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Simplifying the division of integrals
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Is there a way to simplify this division of integrals, or at least find a bound for it, without writing it in terms of CDF of Beta?
I tried to use the idea here but didn't have much luck.
$$fracint_l^u x^a (1-x)^b-1 ;dxint_l^u x^a-1 (1-x)^b-1 ; dx$$
integration
asked Jul 23 at 22:08
mathlover
416
add a comment |Â
up vote
1
down vote
favorite
Is there a way to simplify this division of integrals, or at least find a bound for it, without writing it in terms of CDF of Beta?
I tried to use the idea here but didn't have much luck.
$$fracint_l^u x^a (1-x)^b-1 ;dxint_l^u x^a-1 (1-x)^b-1 ; dx$$
integration
asked Jul 23 at 22:08
mathlover
416
Are ther any restrictions for $l$ and $u$? Because if $l=0$ and $u=1$ you could rewrite this fraction using Beta or Gamma functions. Or, without these conditions, maybe in terms of the incomplete Beta function. I have asked similiar question by myself and I would be interested in an answer too.
– mrtaurho
Jul 23 at 22:14
1
@mrtaurho $l >0$ and $u < 1$.
– mathlover
Jul 23 at 22:19
Doesn't the identity $$int_l^u x^a (1-x)^b-1 , dx=int_l^u x^a-1 (1-x)^b-1 , dx-int_l^u x^a-1 (1-x)^b , dx$$, and that the integrals are all positive, imply that the division of the above integrals is bounded between 0 and 1?
– James Arathoon
Jul 24 at 23:59
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is there a way to simplify this division of integrals, or at least find a bound for it, without writing it in terms of CDF of Beta?
I tried to use the idea here but didn't have much luck.
$$fracint_l^u x^a (1-x)^b-1 ;dxint_l^u x^a-1 (1-x)^b-1 ; dx$$
integration
asked Jul 23 at 22:08
mathlover
416
Is there a way to simplify this division of integrals, or at least find a bound for it, without writing it in terms of CDF of Beta?
I tried to use the idea here but didn't have much luck.
$$fracint_l^u x^a (1-x)^b-1 ;dxint_l^u x^a-1 (1-x)^b-1 ; dx$$
integration
asked Jul 23 at 22:08
mathlover
416
edited Jul 24 at 22:37
asked Jul 23 at 22:08
mathlover
416
asked Jul 23 at 22:08
mathlover
416
asked Jul 23 at 22:08
mathlover
416
416
Are ther any restrictions for $l$ and $u$? Because if $l=0$ and $u=1$ you could rewrite this fraction using Beta or Gamma functions. Or, without these conditions, maybe in terms of the incomplete Beta function. I have asked similiar question by myself and I would be interested in an answer too.
– mrtaurho
Jul 23 at 22:14
1
@mrtaurho $l >0$ and $u < 1$.
– mathlover
Jul 23 at 22:19
Doesn't the identity $$int_l^u x^a (1-x)^b-1 , dx=int_l^u x^a-1 (1-x)^b-1 , dx-int_l^u x^a-1 (1-x)^b , dx$$, and that the integrals are all positive, imply that the division of the above integrals is bounded between 0 and 1?
– James Arathoon
Jul 24 at 23:59
add a comment |Â
Are ther any restrictions for $l$ and $u$? Because if $l=0$ and $u=1$ you could rewrite this fraction using Beta or Gamma functions. Or, without these conditions, maybe in terms of the incomplete Beta function. I have asked similiar question by myself and I would be interested in an answer too.
– mrtaurho
Jul 23 at 22:14
1
@mrtaurho $l >0$ and $u < 1$.
– mathlover
Jul 23 at 22:19
Doesn't the identity $$int_l^u x^a (1-x)^b-1 , dx=int_l^u x^a-1 (1-x)^b-1 , dx-int_l^u x^a-1 (1-x)^b , dx$$, and that the integrals are all positive, imply that the division of the above integrals is bounded between 0 and 1?
