Proof of positive homogeneity of risk measure
$begingroup$
I am stuck with a proof about positive homogeneity and have tried now for several days. Hopefully someone can help me out. Thanks in advance.
I am considering a risk measure $p=(x)$, where $x$ denotes a loss. $p$ satisfies the following properties:
P1: Normalization, $p(0)=0$
P2: Cash additive, $p(x+c)=p(x)+c$
P3: Monotonic, if $x$$geq$$y$ then $p(x)$$geq$$p(y)$
P4: Convexity, if $lambdain(0,1)$ then $p(lambda$$x+(1-lambda$$y$) $leq$ $lambda$$p(x)+(1-lambda$$)p(y)$
P6: Subadditivity, $p(x+y)leq$$p(x)+p(y)$
Also, we have that $lambdain(0,1)$ and $lambdain$Q
I proved that $p(lambda$$x)leq$ $lambda$$p(x)$for $lambdain(0,1)$ and $lambdain$Q
Now, I want to prove that $p(lambda$$x)geq$ $lambda$$p(x)$ for $lambdain(0,1)$ and $lambdain$Q as well.
This should be possible using the properties above and the fact that $lambdain$Q.
Thanks for anyone who can help me with this proof.
measure-theory convex-analysis
asked Dec 6 '18 at 13:31
René WijnenRené Wijnen
112
$endgroup$
add a comment |
$begingroup$
I am stuck with a proof about positive homogeneity and have tried now for several days. Hopefully someone can help me out. Thanks in advance.
I am considering a risk measure $p=(x)$, where $x$ denotes a loss. $p$ satisfies the following properties:
P1: Normalization, $p(0)=0$
P2: Cash additive, $p(x+c)=p(x)+c$
P3: Monotonic, if $x$$geq$$y$ then $p(x)$$geq$$p(y)$
P4: Convexity, if $lambdain(0,1)$ then $p(lambda$$x+(1-lambda$$y$) $leq$ $lambda$$p(x)+(1-lambda$$)p(y)$
P6: Subadditivity, $p(x+y)leq$$p(x)+p(y)$
Also, we have that $lambdain(0,1)$ and $lambdain$Q
I proved that $p(lambda$$x)leq$ $lambda$$p(x)$for $lambdain(0,1)$ and $lambdain$Q
Now, I want to prove that $p(lambda$$x)geq$ $lambda$$p(x)$ for $lambdain(0,1)$ and $lambdain$Q as well.
This should be possible using the properties above and the fact that $lambdain$Q.
Thanks for anyone who can help me with this proof.
measure-theory convex-analysis
asked Dec 6 '18 at 13:31
René WijnenRené Wijnen
112
$endgroup$
$begingroup$
by the way $xin$ R
$endgroup$
– René Wijnen
Dec 6 '18 at 13:39
$begingroup$
I guess $c$ is a fixed constant? If not with $p(c)=p(c+0)=p(0)+c=c$ and $c$ arbitrary, $p$ would be the identity.
$endgroup$
– humanStampedist
Dec 6 '18 at 13:48
1
$begingroup$
c is a fixed constant and x is some loss. the measure p(x) gives the amount of capital that you should hold to cover the loss, x.
$endgroup$
– René Wijnen
Dec 6 '18 at 15:20
$begingroup$
In the second sentence it says a risk measure p = (x), this was a typo. It should be p(x)
$endgroup$
– René Wijnen
Dec 6 '18 at 15:26
add a comment |
$begingroup$
I am stuck with a proof about positive homogeneity and have tried now for several days. Hopefully someone can help me out. Thanks in advance.
I am considering a risk measure $p=(x)$, where $x$ denotes a loss. $p$ satisfies the following properties:
P1: Normalization, $p(0)=0$
P2: Cash additive, $p(x+c)=p(x)+c$
P3: Monotonic, if $x$$geq$$y$ then $p(x)$$geq$$p(y)$
P4: Convexity, if $lambdain(0,1)$ then $p(lambda$$x+(1-lambda$$y$) $leq$ $lambda$$p(x)+(1-lambda$$)p(y)$
P6: Subadditivity, $p(x+y)leq$$p(x)+p(y)$
Also, we have that $lambdain(0,1)$ and $lambdain$Q
I proved that $p(lambda$$x)leq$ $lambda$$p(x)$for $lambdain(0,1)$ and $lambdain$Q
Now, I want to prove that $p(lambda$$x)geq$ $lambda$$p(x)$ for $lambdain(0,1)$ and $lambdain$Q as well.
This should be possible using the properties above and the fact that $lambdain$Q.
Thanks for anyone who can help me with this proof.
measure-theory convex-analysis
asked Dec 6 '18 at 13:31
René WijnenRené Wijnen
112
$endgroup$
I am stuck with a proof about positive homogeneity and have tried now for several days. Hopefully someone can help me out. Thanks in advance.
