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







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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
    2 days ago












  • 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
    2 days ago

















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







share|cite|improve this question







New contributor




Francisco is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • 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
    2 days ago












  • 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
    2 days ago















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







share|cite|improve this question







New contributor




Francisco is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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






share|cite|improve this question







New contributor




Francisco is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Francisco is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Francisco is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









Francisco

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New contributor




Francisco is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Francisco is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Francisco is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • 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
    2 days ago












  • 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
    2 days ago




















  • 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
    2 days ago












  • 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
    2 days ago


















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
2 days ago






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
2 days ago














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
2 days ago






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
2 days ago

















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