Change of variable in Definite Integrals : Shift by value of Integral [closed]
$begingroup$
For an integral of the form(given below) which does not have an anti-derivative
$I = int_{t=0}^infty f(t)dt$
I wish to bound the value of the integral, by using a bound given in the problem definition, $f(t-I)<= g(t)$, such that $g(t)$ has an anti-derivative.
Is a change of variable $u=t-I$ allowed which results in possibility of bounding the function to be integrated?
I.e., Can we shift the variable by the value of integral which we are currently calculating.
What are the conditions on $I$ or $f(t)$ that are required for the Change of variable given above to be allowed?
definite-integrals change-of-variable
asked Dec 28 '18 at 16:59
Dheeraj PrasannaDheeraj Prasanna
92
$endgroup$
closed as off-topic by RRL, Holo, KReiser, José Carlos Santos, metamorphy Dec 29 '18 at 8:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Holo, KReiser, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 5 more comments
$begingroup$
For an integral of the form(given below) which does not have an anti-derivative
$I = int_{t=0}^infty f(t)dt$
I wish to bound the value of the integral, by using a bound given in the problem definition, $f(t-I)<= g(t)$, such that $g(t)$ has an anti-derivative.
Is a change of variable $u=t-I$ allowed which results in possibility of bounding the function to be integrated?
I.e., Can we shift the variable by the value of integral which we are currently calculating.
What are the conditions on $I$ or $f(t)$ that are required for the Change of variable given above to be allowed?
definite-integrals change-of-variable
asked Dec 28 '18 at 16:59
Dheeraj PrasannaDheeraj Prasanna
92
$endgroup$
closed as off-topic by RRL, Holo, KReiser, José Carlos Santos, metamorphy Dec 29 '18 at 8:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Holo, KReiser, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
As long as $I$ exists and is finite, why not
$endgroup$
– Jakobian
Dec 28 '18 at 17:14
$begingroup$
The value $I$ is not known. The change of variable is done to bound the integral value. Finally answer i guess comes out to be finite, but i have not found the exact value.
$endgroup$
– Dheeraj Prasanna
Dec 28 '18 at 17:21
$begingroup$
Then you can't just do that, you might end up with a false positive. Would be better if you shed more light on what you're doing if you want some real discussion about your problem here
$endgroup$
– Jakobian
Dec 28 '18 at 17:22
$begingroup$
The problem statement is to bound Expectation of a Non-Negative Random Variable. The input to the problem gives a bound on $P(Z-E[Z]>t)$. Hence to calculate a bound $E[Z]$, I am using the definition of Expectation as Integration of Complementary CDF. But to bring it to the form of bound given in definition, it is required to shift the integral variable by $E[Z]$ which is the value of the integral.
$endgroup$
– Dheeraj Prasanna
Dec 28 '18 at 17:31
$begingroup$
Do we know that the integral is bounded? If we do, then we know it exists, and there's no problem with substituting $u = t - E[Z]$
$endgroup$
– Jakobian
Dec 28 '18 at 17:39
|
show 5 more comments
$begingroup$
For an integral of the form(given below) which does not have an anti-derivative
$I = int_{t=0}^infty f(t)dt$
I wish to bound the value of the integral, by using a bound given in the problem definition, $f(t-I)<= g(t)$, such that $g(t)$ has an anti-derivative.
Is a change of variable $u=t-I$ allowed which results in possibility of bounding the function to be integrated?
I.e., Can we shift the variable by the value of integral which we are currently calculating.
What are the conditions on $I$ or $f(t)$ that are required for the Change of variable given above to be allowed?
definite-integrals change-of-variable
asked Dec 28 '18 at 16:59
Dheeraj PrasannaDheeraj Prasanna
92
$endgroup$
For an integral of the form(given below) which does not have an anti-derivative
$I = int_{t=0}^infty f(t)dt$
I wish to bound the value of the integral, by using a bound given in the problem definition, $f(t-I)<= g(t)$, such that $g(t)$ has an anti-derivative.
Is a change of variable $u=t-I$ allowed which results in possibility of bounding the function to be integrated?
I.e., Can we shift the variable by the value of integral which we are currently calculating.
