)=P(Xleqslant x)-P(Xleqslant x,Yleqslant y)$$
$endgroup$
– Did
Dec 17 '18 at 0:36


















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