Prove $(b-a) cdot int_a^b alpha(x) beta(x) dx ge int_a^b alpha(x) dx cdot int_a^b beta(x) dx$.
Suppose $alpha: [a, b] to (-infty, infty)$ and $beta:[a,b] to (-infty, infty)$ are both nondecreasing. Then,
$$(b-a) cdot int_a^b alpha(x) beta(x) dx ge int_a^b alpha(x) dx cdot int_a^b beta(x) dx.$$
This is Lemma A.3 from Robson, A.J., 1992. Status, the distribution of wealth, private and social attitudes to risk. Econometrica: Journal of the Econometric Society, pp.837-857.
The author omit this proof since it is straightforward, but I am struggling to prove this. I appreciate if you give some help, and please tell me if more context is needed.
integration economics
edited Dec 10 '18 at 3:11
the_fox
2,43411431
asked Dec 10 '18 at 3:00
Daeseon
818311
add a comment |
Suppose $alpha: [a, b] to (-infty, infty)$ and $beta:[a,b] to (-infty, infty)$ are both nondecreasing. Then,
$$(b-a) cdot int_a^b alpha(x) beta(x) dx ge int_a^b alpha(x) dx cdot int_a^b beta(x) dx.$$
This is Lemma A.3 from Robson, A.J., 1992. Status, the distribution of wealth, private and social attitudes to risk. Econometrica: Journal of the Econometric Society, pp.837-857.
The author omit this proof since it is straightforward, but I am struggling to prove this. I appreciate if you give some help, and please tell me if more context is needed.
integration economics
edited Dec 10 '18 at 3:11
the_fox
2,43411431
asked Dec 10 '18 at 3:00
Daeseon
818311
add a comment |
Suppose $alpha: [a, b] to (-infty, infty)$ and $beta:[a,b] to (-infty, infty)$ are both nondecreasing. Then,
$$(b-a) cdot int_a^b alpha(x) beta(x) dx ge int_a^b alpha(x) dx cdot int_a^b beta(x) dx.$$
This is Lemma A.3 from Robson, A.J., 1992. Status, the distribution of wealth, private and social attitudes to risk. Econometrica: Journal of the Econometric Society, pp.837-857.
The author omit this proof since it is straightforward, but I am struggling to prove this. I appreciate if you give some help, and please tell me if more context is needed.
integration economics
edited Dec 10 '18 at 3:11
the_fox
2,43411431
asked Dec 10 '18 at 3:00
Daeseon
818311
Suppose $alpha: [a, b] to (-infty, infty)$ and $beta:[a,b] to (-infty, infty)$ are both nondecreasing. Then,
$$(b-a) cdot int_a^b alpha(x) beta(x) dx ge int_a^b alpha(x) dx cdot int_a^b beta(x) dx.$$
This is Lemma A.3 from Robson, A.J., 1992. Status, the distribution of wealth, private and social attitudes to risk. Econometrica: Journal of the Econometric Society, pp.837-857.
The author omit this proof since it is straightforward, but I am struggling to prove this. I appreciate if you give some help, and please tell me if more context is needed.
integration economics
integration economics
edited Dec 10 '18 at 3:11
the_fox
2,43411431
asked Dec 10 '18 at 3:00
Daeseon
818311
edited Dec 10 '18 at 3:11
the_fox
2,43411431
asked Dec 10 '18 at 3:00
Daeseon
818311
edited Dec 10 '18 at 3:11
the_fox
2,43411431
edited Dec 10 '18 at 3:11
the_fox
2,43411431
edited Dec 10 '18 at 3:11
the_fox
2,43411431
2,43411431
asked Dec 10 '18 at 3:00
Daeseon
818311
asked Dec 10 '18 at 3:00
Daeseon
818311
asked Dec 10 '18 at 3:00
Daeseon
818311
818311
add a comment |
add a comment |
1 Answer
1
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Proof.