I am considering a risk measure $p=(x)$, where $x$ denotes a loss. $p$ satisfies the following properties:
P1: Normalization, $p(0)=0$
P2: Cash additive, $p(x+c)=p(x)+c$
P3: Monotonic, if $x$$geq$$y$ then $p(x)$$geq$$p(y)$
P4: Convexity, if $lambdain(0,1)$ then $p(lambda$$x+(1-lambda$$y$) $leq$ $lambda$$p(x)+(1-lambda$$)p(y)$
P6: Subadditivity, $p(x+y)leq$$p(x)+p(y)$
Also, we have that $lambdain(0,1)$ and $lambdain$Q
I proved that $p(lambda$$x)leq$ $lambda$$p(x)$for $lambdain(0,1)$ and $lambdain$Q
Now, I want to prove that $p(lambda$$x)geq$ $lambda$$p(x)$ for $lambdain(0,1)$ and $lambdain$Q as well.
This should be possible using the properties above and the fact that $lambdain$Q.
Thanks for anyone who can help me with this proof.
measure-theory convex-analysis
measure-theory convex-analysis
asked Dec 6 '18 at 13:31
René WijnenRené Wijnen
112
asked Dec 6 '18 at 13:31
René WijnenRené Wijnen
112
asked Dec 6 '18 at 13:31
René WijnenRené Wijnen
112
asked Dec 6 '18 at 13:31
René WijnenRené Wijnen
112
asked Dec 6 '18 at 13:31
René WijnenRené Wijnen
112
112
$begingroup$
by the way $xin$ R
$endgroup$
– René Wijnen
Dec 6 '18 at 13:39
$begingroup$
I guess $c$ is a fixed constant? If not with $p(c)=p(c+0)=p(0)+c=c$ and $c$ arbitrary, $p$ would be the identity.
$endgroup$
– humanStampedist
Dec 6 '18 at 13:48
1
$begingroup$
c is a fixed constant and x is some loss. the measure p(x) gives the amount of capital that you should hold to cover the loss, x.
$endgroup$
– René Wijnen
Dec 6 '18 at 15:20
$begingroup$
In the second sentence it says a risk measure p = (x), this was a typo. It should be p(x)
$endgroup$
– René Wijnen
Dec 6 '18 at 15:26
add a comment |
$begingroup$
by the way $xin$ R
$endgroup$
– René Wijnen
Dec 6 '18 at 13:39
$begingroup$
I guess $c$ is a fixed constant? If not with $p(c)=p(c+0)=p(0)+c=c$ and $c$ arbitrary, $p$ would be the identity.
$endgroup$
– humanStampedist
Dec 6 '18 at 13:48
1
$begingroup$
c is a fixed constant and x is some loss. the measure p(x) gives the amount of capital that you should hold to cover the loss, x.
$endgroup$
– René Wijnen
Dec 6 '18 at 15:20
$begingroup$
In the second sentence it says a risk measure p = (x), this was a typo. It should be p(x)
$endgroup$
– René Wijnen
Dec 6 '18 at 15:26
$begingroup$
by the way $xin$ R
$endgroup$
– René Wijnen
Dec 6 '18 at 13:39
$begingroup$
by the way $xin$ R
$endgroup$
– René Wijnen
Dec 6 '18 at 13:39
$begingroup$
I guess $c$ is a fixed constant? If not with $p(c)=p(c+0)=p(0)+c=c$ and $c$ arbitrary, $p$ would be the identity.
$endgroup$
– humanStampedist
Dec 6 '18 at 13:48
$begingroup$
I guess $c$ is a fixed constant? If not with $p(c)=p(c+0)=p(0)+c=c$ and $c$ arbitrary, $p$ would be the identity.
$endgroup$
– humanStampedist
Dec 6 '18 at 13:48
1
1
$begingroup$
c is a fixed constant and x is some loss. the measure p(x) gives the amount of capital that you should hold to cover the loss, x.
$endgroup$
– René Wijnen
Dec 6 '18 at 15:20
$begingroup$
c is a fixed constant and x is some loss. the measure p(x) gives the amount of capital that you should hold to cover the loss, x.
$endgroup$
– René Wijnen
Dec 6 '18 at 15:20
$begingroup$
In the second sentence it says a risk measure p = (x), this was a typo. It should be p(x)
$endgroup$
– René Wijnen
Dec 6 '18 at 15:26
$begingroup$
In the second sentence it says a risk measure p = (x), this was a typo. It should be p(x)
$endgroup$
– René Wijnen
Dec 6 '18 at 15:26
add a comment |
1 Answer
1
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oldest
votes
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From P1 and P2 it follows that
$$
p(y) = p(0+y) = p(0)+y = 0+y = y
$$
has to be true for all $yinmathbb R$.