What are the conditions on $I$ or $f(t)$ that are required for the Change of variable given above to be allowed?
definite-integrals change-of-variable
definite-integrals change-of-variable
asked Dec 28 '18 at 16:59
Dheeraj PrasannaDheeraj Prasanna
92
asked Dec 28 '18 at 16:59
Dheeraj PrasannaDheeraj Prasanna
92
edited Dec 29 '18 at 10:42
Dheeraj Prasanna
asked Dec 28 '18 at 16:59
Dheeraj PrasannaDheeraj Prasanna
92
asked Dec 28 '18 at 16:59
Dheeraj PrasannaDheeraj Prasanna
92
asked Dec 28 '18 at 16:59
Dheeraj PrasannaDheeraj Prasanna
92
92
closed as off-topic by RRL, Holo, KReiser, José Carlos Santos, metamorphy Dec 29 '18 at 8:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Holo, KReiser, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, Holo, KReiser, José Carlos Santos, metamorphy Dec 29 '18 at 8:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Holo, KReiser, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
As long as $I$ exists and is finite, why not
$endgroup$
– Jakobian
Dec 28 '18 at 17:14
$begingroup$
The value $I$ is not known. The change of variable is done to bound the integral value. Finally answer i guess comes out to be finite, but i have not found the exact value.
$endgroup$
– Dheeraj Prasanna
Dec 28 '18 at 17:21
$begingroup$
Then you can't just do that, you might end up with a false positive. Would be better if you shed more light on what you're doing if you want some real discussion about your problem here
$endgroup$
– Jakobian
Dec 28 '18 at 17:22
$begingroup$
The problem statement is to bound Expectation of a Non-Negative Random Variable. The input to the problem gives a bound on $P(Z-E[Z]>t)$. Hence to calculate a bound $E[Z]$, I am using the definition of Expectation as Integration of Complementary CDF. But to bring it to the form of bound given in definition, it is required to shift the integral variable by $E[Z]$ which is the value of the integral.
$endgroup$
– Dheeraj Prasanna
Dec 28 '18 at 17:31
$begingroup$
Do we know that the integral is bounded? If we do, then we know it exists, and there's no problem with substituting $u = t - E[Z]$
$endgroup$
– Jakobian
Dec 28 '18 at 17:39
|
show 5 more comments
$begingroup$
As long as $I$ exists and is finite, why not
$endgroup$
– Jakobian
Dec 28 '18 at 17:14
$begingroup$
The value $I$ is not known. The change of variable is done to bound the integral value. Finally answer i guess comes out to be finite, but i have not found the exact value.
$endgroup$
– Dheeraj Prasanna
Dec 28 '18 at 17:21
$begingroup$
Then you can't just do that, you might end up with a false positive. Would be better if you shed more light on what you're doing if you want some real discussion about your problem here
$endgroup$
– Jakobian
Dec 28 '18 at 17:22
$begingroup$
The problem statement is to bound Expectation of a Non-Negative Random Variable. The input to the problem gives a bound on $P(Z-E[Z]>t)$. Hence to calculate a bound $E[Z]$, I am using the definition of Expectation as Integration of Complementary CDF. But to bring it to the form of bound given in definition, it is required to shift the integral variable by $E[Z]$ which is the value of the integral.
$endgroup$
– Dheeraj Prasanna
Dec 28 '18 at 17:31
$begingroup$
Do we know that the integral is bounded? If we do, then we know it exists, and there's no problem with substituting $u = t - E[Z]$
$endgroup$
– Jakobian
Dec 28 '18 at 17:39
$begingroup$
As long as $I$ exists and is finite, why not
$endgroup$
– Jakobian
Dec 28 '18 at 17:14
$begingroup$
As long as $I$ exists and is finite, why not
$endgroup$
– Jakobian
Dec 28 '18 at 17:14
$begingroup$
The value $I$ is not known. The change of variable is done to bound the integral value. Finally answer i guess comes out to be finite, but i have not found the exact value.
$endgroup$
– Dheeraj Prasanna
Dec 28 '18 at 17:21
$begingroup$
The value $I$ is not known. The change of variable is done to bound the integral value. Finally answer i guess comes out to be finite, but i have not found the exact value.