Let $D = [a,b]^2$. Then
$$
int_a^b 1 int_a^b alphacdot beta- int_a^b alpha int_a^b betastackrel{(1)}= iint_D( alpha(y)beta(y)-alpha(x)beta(y) ),mathrm dx ,mathrm dy stackrel{(2)}= iint_D (alpha (x)beta(x)-alpha(y)beta(x)),mathrm dx,mathrm dy = frac 12 iint_D (alpha(x)beta(x)+alpha(y)beta(y)-alpha (x)beta(y)-alpha(y)beta(x)) ,mathrm dx ,mathrm dy = frac 12 iint_D (alpha(x)-alpha(y))(beta(x)-beta(y)),mathrm dx,mathrm dy geqslant 0,
$$
since both $alpha, beta$ are nondecreasing, i.e. $(alpha(x) - alpha (y)), (beta(x)- beta(y))$ have the same sign, equivalently $(alpha(x)-alpha(y))(beta(x)-beta(y)) geqslant 0$ for all $(x,y)in [a,b]^2$.
EXPLANATION
- For two functions $f,g colon [a,b] to mathbb R$,
$$
int_a^b f int_a^b g = int_a^b f(x),mathrm dx int_a^b g(x),mathrm dx = int_a^b f(x),mathrm dx int_a^b g(y),mathrm dy ;[text{change the integration parameter}] = iint_D f(x)g(y),mathrm dx ,mathrm dy ;[text{transfer to a double integral}].
$$
Hence the $(1)$. - Use a different parameter, we have
$$
int_a^b f int_a^b g = int_a^b f(y),mathrm dy int_a^b g(x), mathrm dx = iint_D f(y)g(x),mathrm dx ,mathrm dy.
$$
Go back to the question, we have
$$
(b-a)int_a^b alphacdot beta - int_a^b alpha int_a^b beta = iint_D (alpha(y)beta(y) - alpha(x)beta(y)),
$$
also
$$
(b-a)int_a^b alphacdot beta - int_a^b alpha int_a^b beta = iint_D (alpha(x)beta(x) - alpha(y)beta(x)),
$$
add these two equations we get the $(2)$.
answered Dec 10 '18 at 3:09
xbh
5,6551522
Thanks for your answer. Could you explain how the first and second equality holds?
– Daeseon
Dec 10 '18 at 3:25
@Daeseon I add the explanation. Feel free to check it.
– xbh
Dec 10 '18 at 3:43
add a comment |
Your Answer
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1 Answer
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1 Answer
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Proof.
Let $D = [a,b]^2$. Then
$$
int_a^b 1 int_a^b alphacdot beta- int_a^b alpha int_a^b betastackrel{(1)}= iint_D( alpha(y)beta(y)-alpha(x)beta(y) ),mathrm dx ,mathrm dy stackrel{(2)}= iint_D (alpha (x)beta(x)-alpha(y)beta(x)),mathrm dx,mathrm dy = frac 12 iint_D (alpha(x)beta(x)+alpha(y)beta(y)-alpha (x)beta(y)-alpha(y)beta(x)) ,mathrm dx ,mathrm dy = frac 12 iint_D (alpha(x)-alpha(y))(beta(x)-beta(y)),mathrm dx,mathrm dy geqslant 0,
$$
since both $alpha, beta$ are nondecreasing, i.e. $(alpha(x) - alpha (y)), (beta(x)- beta(y))$ have the same sign, equivalently $(alpha(x)-alpha(y))(beta(x)-beta(y)) geqslant 0$ for all $(x,y)in [a,b]^2$.
EXPLANATION
- For two functions $f,g colon [a,b] to mathbb R$,
$$
int_a^b f int_a^b g = int_a^b f(x),mathrm dx int_a^b g(x),mathrm dx = int_a^b f(x),mathrm dx int_a^b g(y),mathrm dy ;[text{change the integration parameter}] = iint_D f(x)g(y),mathrm dx ,mathrm dy ;[text{transfer to a double integral}].
$$
Hence the $(1)$. - Use a different parameter, we have
$$
int_a^b f int_a^b g = int_a^b f(y),mathrm dy int_a^b g(x), mathrm dx = iint_D f(y)g(x),mathrm dx ,mathrm dy.
$$
Go back to the question, we have
$$
(b-a)int_a^b alphacdot beta - int_a^b alpha int_a^b beta = iint_D (alpha(y)beta(y) - alpha(x)beta(y)),
$$
also
$$
(b-a)int_a^b alphacdot beta - int_a^b alpha int_a^b beta = iint_D (alpha(x)beta(x) - alpha(y)beta(x)),
$$
add these two equations we get the $(2)$.
answered Dec 10 '18 at 3:09
xbh
5,6551522
Thanks for your answer. Could you explain how the first and second equality holds?