– James Arathoon
Jul 24 at 23:59
Are ther any restrictions for $l$ and $u$? Because if $l=0$ and $u=1$ you could rewrite this fraction using Beta or Gamma functions. Or, without these conditions, maybe in terms of the incomplete Beta function. I have asked similiar question by myself and I would be interested in an answer too.
– mrtaurho
Jul 23 at 22:14
Are ther any restrictions for $l$ and $u$? Because if $l=0$ and $u=1$ you could rewrite this fraction using Beta or Gamma functions. Or, without these conditions, maybe in terms of the incomplete Beta function. I have asked similiar question by myself and I would be interested in an answer too.
– mrtaurho
Jul 23 at 22:14
1
1
@mrtaurho $l >0$ and $u < 1$.
– mathlover
Jul 23 at 22:19
@mrtaurho $l >0$ and $u < 1$.
– mathlover
Jul 23 at 22:19
Doesn't the identity $$int_l^u x^a (1-x)^b-1 , dx=int_l^u x^a-1 (1-x)^b-1 , dx-int_l^u x^a-1 (1-x)^b , dx$$, and that the integrals are all positive, imply that the division of the above integrals is bounded between 0 and 1?
– James Arathoon
Jul 24 at 23:59
Doesn't the identity $$int_l^u x^a (1-x)^b-1 , dx=int_l^u x^a-1 (1-x)^b-1 , dx-int_l^u x^a-1 (1-x)^b , dx$$, and that the integrals are all positive, imply that the division of the above integrals is bounded between 0 and 1?
– James Arathoon
Jul 24 at 23:59
add a comment |Â
1 Answer
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I would try it as follows:
$$int_l^u x^a(1-x)^b-1mathrmdx~=~int_0^u x^a(1-x)^b-1mathrmdx~-~int_0^l x^a(1-x)^b-1mathrmdx$$
$$int_l^u x^a-1(1-x)^b-1mathrmdx~=~int_0^u x^a-1(1-x)^b-1mathrmdx~-~int_0^l x^a-1(1-x)^b-1mathrmdx$$
With the incomplete Beta function
$$B(x; a,b)~=~ int_0^x x^a-1(1-x)^b-1mathrmdx$$
The fraction would become
$$beginalignfracint_l^u x^a(1-x)^b-1mathrmdxint_l^u x^a-1(1-x)^b-1mathrmdx~&=~fracint_0^u x^a(1-x)^b-1mathrmdx~-~int_0^l x^a(1-x)^b-1mathrmdxint_0^u x^a-1(1-x)^b-1mathrmdx~-~int_0^l x^a-1(1-x)^b-1mathrmdx\~&=~fracB(u;a+1,b)-B(l;a+1,b)B(u;a,b)-B(l;a,b)endalign$$
First of all I am not sure if this is even possible to do so in this way and on the other hand I am not sure if this helps.
I do not know another way without the incomplete Beta function.
answered Jul 23 at 22:27
mrtaurho
750219
Thanks. I was aware of this way of derivation(I mentioned in the question without using CDF of beta). I'm interested in another way without using the incomplete beta.
– mathlover
Jul 23 at 23:31
I just realized it afterwards. I am not familiar with another of simplification but as I already mentioned I would be interested in one too.
