How to prove that given quotation is not a distribution function
I have function
$F(x,y)=begin{cases} 1, text{ if } x+y geq 0, \ 0, text{ if } x+y<0 end{cases}$.
I have no idea how can I prove that given quotation is not a distribution function. Only one think I know is $Delta_sF geq 0$.
probability probability-theory probability-distributions
edited Nov 26 '18 at 8:37
Tianlalu
3,09621038
asked Nov 26 '18 at 6:53
Atstovas
937
add a comment |
I have function
$F(x,y)=begin{cases} 1, text{ if } x+y geq 0, \ 0, text{ if } x+y<0 end{cases}$.
I have no idea how can I prove that given quotation is not a distribution function. Only one think I know is $Delta_sF geq 0$.
probability probability-theory probability-distributions
edited Nov 26 '18 at 8:37
Tianlalu
3,09621038
asked Nov 26 '18 at 6:53
Atstovas
937
add a comment |
I have function
$F(x,y)=begin{cases} 1, text{ if } x+y geq 0, \ 0, text{ if } x+y<0 end{cases}$.
I have no idea how can I prove that given quotation is not a distribution function. Only one think I know is $Delta_sF geq 0$.
probability probability-theory probability-distributions
edited Nov 26 '18 at 8:37
Tianlalu
3,09621038
asked Nov 26 '18 at 6:53
Atstovas
937
I have function
$F(x,y)=begin{cases} 1, text{ if } x+y geq 0, \ 0, text{ if } x+y<0 end{cases}$.
I have no idea how can I prove that given quotation is not a distribution function. Only one think I know is $Delta_sF geq 0$.
probability probability-theory probability-distributions
probability probability-theory probability-distributions
edited Nov 26 '18 at 8:37
Tianlalu
3,09621038
asked Nov 26 '18 at 6:53
Atstovas
937
edited Nov 26 '18 at 8:37
Tianlalu
3,09621038
asked Nov 26 '18 at 6:53
Atstovas
937
edited Nov 26 '18 at 8:37
Tianlalu
3,09621038
edited Nov 26 '18 at 8:37
Tianlalu
3,09621038
edited Nov 26 '18 at 8:37
Tianlalu
3,09621038
3,09621038
asked Nov 26 '18 at 6:53
Atstovas
937
asked Nov 26 '18 at 6:53
Atstovas
937
asked Nov 26 '18 at 6:53
Atstovas
937
937
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$F(1,1)-F(1,-1)-F(-1,1)+F(-1,-1)=-1 <0$ so $F$ is not a distribution function. [In terms of random variable this means $P{-1<Xleq 1,-1<Yleq 1} <0$!].
answered Nov 26 '18 at 7:19
Kavi Rama Murthy
50.7k31854
One more quostion. Why we take $-1$ instead $0$?
– Atstovas
Nov 26 '18 at 13:37
I know that you used $Delta_s G= G(x_2,y_2)-G(x_1,y_2)-G(x_2,y_1)+G(x_1,y_1)$ From where I have to know how much $x_1,x_2,y_1,y_2$ is?
– Atstovas
Nov 26 '18 at 13:47
@Atstovas If you take $0$ instead of $1$ you won't get a contradiction.
– Kavi Rama Murthy
Nov 26 '18 at 23:17
so how to determine $x_1, x_2, y_1, y_2$ from given function $F(x,y)$?
– Atstovas
Nov 27 '18 at 4:29
No thumb rule. You have to just play around with the function.
– Kavi Rama Murthy
Nov 27 '18 at 5:27
add a comment |
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$F(1,1)-F(1,-1)-F(-1,1)+F(-1,-1)=-1 <0$ so $F$ is not a distribution function. [In terms of random variable this means $P{-1<Xleq 1,-1<Yleq 1} <0$!].
answered Nov 26 '18 at 7:19
Kavi Rama Murthy
50.7k31854
One more quostion. Why we take $-1$ instead $0$?
– Atstovas
Nov 26 '18 at 13:37
I know that you used $Delta_s G= G(x_2,y_2)-G(x_1,y_2)-G(x_2,y_1)+G(x_1,y_1)$ From where I have to know how much $x_1,x_2,y_1,y_2$ is?
– Atstovas
Nov 26 '18 at 13:47
@Atstovas If you take $0$ instead of $1$ you won't get a contradiction.
– Kavi Rama Murthy
Nov 26 '18 at 23:17
so how to determine $x_1, x_2, y_1, y_2$ from given function $F(x,y)$?
– Atstovas
Nov 27 '18 at 4:29
No thumb rule. You have to just play around with the function.
– Kavi Rama Murthy
Nov 27 '18 at 5:27
add a comment |
$F(1,1)-F(1,-1)-F(-1,1)+F(-1,-1)=-1 <0$ so $F$ is not a distribution function. [In terms of random variable this means $P{-1<Xleq 1,-1<Yleq 1} <0$!].
answered Nov 26 '18 at 7:19
Kavi Rama Murthy
50.7k31854
One more quostion. Why we take $-1$ instead $0$?