– Daeseon
Dec 10 '18 at 3:25
@Daeseon I add the explanation. Feel free to check it.
– xbh
Dec 10 '18 at 3:43
add a comment |
Proof.
Let $D = [a,b]^2$. Then
$$
int_a^b 1 int_a^b alphacdot beta- int_a^b alpha int_a^b betastackrel{(1)}= iint_D( alpha(y)beta(y)-alpha(x)beta(y) ),mathrm dx ,mathrm dy stackrel{(2)}= iint_D (alpha (x)beta(x)-alpha(y)beta(x)),mathrm dx,mathrm dy = frac 12 iint_D (alpha(x)beta(x)+alpha(y)beta(y)-alpha (x)beta(y)-alpha(y)beta(x)) ,mathrm dx ,mathrm dy = frac 12 iint_D (alpha(x)-alpha(y))(beta(x)-beta(y)),mathrm dx,mathrm dy geqslant 0,
$$
since both $alpha, beta$ are nondecreasing, i.e. $(alpha(x) - alpha (y)), (beta(x)- beta(y))$ have the same sign, equivalently $(alpha(x)-alpha(y))(beta(x)-beta(y)) geqslant 0$ for all $(x,y)in [a,b]^2$.
EXPLANATION
- For two functions $f,g colon [a,b] to mathbb R$,
$$
int_a^b f int_a^b g = int_a^b f(x),mathrm dx int_a^b g(x),mathrm dx = int_a^b f(x),mathrm dx int_a^b g(y),mathrm dy ;[text{change the integration parameter}] = iint_D f(x)g(y),mathrm dx ,mathrm dy ;[text{transfer to a double integral}].
$$
Hence the $(1)$. - Use a different parameter, we have
$$
int_a^b f int_a^b g = int_a^b f(y),mathrm dy int_a^b g(x), mathrm dx = iint_D f(y)g(x),mathrm dx ,mathrm dy.
$$
Go back to the question, we have
$$
(b-a)int_a^b alphacdot beta - int_a^b alpha int_a^b beta = iint_D (alpha(y)beta(y) - alpha(x)beta(y)),
$$
also
$$
(b-a)int_a^b alphacdot beta - int_a^b alpha int_a^b beta = iint_D (alpha(x)beta(x) - alpha(y)beta(x)),
$$
add these two equations we get the $(2)$.
answered Dec 10 '18 at 3:09
xbh
5,6551522
Thanks for your answer. Could you explain how the first and second equality holds?
– Daeseon
Dec 10 '18 at 3:25
@Daeseon I add the explanation. Feel free to check it.
– xbh
Dec 10 '18 at 3:43
add a comment |
Proof.
Let $D = [a,b]^2$. Then
$$
int_a^b 1 int_a^b alphacdot beta- int_a^b alpha int_a^b betastackrel{(1)}= iint_D( alpha(y)beta(y)-alpha(x)beta(y) ),mathrm dx ,mathrm dy stackrel{(2)}= iint_D (alpha (x)beta(x)-alpha(y)beta(x)),mathrm dx,mathrm dy = frac 12 iint_D (alpha(x)beta(x)+alpha(y)beta(y)-alpha (x)beta(y)-alpha(y)beta(x)) ,mathrm dx ,mathrm dy = frac 12 iint_D (alpha(x)-alpha(y))(beta(x)-beta(y)),mathrm dx,mathrm dy geqslant 0,
$$
since both $alpha, beta$ are nondecreasing, i.e. $(alpha(x) - alpha (y)), (beta(x)- beta(y))$ have the same sign, equivalently $(alpha(x)-alpha(y))(beta(x)-beta(y)) geqslant 0$ for all $(x,y)in [a,b]^2$.
