How to get distribution function
I have a density function $g(x,y)=4xy mathbb{I}_{[0,1]times[0,1] }(x,y)$. How can I get a distribution function $G(x,y)$?
I found that $G(x,y)= int _{-infty}^x int _{-infty}^y f(u,v) du dv$ where $f(u,v)$ is density function and $G(x,y)$ distribution function. I this right? How can I use it
probability integration probability-theory probability-distributions density-function
asked Nov 26 at 9:48
Atstovas
697
add a comment |
I have a density function $g(x,y)=4xy mathbb{I}_{[0,1]times[0,1] }(x,y)$. How can I get a distribution function $G(x,y)$?
I found that $G(x,y)= int _{-infty}^x int _{-infty}^y f(u,v) du dv$ where $f(u,v)$ is density function and $G(x,y)$ distribution function. I this right? How can I use it
probability integration probability-theory probability-distributions density-function
asked Nov 26 at 9:48
Atstovas
697
add a comment |
I have a density function $g(x,y)=4xy mathbb{I}_{[0,1]times[0,1] }(x,y)$. How can I get a distribution function $G(x,y)$?
I found that $G(x,y)= int _{-infty}^x int _{-infty}^y f(u,v) du dv$ where $f(u,v)$ is density function and $G(x,y)$ distribution function. I this right? How can I use it
probability integration probability-theory probability-distributions density-function
asked Nov 26 at 9:48
Atstovas
697
I have a density function $g(x,y)=4xy mathbb{I}_{[0,1]times[0,1] }(x,y)$. How can I get a distribution function $G(x,y)$?
I found that $G(x,y)= int _{-infty}^x int _{-infty}^y f(u,v) du dv$ where $f(u,v)$ is density function and $G(x,y)$ distribution function. I this right? How can I use it
probability integration probability-theory probability-distributions density-function
probability integration probability-theory probability-distributions density-function
asked Nov 26 at 9:48
Atstovas
697
asked Nov 26 at 9:48
Atstovas
697
asked Nov 26 at 9:48
Atstovas
697
asked Nov 26 at 9:48
Atstovas
697
asked Nov 26 at 9:48
Atstovas
697
697
add a comment |
add a comment |
1 Answer
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$G(x,y)=int_0^{x}int_0^{y} 4uv du dv=x^{2}y^{2}$ for $0leq x,y leq 1$. For $0leq x leq 1$ and $y >1$ we have $G(x,y)=int_0^{x}int_0^{1} 4uv du dv=x^{2}$ and for $0leq y leq 1$ and $x >1$ we have $G(x,y)=y^{2}$. Finally $G(x,y) =1$ if $x,y >1$ and $0$ of $x,y<0$.
answered Nov 26 at 9:54
Kavi Rama Murthy
49.1k31854
Could you explain me why you took those intervals for example if $0geq x, y geq 1$ you took integrals $int_0^x int_0^y$?
– Atstovas
Nov 26 at 10:07
1
Since $g(x,y)=0$ if $x<0$ or $y<0$ all integrals start from $0$. If $x>1$ then the integral from $0$ to $x$ becomes integral from $0$ to $1$ because the density function vanishes for outside $[0,1]$ so it makes no contribution to the integral from $1$ to $x$, etc.
– Kavi Rama Murthy
Nov 26 at 10:12
add a comment |
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$G(x,y)=int_0^{x}int_0^{y} 4uv du dv=x^{2}y^{2}$ for $0leq x,y leq 1$. For $0leq x leq 1$ and $y >1$ we have $G(x,y)=int_0^{x}int_0^{1} 4uv du dv=x^{2}$ and for $0leq y leq 1$ and $x >1$ we have $G(x,y)=y^{2}$. Finally $G(x,y) =1$ if $x,y >1$ and $0$ of $x,y<0$.
answered Nov 26 at 9:54
Kavi Rama Murthy
49.1k31854
Could you explain me why you took those intervals for example if $0geq x, y geq 1$ you took integrals $int_0^x int_0^y$?
– Atstovas
Nov 26 at 10:07
1
Since $g(x,y)=0$ if $x<0$ or $y<0$ all integrals start from $0$. If $x>1$ then the integral from $0$ to $x$ becomes integral from $0$ to $1$ because the density function vanishes for outside $[0,1]$ so it makes no contribution to the integral from $1$ to $x$, etc.
