Mixing
Definition of monotonically non-decreasing property for 2-dimensional CDF
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I've learned in a lecture that for every function $f(x)$ that is:
- Right-continuous
- Monotonically non-decreasing
$lim_{xto -infty}=0, lim_{xto infty}=1$.
we can define a random variable $X$ such that $f(x)$ is the CDF of $X$.
We've also learned that the same holds for two-dimensional functions. Our lecturer said that the 2-dimensional function have to be monotonically non-decreasing for each of its variables.
After the lecture, I've came up with what I thought was a counter-example:
$f(x,y) = 1$ for $xge0$ and $yge0$ and $(xge1$ or $yge1)$,
$f(x,y) = 0$ otherwise.
If we'll find a corresponding random variable $X$, it'll have 2 atoms in $(0,1)$ and $(1,0)$, both with probability $1$.
When I asked my lecturer, she said that the function isn't monotonically increasing because if I'd look at the square $ABCD$ when $A=(0.5,0.5), B=(1.5,0.5), C=(0.5,1.5), D=(1.5,1.5)$ I'll get that $f(A)-f(B)-f(C)+f(D) = -1 < 0$, and all squares on the $XY$ plane must satisfy $f(A)-f(B)-f(C)+f(D)>0$.
How does being "monotonically non-decreasing for each of its variables" implies that? is it a different definition for being monotonically non-decreasing in 2-dimensional functions? Did she even mean squares or did she mean rectangles?
What are in fact the correct properties that a function must have to be a CDF?
probability-theory
asked Jan 18 at 15:35
Rei HenigmanRei Henigman
717
$endgroup$
add a comment |
$begingroup$
I've learned in a lecture that for every function $f(x)$ that is:
- Right-continuous
- Monotonically non-decreasing
$lim_{xto -infty}=0, lim_{xto infty}=1$.
we can define a random variable $X$ such that $f(x)$ is the CDF of $X$.
We've also learned that the same holds for two-dimensional functions. Our lecturer said that the 2-dimensional function have to be monotonically non-decreasing for each of its variables.
After the lecture, I've came up with what I thought was a counter-example:
$f(x,y) = 1$ for $xge0$ and $yge0$ and $(xge1$ or $yge1)$,
$f(x,y) = 0$ otherwise.
If we'll find a corresponding random variable $X$, it'll have 2 atoms in $(0,1)$ and $(1,0)$, both with probability $1$.
When I asked my lecturer, she said that the function isn't monotonically increasing because if I'd look at the square $ABCD$ when $A=(0.5,0.5), B=(1.5,0.5), C=(0.5,1.5), D=(1.5,1.5)$ I'll get that $f(A)-f(B)-f(C)+f(D) = -1 < 0$, and all squares on the $XY$ plane must satisfy $f(A)-f(B)-f(C)+f(D)>0$.
How does being "monotonically non-decreasing for each of its variables" implies that? is it a different definition for being monotonically non-decreasing in 2-dimensional functions? Did she even mean squares or did she mean rectangles?
What are in fact the correct properties that a function must have to be a CDF?
probability-theory
asked Jan 18 at 15:35
Rei HenigmanRei Henigman
717
$endgroup$
add a comment |
$begingroup$
I've learned in a lecture that for every function $f(x)$ that is:
- Right-continuous
- Monotonically non-decreasing
$lim_{xto -infty}=0, lim_{xto infty}=1$.
we can define a random variable $X$ such that $f(x)$ is the CDF of $X$.
We've also learned that the same holds for two-dimensional functions. Our lecturer said that the 2-dimensional function have to be monotonically non-decreasing for each of its variables.
After the lecture, I've came up with what I thought was a counter-example:
$f(x,y) = 1$ for $xge0$ and $yge0$ and $(xge1$ or $yge1)$,
$f(x,y) = 0$ otherwise.
If we'll find a corresponding random variable $X$, it'll have 2 atoms in $(0,1)$ and $(1,0)$, both with probability $1$.
When I asked my lecturer, she said that the function isn't monotonically increasing because if I'd look at the square $ABCD$ when $A=(0.5,0.5), B=(1.5,0.5), C=(0.5,1.5), D=(1.5,1.5)$ I'll get that $f(A)-f(B)-f(C)+f(D) = -1 < 0$, and all squares on the $XY$ plane must satisfy $f(A)-f(B)-f(C)+f(D)>0$.
How does being "monotonically non-decreasing for each of its variables" implies that? is it a different definition for being monotonically non-decreasing in 2-dimensional functions? Did she even mean squares or did she mean rectangles?
What are in fact the correct properties that a function must have to be a CDF?
probability-theory
asked Jan 18 at 15:35
Rei HenigmanRei Henigman
717
$endgroup$
I've learned in a lecture that for every function $f(x)$ that is:
- Right-continuous
- Monotonically non-decreasing
$lim_{xto -infty}=0, lim_{xto infty}=1$.
we can define a random variable $X$ such that $f(x)$ is the CDF of $X$.
We've also learned that the same holds for two-dimensional functions. Our lecturer said that the 2-dimensional function have to be monotonically non-decreasing for each of its variables.
After the lecture, I've came up with what I thought was a counter-example:
$f(x,y) = 1$ for $xge0$ and $yge0$ and $(xge1$ or $yge1)$,
$f(x,y) = 0$ otherwise.
If we'll find a corresponding random variable $X$, it'll have 2 atoms in $(0,1)$ and $(1,0)$, both with probability $1$.
When I asked my lecturer, she said that the function isn't monotonically increasing because if I'd look at the square $ABCD$ when $A=(0.5,0.5), B=(1.5,0.5), C=(0.5,1.5), D=(1.5,1.5)$ I'll get that $f(A)-f(B)-f(C)+f(D) = -1 < 0$, and all squares on the $XY$ plane must satisfy $f(A)-f(B)-f(C)+f(D)>0$.
How does being "monotonically non-decreasing for each of its variables" implies that? is it a different definition for being monotonically non-decreasing in 2-dimensional functions? Did she even mean squares or did she mean rectangles?
What are in fact the correct properties that a function must have to be a CDF?
probability-theory
probability-theory
asked Jan 18 at 15:35
Rei HenigmanRei Henigman
717
asked Jan 18 at 15:35
Rei HenigmanRei Henigman
717
asked Jan 18 at 15:35
Rei HenigmanRei Henigman
717
asked Jan 18 at 15:35
Rei HenigmanRei Henigman
717
asked Jan 18 at 15:35
Rei HenigmanRei Henigman
717
717
add a comment |
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