– mrtaurho
Jul 23 at 23:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I would try it as follows:
$$int_l^u x^a(1-x)^b-1mathrmdx~=~int_0^u x^a(1-x)^b-1mathrmdx~-~int_0^l x^a(1-x)^b-1mathrmdx$$
$$int_l^u x^a-1(1-x)^b-1mathrmdx~=~int_0^u x^a-1(1-x)^b-1mathrmdx~-~int_0^l x^a-1(1-x)^b-1mathrmdx$$
With the incomplete Beta function
$$B(x; a,b)~=~ int_0^x x^a-1(1-x)^b-1mathrmdx$$
The fraction would become
$$beginalignfracint_l^u x^a(1-x)^b-1mathrmdxint_l^u x^a-1(1-x)^b-1mathrmdx~&=~fracint_0^u x^a(1-x)^b-1mathrmdx~-~int_0^l x^a(1-x)^b-1mathrmdxint_0^u x^a-1(1-x)^b-1mathrmdx~-~int_0^l x^a-1(1-x)^b-1mathrmdx\~&=~fracB(u;a+1,b)-B(l;a+1,b)B(u;a,b)-B(l;a,b)endalign$$
First of all I am not sure if this is even possible to do so in this way and on the other hand I am not sure if this helps.
I do not know another way without the incomplete Beta function.
answered Jul 23 at 22:27
mrtaurho
750219
Thanks. I was aware of this way of derivation(I mentioned in the question without using CDF of beta). I'm interested in another way without using the incomplete beta.
– mathlover
Jul 23 at 23:31
I just realized it afterwards. I am not familiar with another of simplification but as I already mentioned I would be interested in one too.
– mrtaurho
Jul 23 at 23:34
add a comment |Â
up vote
1
down vote
I would try it as follows:
$$int_l^u x^a(1-x)^b-1mathrmdx~=~int_0^u x^a(1-x)^b-1mathrmdx~-~int_0^l x^a(1-x)^b-1mathrmdx$$
$$int_l^u x^a-1(1-x)^b-1mathrmdx~=~int_0^u x^a-1(1-x)^b-1mathrmdx~-~int_0^l x^a-1(1-x)^b-1mathrmdx$$
With the incomplete Beta function
$$B(x; a,b)~=~ int_0^x x^a-1(1-x)^b-1mathrmdx$$
The fraction would become
$$beginalignfracint_l^u x^a(1-x)^b-1mathrmdxint_l^u x^a-1(1-x)^b-1mathrmdx~&=~fracint_0^u x^a(1-x)^b-1mathrmdx~-~int_0^l x^a(1-x)^b-1mathrmdxint_0^u x^a-1(1-x)^b-1mathrmdx~-~int_0^l x^a-1(1-x)^b-1mathrmdx\~&=~fracB(u;a+1,b)-B(l;a+1,b)B(u;a,b)-B(l;a,b)endalign$$
First of all I am not sure if this is even possible to do so in this way and on the other hand I am not sure if this helps.
I do not know another way without the incomplete Beta function.
answered Jul 23 at 22:27
mrtaurho
750219
Thanks. I was aware of this way of derivation(I mentioned in the question without using CDF of beta). I'm interested in another way without using the incomplete beta.
– mathlover
Jul 23 at 23:31
I just realized it afterwards. I am not familiar with another of simplification but as I already mentioned I would be interested in one too.
– mrtaurho
Jul 23 at 23:34
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I would try it as follows:
$$int_l^u x^a(1-x)^b-1mathrmdx~=~int_0^u x^a(1-x)^b-1mathrmdx~-~int_0^l x^a(1-x)^b-1mathrmdx$$
$$int_l^u x^a-1(1-x)^b-1mathrmdx~=~int_0^u x^a-1(1-x)^b-1mathrmdx~-~int_0^l x^a-1(1-x)^b-1mathrmdx$$
With the incomplete Beta function
$$B(x; a,b)~=~ int_0^x x^a-1(1-x)^b-1mathrmdx$$
The fraction would become
$$beginalignfracint_l^u x^a(1-x)^b-1mathrmdxint_l^u x^a-1(1-x)^b-1mathrmdx~&=~fracint_0^u x^a(1-x)^b-1mathrmdx~-~int_0^l x^a(1-x)^b-1mathrmdxint_0^u x^a-1(1-x)^b-1mathrmdx~-~int_0^l x^a-1(1-x)^b-1mathrmdx\~&=~fracB(u;a+1,b)-B(l;a+1,b)B(u;a,b)-B(l;a,b)endalign$$
First of all I am not sure if this is even possible to do so in this way and on the other hand I am not sure if this helps.