– Kavi Rama Murthy
Nov 26 at 10:12
add a comment |
$G(x,y)=int_0^{x}int_0^{y} 4uv du dv=x^{2}y^{2}$ for $0leq x,y leq 1$. For $0leq x leq 1$ and $y >1$ we have $G(x,y)=int_0^{x}int_0^{1} 4uv du dv=x^{2}$ and for $0leq y leq 1$ and $x >1$ we have $G(x,y)=y^{2}$. Finally $G(x,y) =1$ if $x,y >1$ and $0$ of $x,y<0$.
answered Nov 26 at 9:54
Kavi Rama Murthy
49.1k31854
Could you explain me why you took those intervals for example if $0geq x, y geq 1$ you took integrals $int_0^x int_0^y$?
– Atstovas
Nov 26 at 10:07
1
Since $g(x,y)=0$ if $x<0$ or $y<0$ all integrals start from $0$. If $x>1$ then the integral from $0$ to $x$ becomes integral from $0$ to $1$ because the density function vanishes for outside $[0,1]$ so it makes no contribution to the integral from $1$ to $x$, etc.
– Kavi Rama Murthy
Nov 26 at 10:12
add a comment |
$G(x,y)=int_0^{x}int_0^{y} 4uv du dv=x^{2}y^{2}$ for $0leq x,y leq 1$. For $0leq x leq 1$ and $y >1$ we have $G(x,y)=int_0^{x}int_0^{1} 4uv du dv=x^{2}$ and for $0leq y leq 1$ and $x >1$ we have $G(x,y)=y^{2}$. Finally $G(x,y) =1$ if $x,y >1$ and $0$ of $x,y<0$.
answered Nov 26 at 9:54
Kavi Rama Murthy
49.1k31854
$G(x,y)=int_0^{x}int_0^{y} 4uv du dv=x^{2}y^{2}$ for $0leq x,y leq 1$. For $0leq x leq 1$ and $y >1$ we have $G(x,y)=int_0^{x}int_0^{1} 4uv du dv=x^{2}$ and for $0leq y leq 1$ and $x >1$ we have $G(x,y)=y^{2}$. Finally $G(x,y) =1$ if $x,y >1$ and $0$ of $x,y<0$.
answered Nov 26 at 9:54
Kavi Rama Murthy
49.1k31854
answered Nov 26 at 9:54
Kavi Rama Murthy
49.1k31854
answered Nov 26 at 9:54
Kavi Rama Murthy
49.1k31854
answered Nov 26 at 9:54
Kavi Rama Murthy
49.1k31854
49.1k31854
Could you explain me why you took those intervals for example if $0geq x, y geq 1$ you took integrals $int_0^x int_0^y$?
– Atstovas
Nov 26 at 10:07
1
Since $g(x,y)=0$ if $x<0$ or $y<0$ all integrals start from $0$. If $x>1$ then the integral from $0$ to $x$ becomes integral from $0$ to $1$ because the density function vanishes for outside $[0,1]$ so it makes no contribution to the integral from $1$ to $x$, etc.
– Kavi Rama Murthy
Nov 26 at 10:12
add a comment |
Could you explain me why you took those intervals for example if $0geq x, y geq 1$ you took integrals $int_0^x int_0^y$?
– Atstovas
Nov 26 at 10:07
1
Since $g(x,y)=0$ if $x<0$ or $y<0$ all integrals start from $0$. If $x>1$ then the integral from $0$ to $x$ becomes integral from $0$ to $1$ because the density function vanishes for outside $[0,1]$ so it makes no contribution to the integral from $1$ to $x$, etc.
– Kavi Rama Murthy
Nov 26 at 10:12
Could you explain me why you took those intervals for example if $0geq x, y geq 1$ you took integrals $int_0^x int_0^y$?
– Atstovas
Nov 26 at 10:07
Could you explain me why you took those intervals for example if $0geq x, y geq 1$ you took integrals $int_0^x int_0^y$?
– Atstovas
Nov 26 at 10:07
1
1
Since $g(x,y)=0$ if $x<0$ or $y<0$ all integrals start from $0$. If $x>1$ then the integral from $0$ to $x$ becomes integral from $0$ to $1$ because the density function vanishes for outside $[0,1]$ so it makes no contribution to the integral from $1$ to $x$, etc.
– Kavi Rama Murthy
Nov 26 at 10:12
Since $g(x,y)=0$ if $x<0$ or $y<0$ all integrals start from $0$. If $x>1$ then the integral from $0$ to $x$ becomes integral from $0$ to $1$ because the density function vanishes for outside $[0,1]$ so it makes no contribution to the integral from $1$ to $x$, etc.
– Kavi Rama Murthy
Nov 26 at 10:12
add a comment |
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