– Atstovas
Nov 26 '18 at 13:37
I know that you used $Delta_s G= G(x_2,y_2)-G(x_1,y_2)-G(x_2,y_1)+G(x_1,y_1)$ From where I have to know how much $x_1,x_2,y_1,y_2$ is?
– Atstovas
Nov 26 '18 at 13:47
@Atstovas If you take $0$ instead of $1$ you won't get a contradiction.
– Kavi Rama Murthy
Nov 26 '18 at 23:17
so how to determine $x_1, x_2, y_1, y_2$ from given function $F(x,y)$?
– Atstovas
Nov 27 '18 at 4:29
No thumb rule. You have to just play around with the function.
– Kavi Rama Murthy
Nov 27 '18 at 5:27
add a comment |
$F(1,1)-F(1,-1)-F(-1,1)+F(-1,-1)=-1 <0$ so $F$ is not a distribution function. [In terms of random variable this means $P{-1<Xleq 1,-1<Yleq 1} <0$!].
answered Nov 26 '18 at 7:19
Kavi Rama Murthy
50.7k31854
$F(1,1)-F(1,-1)-F(-1,1)+F(-1,-1)=-1 <0$ so $F$ is not a distribution function. [In terms of random variable this means $P{-1<Xleq 1,-1<Yleq 1} <0$!].
answered Nov 26 '18 at 7:19
Kavi Rama Murthy
50.7k31854
answered Nov 26 '18 at 7:19
Kavi Rama Murthy
50.7k31854
answered Nov 26 '18 at 7:19
Kavi Rama Murthy
50.7k31854
answered Nov 26 '18 at 7:19
Kavi Rama Murthy
50.7k31854
50.7k31854
One more quostion. Why we take $-1$ instead $0$?
– Atstovas
Nov 26 '18 at 13:37
I know that you used $Delta_s G= G(x_2,y_2)-G(x_1,y_2)-G(x_2,y_1)+G(x_1,y_1)$ From where I have to know how much $x_1,x_2,y_1,y_2$ is?
– Atstovas
Nov 26 '18 at 13:47
@Atstovas If you take $0$ instead of $1$ you won't get a contradiction.
– Kavi Rama Murthy
Nov 26 '18 at 23:17
so how to determine $x_1, x_2, y_1, y_2$ from given function $F(x,y)$?
– Atstovas
Nov 27 '18 at 4:29
No thumb rule. You have to just play around with the function.
– Kavi Rama Murthy
Nov 27 '18 at 5:27
add a comment |
One more quostion. Why we take $-1$ instead $0$?
– Atstovas
Nov 26 '18 at 13:37
I know that you used $Delta_s G= G(x_2,y_2)-G(x_1,y_2)-G(x_2,y_1)+G(x_1,y_1)$ From where I have to know how much $x_1,x_2,y_1,y_2$ is?
– Atstovas
Nov 26 '18 at 13:47
@Atstovas If you take $0$ instead of $1$ you won't get a contradiction.
– Kavi Rama Murthy
Nov 26 '18 at 23:17
so how to determine $x_1, x_2, y_1, y_2$ from given function $F(x,y)$?
– Atstovas
Nov 27 '18 at 4:29
No thumb rule. You have to just play around with the function.
– Kavi Rama Murthy
Nov 27 '18 at 5:27
One more quostion. Why we take $-1$ instead $0$?
– Atstovas
Nov 26 '18 at 13:37
One more quostion. Why we take $-1$ instead $0$?
– Atstovas
Nov 26 '18 at 13:37
I know that you used $Delta_s G= G(x_2,y_2)-G(x_1,y_2)-G(x_2,y_1)+G(x_1,y_1)$ From where I have to know how much $x_1,x_2,y_1,y_2$ is?
– Atstovas
Nov 26 '18 at 13:47
I know that you used $Delta_s G= G(x_2,y_2)-G(x_1,y_2)-G(x_2,y_1)+G(x_1,y_1)$ From where I have to know how much $x_1,x_2,y_1,y_2$ is?
– Atstovas
Nov 26 '18 at 13:47
@Atstovas If you take $0$ instead of $1$ you won't get a contradiction.
– Kavi Rama Murthy
Nov 26 '18 at 23:17
@Atstovas If you take $0$ instead of $1$ you won't get a contradiction.
– Kavi Rama Murthy
Nov 26 '18 at 23:17
so how to determine $x_1, x_2, y_1, y_2$ from given function $F(x,y)$?
– Atstovas
Nov 27 '18 at 4:29
so how to determine $x_1, x_2, y_1, y_2$ from given function $F(x,y)$?
– Atstovas
Nov 27 '18 at 4:29
No thumb rule. You have to just play around with the function.
– Kavi Rama Murthy
Nov 27 '18 at 5:27
No thumb rule. You have to just play around with the function.
– Kavi Rama Murthy
Nov 27 '18 at 5:27
add a comment |
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