I do not know another way without the incomplete Beta function.
answered Jul 23 at 22:27
mrtaurho
750219
I would try it as follows:
$$int_l^u x^a(1-x)^b-1mathrmdx~=~int_0^u x^a(1-x)^b-1mathrmdx~-~int_0^l x^a(1-x)^b-1mathrmdx$$
$$int_l^u x^a-1(1-x)^b-1mathrmdx~=~int_0^u x^a-1(1-x)^b-1mathrmdx~-~int_0^l x^a-1(1-x)^b-1mathrmdx$$
With the incomplete Beta function
$$B(x; a,b)~=~ int_0^x x^a-1(1-x)^b-1mathrmdx$$
The fraction would become
$$beginalignfracint_l^u x^a(1-x)^b-1mathrmdxint_l^u x^a-1(1-x)^b-1mathrmdx~&=~fracint_0^u x^a(1-x)^b-1mathrmdx~-~int_0^l x^a(1-x)^b-1mathrmdxint_0^u x^a-1(1-x)^b-1mathrmdx~-~int_0^l x^a-1(1-x)^b-1mathrmdx\~&=~fracB(u;a+1,b)-B(l;a+1,b)B(u;a,b)-B(l;a,b)endalign$$
First of all I am not sure if this is even possible to do so in this way and on the other hand I am not sure if this helps.
I do not know another way without the incomplete Beta function.
answered Jul 23 at 22:27
mrtaurho
750219
answered Jul 23 at 22:27
mrtaurho
750219
answered Jul 23 at 22:27
mrtaurho
750219
answered Jul 23 at 22:27
mrtaurho
750219
750219
Thanks. I was aware of this way of derivation(I mentioned in the question without using CDF of beta). I'm interested in another way without using the incomplete beta.
– mathlover
Jul 23 at 23:31
I just realized it afterwards. I am not familiar with another of simplification but as I already mentioned I would be interested in one too.
– mrtaurho
Jul 23 at 23:34
add a comment |Â
Thanks. I was aware of this way of derivation(I mentioned in the question without using CDF of beta). I'm interested in another way without using the incomplete beta.
– mathlover
Jul 23 at 23:31
I just realized it afterwards. I am not familiar with another of simplification but as I already mentioned I would be interested in one too.
– mrtaurho
Jul 23 at 23:34
Thanks. I was aware of this way of derivation(I mentioned in the question without using CDF of beta). I'm interested in another way without using the incomplete beta.
– mathlover
Jul 23 at 23:31
Thanks. I was aware of this way of derivation(I mentioned in the question without using CDF of beta). I'm interested in another way without using the incomplete beta.
– mathlover
Jul 23 at 23:31
I just realized it afterwards. I am not familiar with another of simplification but as I already mentioned I would be interested in one too.
– mrtaurho
Jul 23 at 23:34
I just realized it afterwards. I am not familiar with another of simplification but as I already mentioned I would be interested in one too.
– mrtaurho
Jul 23 at 23:34
add a comment |Â
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Are ther any restrictions for $l$ and $u$? Because if $l=0$ and $u=1$ you could rewrite this fraction using Beta or Gamma functions. Or, without these conditions, maybe in terms of the incomplete Beta function. I have asked similiar question by myself and I would be interested in an answer too.
– mrtaurho
Jul 23 at 22:14
1
@mrtaurho $l >0$ and $u < 1$.
– mathlover
Jul 23 at 22:19
Doesn't the identity $$int_l^u x^a (1-x)^b-1 , dx=int_l^u x^a-1 (1-x)^b-1 , dx-int_l^u x^a-1 (1-x)^b , dx$$, and that the integrals are all positive, imply that the division of the above integrals is bounded between 0 and 1?
– James Arathoon
Jul 24 at 23:59