Prove $(b-a) cdot int_a^b alpha(x) beta(x) dx ge int_a^b alpha(x) dx cdot int_a^b beta(x) dx$.












2















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







share|cite|improve this question





























    2















    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







    share|cite|improve this question



























      2












      2








      2








      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







      share|cite|improve this question
















      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 3:11









      the_fox

      2,43411431




      2,43411431










      asked Dec 10 '18 at 3:00









      Daeseon

      818311




      818311






















          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




          1. 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 $(1)$.

          2. 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 $(2)$.






          share|cite|improve this answer























          • 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











          Your 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




          1. 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 $(1)$.

          2. 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 $(2)$.






          share|cite|improve this answer























          • 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
















          1














          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




          1. 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 $(1)$.

          2. 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 $(2)$.






          share|cite|improve this answer























          • 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














          1












          1








          1






          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




          1. 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 $(1)$.

          2. 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 $(2)$.






          share|cite|improve this answer














          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




          1. 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 $(1)$.

          2. 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 $(2)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 3:39

























          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


















          • 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


















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