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Help to prove that if C is a copula such that $C(t,1-t) = 0$ for all $t$ in $[0,1]$ then $C = max(u+v-1,0)$
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I'm asked to prove that if C is a copula such that $C(t,1-t) = 0$ for all $t$ in $[0,1]$ then $C = max(u+v-1,0)$. So far I have this:
Let $u,v$ be in $[0,1]$. If $u+v-1 < 0$, i.e. $u < 1-v$. Given that $C$ is nondecreasing
$$0 leq C(u,v) leq C(1-v,v) = 0$$
Therefore $C(u,v) = 0$ when $u+v-1 < 0$.
On the other hand, suppose $u+v-1 geq 0$, then $u geq v-1$. It follows from the Fréchet-Hoeeffding theorem that
$$max(u+v-1,0)=u+v-1 leq C(u,v)$$
That's all I have come up with. Obviously, I need to show that $u+v-1$ is also an upper bound of C(u,v) but I can't find a way to do that. My best attempt has been:
It follows from the nondeacreasing property of C that
$$C(u,v) = C(u,v)-C(u, 1-u) leq C(u,v) - C(1-v,1-u)$$
Additionally, I noticed that $min(u,v) - min(1-u,1-v) = u + v -1$.
So, my main focus has been trying to proof
$$C(u,v) - C(1-v, 1-u) leq min(u,v) - min(1-u,1-v)$$
I don't know if I should try to solve this problem using a different approach or maybe I'm ignoring some known properties of copulas that might help me in this particular situation. Any suggestions?
probability-theory statistics inequality
asked Jan 22 at 7:57
R. CruzR. Cruz
63
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add a comment |
$begingroup$
I'm asked to prove that if C is a copula such that $C(t,1-t) = 0$ for all $t$ in $[0,1]$ then $C = max(u+v-1,0)$. So far I have this:
Let $u,v$ be in $[0,1]$. If $u+v-1 < 0$, i.e. $u < 1-v$. Given that $C$ is nondecreasing
$$0 leq C(u,v) leq C(1-v,v) = 0$$
Therefore $C(u,v) = 0$ when $u+v-1 < 0$.
On the other hand, suppose $u+v-1 geq 0$, then $u geq v-1$. It follows from the Fréchet-Hoeeffding theorem that
$$max(u+v-1,0)=u+v-1 leq C(u,v)$$
That's all I have come up with. Obviously, I need to show that $u+v-1$ is also an upper bound of C(u,v) but I can't find a way to do that. My best attempt has been:
It follows from the nondeacreasing property of C that
$$C(u,v) = C(u,v)-C(u, 1-u) leq C(u,v) - C(1-v,1-u)$$
Additionally, I noticed that $min(u,v) - min(1-u,1-v) = u + v -1$.
So, my main focus has been trying to proof
$$C(u,v) - C(1-v, 1-u) leq min(u,v) - min(1-u,1-v)$$
I don't know if I should try to solve this problem using a different approach or maybe I'm ignoring some known properties of copulas that might help me in this particular situation. Any suggestions?
probability-theory statistics inequality
asked Jan 22 at 7:57
R. CruzR. Cruz
63
$endgroup$
add a comment |
$begingroup$
I'm asked to prove that if C is a copula such that $C(t,1-t) = 0$ for all $t$ in $[0,1]$ then $C = max(u+v-1,0)$. So far I have this:
Let $u,v$ be in $[0,1]$. If $u+v-1 < 0$, i.e. $u < 1-v$. Given that $C$ is nondecreasing
$$0 leq C(u,v) leq C(1-v,v) = 0$$
Therefore $C(u,v) = 0$ when $u+v-1 < 0$.
On the other hand, suppose $u+v-1 geq 0$, then $u geq v-1$. It follows from the Fréchet-Hoeeffding theorem that
$$max(u+v-1,0)=u+v-1 leq C(u,v)$$
That's all I have come up with. Obviously, I need to show that $u+v-1$ is also an upper bound of C(u,v) but I can't find a way to do that. My best attempt has been:
It follows from the nondeacreasing property of C that
$$C(u,v) = C(u,v)-C(u, 1-u) leq C(u,v) - C(1-v,1-u)$$
Additionally, I noticed that $min(u,v) - min(1-u,1-v) = u + v -1$.
So, my main focus has been trying to proof
$$C(u,v) - C(1-v, 1-u) leq min(u,v) - min(1-u,1-v)$$
I don't know if I should try to solve this problem using a different approach or maybe I'm ignoring some known properties of copulas that might help me in this particular situation. Any suggestions?
probability-theory statistics inequality
asked Jan 22 at 7:57
R. CruzR. Cruz
63
$endgroup$
I'm asked to prove that if C is a copula such that $C(t,1-t) = 0$ for all $t$ in $[0,1]$ then $C = max(u+v-1,0)$. So far I have this:
Let $u,v$ be in $[0,1]$. If $u+v-1 < 0$, i.e. $u < 1-v$. Given that $C$ is nondecreasing
$$0 leq C(u,v) leq C(1-v,v) = 0$$
Therefore $C(u,v) = 0$ when $u+v-1 < 0$.
On the other hand, suppose $u+v-1 geq 0$, then $u geq v-1$. It follows from the Fréchet-Hoeeffding theorem that
$$max(u+v-1,0)=u+v-1 leq C(u,v)$$
That's all I have come up with. Obviously, I need to show that $u+v-1$ is also an upper bound of C(u,v) but I can't find a way to do that. My best attempt has been:
It follows from the nondeacreasing property of C that
$$C(u,v) = C(u,v)-C(u, 1-u) leq C(u,v) - C(1-v,1-u)$$
Additionally, I noticed that $min(u,v) - min(1-u,1-v) = u + v -1$.
So, my main focus has been trying to proof
$$C(u,v) - C(1-v, 1-u) leq min(u,v) - min(1-u,1-v)$$
I don't know if I should try to solve this problem using a different approach or maybe I'm ignoring some known properties of copulas that might help me in this particular situation. Any suggestions?
probability-theory statistics inequality
probability-theory statistics inequality
asked Jan 22 at 7:57
R. CruzR. Cruz
63
asked Jan 22 at 7:57
R. CruzR. Cruz
63
asked Jan 22 at 7:57
R. CruzR. Cruz
63
asked Jan 22 at 7:57
R. CruzR. Cruz
63
asked Jan 22 at 7:57
R. CruzR. Cruz
63
63
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