Get Joint PDF from Joint CDF
$begingroup$
Let joint cumulative probability density function of Random variable X,Y
$$F_{1,2}(x,y) = x^2(1-e^{-2y});; text{when};;0le xlt1, yge0$$ and
$$= (1-e^{-2y}) ;; text{when};; xge 1, yge0$$and
$$=0 ;; text{when} ;;y lt 0$$
in this case, I'd like to reversely get the joint pdf of X,Y.
Is there any typical way or algorithm to get the joint pdf from joint cdf?
probability-distributions
asked Jun 21 '17 at 2:35
BeverlieBeverlie
1,141321
$endgroup$
add a comment |
$begingroup$
Let joint cumulative probability density function of Random variable X,Y
$$F_{1,2}(x,y) = x^2(1-e^{-2y});; text{when};;0le xlt1, yge0$$ and
$$= (1-e^{-2y}) ;; text{when};; xge 1, yge0$$and
$$=0 ;; text{when} ;;y lt 0$$
in this case, I'd like to reversely get the joint pdf of X,Y.
Is there any typical way or algorithm to get the joint pdf from joint cdf?
probability-distributions
asked Jun 21 '17 at 2:35
BeverlieBeverlie
1,141321
$endgroup$
add a comment |
$begingroup$
Let joint cumulative probability density function of Random variable X,Y
$$F_{1,2}(x,y) = x^2(1-e^{-2y});; text{when};;0le xlt1, yge0$$ and
$$= (1-e^{-2y}) ;; text{when};; xge 1, yge0$$and
$$=0 ;; text{when} ;;y lt 0$$
in this case, I'd like to reversely get the joint pdf of X,Y.
Is there any typical way or algorithm to get the joint pdf from joint cdf?
probability-distributions
asked Jun 21 '17 at 2:35
BeverlieBeverlie
1,141321
$endgroup$
Let joint cumulative probability density function of Random variable X,Y
$$F_{1,2}(x,y) = x^2(1-e^{-2y});; text{when};;0le xlt1, yge0$$ and
$$= (1-e^{-2y}) ;; text{when};; xge 1, yge0$$and
$$=0 ;; text{when} ;;y lt 0$$
in this case, I'd like to reversely get the joint pdf of X,Y.
Is there any typical way or algorithm to get the joint pdf from joint cdf?
probability-distributions
probability-distributions
asked Jun 21 '17 at 2:35
BeverlieBeverlie
1,141321
asked Jun 21 '17 at 2:35
BeverlieBeverlie
1,141321
asked Jun 21 '17 at 2:35
BeverlieBeverlie
1,141321
asked Jun 21 '17 at 2:35
BeverlieBeverlie
1,141321
asked Jun 21 '17 at 2:35
BeverlieBeverlie
1,141321
1,141321
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes the typical way is differentiation: $$ f(x,y) = partial_xpartial_y F(x,y).$$ One must be careful in general cause a PDF doesn't always exist, but here taking this derivative will do the trick. (The discontinuity across the line $x=1$ isn't a big deal. The support of the PDF just drops suddenly to zero when you cross into the half plane $x>1$.)
answered Jun 21 '17 at 4:08
spaceisdarkgreenspaceisdarkgreen
32.8k21753
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes the typical way is differentiation: $$ f(x,y) = partial_xpartial_y F(x,y).$$ One must be careful in general cause a PDF doesn't always exist, but here taking this derivative will do the trick. (The discontinuity across the line $x=1$ isn't a big deal. The support of the PDF just drops suddenly to zero when you cross into the half plane $x>1$.)
answered Jun 21 '17 at 4:08
spaceisdarkgreenspaceisdarkgreen
32.8k21753
$endgroup$
add a comment |
$begingroup$
Yes the typical way is differentiation: $$ f(x,y) = partial_xpartial_y F(x,y).$$ One must be careful in general cause a PDF doesn't always exist, but here taking this derivative will do the trick. (The discontinuity across the line $x=1$ isn't a big deal. The support of the PDF just drops suddenly to zero when you cross into the half plane $x>1$.)
answered Jun 21 '17 at 4:08
spaceisdarkgreenspaceisdarkgreen
32.8k21753
$endgroup$
add a comment |
$begingroup$
Yes the typical way is differentiation: $$ f(x,y) = partial_xpartial_y F(x,y).$$ One must be careful in general cause a PDF doesn't always exist, but here taking this derivative will do the trick. (The discontinuity across the line $x=1$ isn't a big deal. The support of the PDF just drops suddenly to zero when you cross into the half plane $x>1$.)
answered Jun 21 '17 at 4:08
spaceisdarkgreenspaceisdarkgreen
32.8k21753
$endgroup$
Yes the typical way is differentiation: $$ f(x,y) = partial_xpartial_y F(x,y).$$ One must be careful in general cause a PDF doesn't always exist, but here taking this derivative will do the trick. (The discontinuity across the line $x=1$ isn't a big deal. The support of the PDF just drops suddenly to zero when you cross into the half plane $x>1$.)
answered Jun 21 '17 at 4:08
spaceisdarkgreenspaceisdarkgreen
32.8k21753
answered Jun 21 '17 at 4:08
spaceisdarkgreenspaceisdarkgreen
32.8k21753
answered Jun 21 '17 at 4:08
spaceisdarkgreenspaceisdarkgreen
32.8k21753
answered Jun 21 '17 at 4:08
spaceisdarkgreenspaceisdarkgreen
32.8k21753
32.8k21753
add a comment |
add a comment |
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