EXPLANATION
- For two functions $f,g colon [a,b] to mathbb R$,
$$
int_a^b f int_a^b g = int_a^b f(x),mathrm dx int_a^b g(x),mathrm dx = int_a^b f(x),mathrm dx int_a^b g(y),mathrm dy ;[text{change the integration parameter}] = iint_D f(x)g(y),mathrm dx ,mathrm dy ;[text{transfer to a double integral}].
$$
Hence the $(1)$. - Use a different parameter, we have
$$
int_a^b f int_a^b g = int_a^b f(y),mathrm dy int_a^b g(x), mathrm dx = iint_D f(y)g(x),mathrm dx ,mathrm dy.
$$
Go back to the question, we have
$$
(b-a)int_a^b alphacdot beta - int_a^b alpha int_a^b beta = iint_D (alpha(y)beta(y) - alpha(x)beta(y)),
$$
also
$$
(b-a)int_a^b alphacdot beta - int_a^b alpha int_a^b beta = iint_D (alpha(x)beta(x) - alpha(y)beta(x)),
$$
add these two equations we get the $(2)$.
answered Dec 10 '18 at 3:09
xbh
5,6551522
Proof.
Let $D = [a,b]^2$. Then
$$
int_a^b 1 int_a^b alphacdot beta- int_a^b alpha int_a^b betastackrel{(1)}= iint_D( alpha(y)beta(y)-alpha(x)beta(y) ),mathrm dx ,mathrm dy stackrel{(2)}= iint_D (alpha (x)beta(x)-alpha(y)beta(x)),mathrm dx,mathrm dy = frac 12 iint_D (alpha(x)beta(x)+alpha(y)beta(y)-alpha (x)beta(y)-alpha(y)beta(x)) ,mathrm dx ,mathrm dy = frac 12 iint_D (alpha(x)-alpha(y))(beta(x)-beta(y)),mathrm dx,mathrm dy geqslant 0,
$$
since both $alpha, beta$ are nondecreasing, i.e. $(alpha(x) - alpha (y)), (beta(x)- beta(y))$ have the same sign, equivalently $(alpha(x)-alpha(y))(beta(x)-beta(y)) geqslant 0$ for all $(x,y)in [a,b]^2$.
EXPLANATION
- For two functions $f,g colon [a,b] to mathbb R$,
$$
int_a^b f int_a^b g = int_a^b f(x),mathrm dx int_a^b g(x),mathrm dx = int_a^b f(x),mathrm dx int_a^b g(y),mathrm dy ;[text{change the integration parameter}] = iint_D f(x)g(y),mathrm dx ,mathrm dy ;[text{transfer to a double integral}].
$$
Hence the $(1)$. - Use a different parameter, we have
$$
int_a^b f int_a^b g = int_a^b f(y),mathrm dy int_a^b g(x), mathrm dx = iint_D f(y)g(x),mathrm dx ,mathrm dy.
$$
Go back to the question, we have
$$
(b-a)int_a^b alphacdot beta - int_a^b alpha int_a^b beta = iint_D (alpha(y)beta(y) - alpha(x)beta(y)),
$$
also
$$
(b-a)int_a^b alphacdot beta - int_a^b alpha int_a^b beta = iint_D (alpha(x)beta(x) - alpha(y)beta(x)),
$$
add these two equations we get the $(2)$.
answered Dec 10 '18 at 3:09
xbh
5,6551522
edited Dec 10 '18 at 3:39
answered Dec 10 '18 at 3:09
xbh
5,6551522
answered Dec 10 '18 at 3:09
xbh
5,6551522
answered Dec 10 '18 at 3:09
xbh
5,6551522
5,6551522
Thanks for your answer. Could you explain how the first and second equality holds?
– Daeseon
Dec 10 '18 at 3:25
@Daeseon I add the explanation. Feel free to check it.
– xbh
Dec 10 '18 at 3:43
add a comment |
Thanks for your answer. Could you explain how the first and second equality holds?
– Daeseon
Dec 10 '18 at 3:25
@Daeseon I add the explanation. Feel free to check it.
– xbh
Dec 10 '18 at 3:43
Thanks for your answer. Could you explain how the first and second equality holds?
– Daeseon
Dec 10 '18 at 3:25
Thanks for your answer. Could you explain how the first and second equality holds?
– Daeseon
Dec 10 '18 at 3:25
@Daeseon I add the explanation. Feel free to check it.
– xbh
Dec 10 '18 at 3:43
@Daeseon I add the explanation. Feel free to check it.
– xbh
Dec 10 '18 at 3:43
add a comment |
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