$endgroup$
– Dheeraj Prasanna
Dec 28 '18 at 17:21
$begingroup$
Then you can't just do that, you might end up with a false positive. Would be better if you shed more light on what you're doing if you want some real discussion about your problem here
$endgroup$
– Jakobian
Dec 28 '18 at 17:22
$begingroup$
Then you can't just do that, you might end up with a false positive. Would be better if you shed more light on what you're doing if you want some real discussion about your problem here
$endgroup$
– Jakobian
Dec 28 '18 at 17:22
$begingroup$
The problem statement is to bound Expectation of a Non-Negative Random Variable. The input to the problem gives a bound on $P(Z-E[Z]>t)$. Hence to calculate a bound $E[Z]$, I am using the definition of Expectation as Integration of Complementary CDF. But to bring it to the form of bound given in definition, it is required to shift the integral variable by $E[Z]$ which is the value of the integral.
$endgroup$
– Dheeraj Prasanna
Dec 28 '18 at 17:31
$begingroup$
The problem statement is to bound Expectation of a Non-Negative Random Variable. The input to the problem gives a bound on $P(Z-E[Z]>t)$. Hence to calculate a bound $E[Z]$, I am using the definition of Expectation as Integration of Complementary CDF. But to bring it to the form of bound given in definition, it is required to shift the integral variable by $E[Z]$ which is the value of the integral.
$endgroup$
– Dheeraj Prasanna
Dec 28 '18 at 17:31
$begingroup$
Do we know that the integral is bounded? If we do, then we know it exists, and there's no problem with substituting $u = t - E[Z]$
$endgroup$
– Jakobian
Dec 28 '18 at 17:39
$begingroup$
Do we know that the integral is bounded? If we do, then we know it exists, and there's no problem with substituting $u = t - E[Z]$
$endgroup$
– Jakobian
Dec 28 '18 at 17:39
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If $I$ is a real.
Let $g(u)=u+I$.
$g$ is a $C^1$ bijection from $[-I,+infty)$ to $[0,+infty)$ then
$$I=int_{-I}^{+infty}f(u+I)du$$
answered Dec 28 '18 at 17:21
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $I$ is a real.
Let $g(u)=u+I$.
$g$ is a $C^1$ bijection from $[-I,+infty)$ to $[0,+infty)$ then
$$I=int_{-I}^{+infty}f(u+I)du$$
answered Dec 28 '18 at 17:21
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
$begingroup$
If $I$ is a real.
Let $g(u)=u+I$.
$g$ is a $C^1$ bijection from $[-I,+infty)$ to $[0,+infty)$ then
$$I=int_{-I}^{+infty}f(u+I)du$$
answered Dec 28 '18 at 17:21
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
$begingroup$
If $I$ is a real.
Let $g(u)=u+I$.
$g$ is a $C^1$ bijection from $[-I,+infty)$ to $[0,+infty)$ then
$$I=int_{-I}^{+infty}f(u+I)du$$
answered Dec 28 '18 at 17:21
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
If $I$ is a real.
Let $g(u)=u+I$.
$g$ is a $C^1$ bijection from $[-I,+infty)$ to $[0,+infty)$ then
$$I=int_{-I}^{+infty}f(u+I)du$$
answered Dec 28 '18 at 17:21
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Dec 28 '18 at 17:21
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Dec 28 '18 at 17:21
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Dec 28 '18 at 17:21
hamam_Abdallahhamam_Abdallah
38.2k21634
38.2k21634
add a comment |
add a comment |
$begingroup$
As long as $I$ exists and is finite, why not
$endgroup$
– Jakobian
Dec 28 '18 at 17:14
$begingroup$
The value $I$ is not known. The change of variable is done to bound the integral value. Finally answer i guess comes out to be finite, but i have not found the exact value.
$endgroup$
– Dheeraj Prasanna
Dec 28 '18 at 17:21
$begingroup$
Then you can't just do that, you might end up with a false positive. Would be better if you shed more light on what you're doing if you want some real discussion about your problem here
$endgroup$
– Jakobian
Dec 28 '18 at 17:22
$begingroup$
The problem statement is to bound Expectation of a Non-Negative Random Variable. The input to the problem gives a bound on $P(Z-E[Z]>t)$. Hence to calculate a bound $E[Z]$, I am using the definition of Expectation as Integration of Complementary CDF. But to bring it to the form of bound given in definition, it is required to shift the integral variable by $E[Z]$ which is the value of the integral.
$endgroup$
– Dheeraj Prasanna
Dec 28 '18 at 17:31
$begingroup$
Do we know that the integral is bounded? If we do, then we know it exists, and there's no problem with substituting $u = t - E[Z]$
$endgroup$
– Jakobian
Dec 28 '18 at 17:39