Determining normalizing constant of joint density function
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Let $T_x$ and $T_y$ be two random variables that desribe the remaining lifetimes of two persons, aged $x$ and $y$ respectively. The joint density of $T_x$ and $T_y$ is given by
$$f(s,t)=begin{cases}C(50^2-(s-t)^2),& s,t in [0,50]\0, &mathrm{otherwise}end{cases}$$
($T_x$ and $T_y$ are not independent). $C>0$ is a normalizing constant.
How can I determine $C$?
probability probability-theory measure-theory probability-distributions
asked Jan 25 at 10:02
user636610user636610
82
$endgroup$
add a comment |
$begingroup$
Let $T_x$ and $T_y$ be two random variables that desribe the remaining lifetimes of two persons, aged $x$ and $y$ respectively. The joint density of $T_x$ and $T_y$ is given by
$$f(s,t)=begin{cases}C(50^2-(s-t)^2),& s,t in [0,50]\0, &mathrm{otherwise}end{cases}$$
($T_x$ and $T_y$ are not independent). $C>0$ is a normalizing constant.
How can I determine $C$?
probability probability-theory measure-theory probability-distributions
asked Jan 25 at 10:02
user636610user636610
82
$endgroup$
3
$begingroup$
Choose $C$ such that $f$ integrates to $1$. The computation involved here is very simple.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 10:05
$begingroup$
Thanks, didn't know it must equal $1$, then it is very simple indeed
$endgroup$
– user636610
Jan 25 at 10:16
$begingroup$
Would you consider posting an answer and accept it yourself (self-answering is encouraged on the site), or deleting this question post to wrap it up? It seems that you have already got what you wanted here or elsewhere and have moved on.
$endgroup$
– Lee David Chung Lin
30 mins ago
add a comment |
$begingroup$
Let $T_x$ and $T_y$ be two random variables that desribe the remaining lifetimes of two persons, aged $x$ and $y$ respectively. The joint density of $T_x$ and $T_y$ is given by
$$f(s,t)=begin{cases}C(50^2-(s-t)^2),& s,t in [0,50]\0, &mathrm{otherwise}end{cases}$$
($T_x$ and $T_y$ are not independent). $C>0$ is a normalizing constant.
How can I determine $C$?
probability probability-theory measure-theory probability-distributions
asked Jan 25 at 10:02
user636610user636610
82
$endgroup$
Let $T_x$ and $T_y$ be two random variables that desribe the remaining lifetimes of two persons, aged $x$ and $y$ respectively. The joint density of $T_x$ and $T_y$ is given by
$$f(s,t)=begin{cases}C(50^2-(s-t)^2),& s,t in [0,50]\0, &mathrm{otherwise}end{cases}$$
($T_x$ and $T_y$ are not independent). $C>0$ is a normalizing constant.
How can I determine $C$?
probability probability-theory measure-theory probability-distributions
probability probability-theory measure-theory probability-distributions
asked Jan 25 at 10:02
user636610user636610
82
asked Jan 25 at 10:02
user636610user636610
82
asked Jan 25 at 10:02
user636610user636610
82
asked Jan 25 at 10:02
user636610user636610
82
asked Jan 25 at 10:02
user636610user636610
82
82
3
$begingroup$
Choose $C$ such that $f$ integrates to $1$. The computation involved here is very simple.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 10:05
$begingroup$
Thanks, didn't know it must equal $1$, then it is very simple indeed
$endgroup$
– user636610
Jan 25 at 10:16
$begingroup$
Would you consider posting an answer and accept it yourself (self-answering is encouraged on the site), or deleting this question post to wrap it up? It seems that you have already got what you wanted here or elsewhere and have moved on.
$endgroup$
– Lee David Chung Lin
30 mins ago
add a comment |
3
$begingroup$
Choose $C$ such that $f$ integrates to $1$. The computation involved here is very simple.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 10:05
$begingroup$
Thanks, didn't know it must equal $1$, then it is very simple indeed
$endgroup$
– user636610
Jan 25 at 10:16
$begingroup$
Would you consider posting an answer and accept it yourself (self-answering is encouraged on the site), or deleting this question post to wrap it up? It seems that you have already got what you wanted here or elsewhere and have moved on.
$endgroup$
– Lee David Chung Lin
30 mins ago
3
3
$begingroup$
Choose $C$ such that $f$ integrates to $1$. The computation involved here is very simple.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 10:05
$begingroup$
Choose $C$ such that $f$ integrates to $1$. The computation involved here is very simple.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 10:05
$begingroup$
Thanks, didn't know it must equal $1$, then it is very simple indeed
$endgroup$
– user636610
Jan 25 at 10:16
$begingroup$
Thanks, didn't know it must equal $1$, then it is very simple indeed
$endgroup$
– user636610
Jan 25 at 10:16
$begingroup$
Would you consider posting an answer and accept it yourself (self-answering is encouraged on the site), or deleting this question post to wrap it up? It seems that you have already got what you wanted here or elsewhere and have moved on.
$endgroup$
– Lee David Chung Lin
30 mins ago
$begingroup$
Would you consider posting an answer and accept it yourself (self-answering is encouraged on the site), or deleting this question post to wrap it up? It seems that you have already got what you wanted here or elsewhere and have moved on.
$endgroup$
– Lee David Chung Lin
30 mins ago
add a comment |
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3
$begingroup$
Choose $C$ such that $f$ integrates to $1$. The computation involved here is very simple.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 10:05
$begingroup$
Thanks, didn't know it must equal $1$, then it is very simple indeed
$endgroup$
– user636610
Jan 25 at 10:16
$begingroup$
Would you consider posting an answer and accept it yourself (self-answering is encouraged on the site), or deleting this question post to wrap it up? It seems that you have already got what you wanted here or elsewhere and have moved on.
$endgroup$
– Lee David Chung Lin
30 mins ago