Now that you know the value of $p$ for all inputs, it is easy to see that
$$p(lambda x) = lambda x = lambda p(x)$$
has to hold for all $lambdain(0,1), xin mathbb Q$.
answered Dec 6 '18 at 13:49
supinfsupinf
6,1391028
$endgroup$
$begingroup$
I noticed that I forgot to mention that P2 only holds for some constant c, my apologies. P2 does not apply for the variable x in the way you suggest in the answer above.
$endgroup$
– René Wijnen
Dec 6 '18 at 15:23
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
From P1 and P2 it follows that
$$
p(y) = p(0+y) = p(0)+y = 0+y = y
$$
has to be true for all $yinmathbb R$.
Now that you know the value of $p$ for all inputs, it is easy to see that
$$p(lambda x) = lambda x = lambda p(x)$$
has to hold for all $lambdain(0,1), xin mathbb Q$.
answered Dec 6 '18 at 13:49
supinfsupinf
6,1391028
$endgroup$
$begingroup$
I noticed that I forgot to mention that P2 only holds for some constant c, my apologies. P2 does not apply for the variable x in the way you suggest in the answer above.
$endgroup$
– René Wijnen
Dec 6 '18 at 15:23
add a comment |
$begingroup$
From P1 and P2 it follows that
$$
p(y) = p(0+y) = p(0)+y = 0+y = y
$$
has to be true for all $yinmathbb R$.
Now that you know the value of $p$ for all inputs, it is easy to see that
$$p(lambda x) = lambda x = lambda p(x)$$
has to hold for all $lambdain(0,1), xin mathbb Q$.
answered Dec 6 '18 at 13:49
supinfsupinf
6,1391028
$endgroup$
$begingroup$
I noticed that I forgot to mention that P2 only holds for some constant c, my apologies. P2 does not apply for the variable x in the way you suggest in the answer above.
$endgroup$
– René Wijnen
Dec 6 '18 at 15:23
add a comment |
$begingroup$
From P1 and P2 it follows that
$$
p(y) = p(0+y) = p(0)+y = 0+y = y
$$
has to be true for all $yinmathbb R$.
Now that you know the value of $p$ for all inputs, it is easy to see that
$$p(lambda x) = lambda x = lambda p(x)$$
has to hold for all $lambdain(0,1), xin mathbb Q$.
answered Dec 6 '18 at 13:49
supinfsupinf
6,1391028
$endgroup$
From P1 and P2 it follows that
$$
p(y) = p(0+y) = p(0)+y = 0+y = y
$$
has to be true for all $yinmathbb R$.
Now that you know the value of $p$ for all inputs, it is easy to see that
$$p(lambda x) = lambda x = lambda p(x)$$
has to hold for all $lambdain(0,1), xin mathbb Q$.
answered Dec 6 '18 at 13:49
supinfsupinf
6,1391028
answered Dec 6 '18 at 13:49
supinfsupinf
6,1391028
answered Dec 6 '18 at 13:49
supinfsupinf
6,1391028
answered Dec 6 '18 at 13:49
supinfsupinf
6,1391028
6,1391028
$begingroup$
I noticed that I forgot to mention that P2 only holds for some constant c, my apologies. P2 does not apply for the variable x in the way you suggest in the answer above.
$endgroup$
– René Wijnen
Dec 6 '18 at 15:23
add a comment |
$begingroup$
I noticed that I forgot to mention that P2 only holds for some constant c, my apologies. P2 does not apply for the variable x in the way you suggest in the answer above.
$endgroup$
– René Wijnen
Dec 6 '18 at 15:23
$begingroup$
I noticed that I forgot to mention that P2 only holds for some constant c, my apologies. P2 does not apply for the variable x in the way you suggest in the answer above.
$endgroup$
– René Wijnen
Dec 6 '18 at 15:23
$begingroup$
I noticed that I forgot to mention that P2 only holds for some constant c, my apologies. P2 does not apply for the variable x in the way you suggest in the answer above.
$endgroup$
– René Wijnen
Dec 6 '18 at 15:23
add a comment |
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$begingroup$
by the way $xin$ R
$endgroup$
– René Wijnen
Dec 6 '18 at 13:39
$begingroup$
I guess $c$ is a fixed constant? If not with $p(c)=p(c+0)=p(0)+c=c$ and $c$ arbitrary, $p$ would be the identity.
$endgroup$
– humanStampedist
Dec 6 '18 at 13:48
1
$begingroup$
c is a fixed constant and x is some loss. the measure p(x) gives the amount of capital that you should hold to cover the loss, x.
$endgroup$
– René Wijnen
Dec 6 '18 at 15:20
$begingroup$
In the second sentence it says a risk measure p = (x), this was a typo. It should be p(x)
$endgroup$
– René Wijnen
Dec 6 '18 at 15:26