Random vector and CDF
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I have to compute CDF of random vector (X,Y) using the following probability density function
c x>=0, y>=0 and x+y<=2
f(x,y) = 0 otherwise
The solution has several cases (x>=0,y>=0, x+y<=2) (x>=2,0<=y<2), (y>=2,0<=x<2), (0<=x<2, 0<=y<2,x+y>=2). Why I should be interested to find these cases?
Could someone help me to understand this exercise?
thank you in advance.
probability-distributions random-variables
asked Nov 19 at 19:33
Francisco
82
add a comment |
up vote
0
down vote
favorite
I have to compute CDF of random vector (X,Y) using the following probability density function
c x>=0, y>=0 and x+y<=2
f(x,y) = 0 otherwise
The solution has several cases (x>=0,y>=0, x+y<=2) (x>=2,0<=y<2), (y>=2,0<=x<2), (0<=x<2, 0<=y<2,x+y>=2). Why I should be interested to find these cases?
Could someone help me to understand this exercise?
thank you in advance.
probability-distributions random-variables
asked Nov 19 at 19:33
Francisco
82
The joint CDF is the probability $F(x,y)=P(Xle x,Yle y)=int_{-infty}^yint_{-infty}^x f(u,v),du,dv$ for all $(x,y)$. We should consider the different cases you mention because we need them to specify the CDF completely. It is only natural that $F(x,y)$ turns out to be a piecewise function, completely defined for all possible values that $x$ and $y$ can take. Note that $F:mathbb R^2to mathbb R$.
– StubbornAtom
Nov 19 at 19:52
For example if I want to compute F(x,y) for this case (0<=x<2, 0<=y<2,x+y>=2), what are the endpoints of the integrals?
– Francisco
Nov 19 at 19:56
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have to compute CDF of random vector (X,Y) using the following probability density function
c x>=0, y>=0 and x+y<=2
f(x,y) = 0 otherwise
The solution has several cases (x>=0,y>=0, x+y<=2) (x>=2,0<=y<2), (y>=2,0<=x<2), (0<=x<2, 0<=y<2,x+y>=2). Why I should be interested to find these cases?
Could someone help me to understand this exercise?
thank you in advance.
probability-distributions random-variables
asked Nov 19 at 19:33
Francisco
82
I have to compute CDF of random vector (X,Y) using the following probability density function
c x>=0, y>=0 and x+y<=2
f(x,y) = 0 otherwise
The solution has several cases (x>=0,y>=0, x+y<=2) (x>=2,0<=y<2), (y>=2,0<=x<2), (0<=x<2, 0<=y<2,x+y>=2). Why I should be interested to find these cases?
Could someone help me to understand this exercise?
thank you in advance.
probability-distributions random-variables
probability-distributions random-variables
asked Nov 19 at 19:33
Francisco
82
asked Nov 19 at 19:33
Francisco
82
asked Nov 19 at 19:33
Francisco
82
asked Nov 19 at 19:33
Francisco
82
asked Nov 19 at 19:33
Francisco
82
82
The joint CDF is the probability $F(x,y)=P(Xle x,Yle y)=int_{-infty}^yint_{-infty}^x f(u,v),du,dv$ for all $(x,y)$. We should consider the different cases you mention because we need them to specify the CDF completely. It is only natural that $F(x,y)$ turns out to be a piecewise function, completely defined for all possible values that $x$ and $y$ can take. Note that $F:mathbb R^2to mathbb R$.
– StubbornAtom
Nov 19 at 19:52
For example if I want to compute F(x,y) for this case (0<=x<2, 0<=y<2,x+y>=2), what are the endpoints of the integrals?
– Francisco
Nov 19 at 19:56
add a comment |
The joint CDF is the probability $F(x,y)=P(Xle x,Yle y)=int_{-infty}^yint_{-infty}^x f(u,v),du,dv$ for all $(x,y)$. We should consider the different cases you mention because we need them to specify the CDF completely. It is only natural that $F(x,y)$ turns out to be a piecewise function, completely defined for all possible values that $x$ and $y$ can take. Note that $F:mathbb R^2to mathbb R$.
– StubbornAtom
Nov 19 at 19:52
For example if I want to compute F(x,y) for this case (0<=x<2, 0<=y<2,x+y>=2), what are the endpoints of the integrals?
– Francisco
Nov 19 at 19:56
The joint CDF is the probability $F(x,y)=P(Xle x,Yle y)=int_{-infty}^yint_{-infty}^x f(u,v),du,dv$ for all $(x,y)$. We should consider the different cases you mention because we need them to specify the CDF completely. It is only natural that $F(x,y)$ turns out to be a piecewise function, completely defined for all possible values that $x$ and $y$ can take. Note that $F:mathbb R^2to mathbb R$.
– StubbornAtom
Nov 19 at 19:52
The joint CDF is the probability $F(x,y)=P(Xle x,Yle y)=int_{-infty}^yint_{-infty}^x f(u,v),du,dv$ for all $(x,y)$. We should consider the different cases you mention because we need them to specify the CDF completely. It is only natural that $F(x,y)$ turns out to be a piecewise function, completely defined for all possible values that $x$ and $y$ can take. Note that $F:mathbb R^2to mathbb R$.
– StubbornAtom
Nov 19 at 19:52
For example if I want to compute F(x,y) for this case (0<=x<2, 0<=y<2,x+y>=2), what are the endpoints of the integrals?
– Francisco
Nov 19 at 19:56
For example if I want to compute F(x,y) for this case (0<=x<2, 0<=y<2,x+y>=2), what are the endpoints of the integrals?
– Francisco
Nov 19 at 19:56
add a comment |
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The joint CDF is the probability $F(x,y)=P(Xle x,Yle y)=int_{-infty}^yint_{-infty}^x f(u,v),du,dv$ for all $(x,y)$. We should consider the different cases you mention because we need them to specify the CDF completely. It is only natural that $F(x,y)$ turns out to be a piecewise function, completely defined for all possible values that $x$ and $y$ can take. Note that $F:mathbb R^2to mathbb R$.
– StubbornAtom
Nov 19 at 19:52
For example if I want to compute F(x,y) for this case (0<=x<2, 0<=y<2,x+y>=2), what are the endpoints of the integrals?
– Francisco
Nov 19 at 19:56