Mixing
Fundamental Theorem of Calculus with Inverse function. Explanation, intution and proof please
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I'm an undergraduate student studying for the actuarial exams and was wondering if someone could please walk me through the proof and intuition of this please? I haven't taken an analysis course yet, I'm a senior and have taken calculus 1 through 3, linear algebra, differential equations, probability, theory of interest, and this year taking statistical inference, mathematics of finance, and number theory. I wanted to put my math background here so that someone can help me knowing the level of mathematics I have. I understand the FTC but for some reason this stumped me and can't seem to find a reference for proofing this. I linked the pictures. Thank you to anyone who has the time to help me with this, your time is much appreciated. The Y and X here are random variables but I don't think that matters since random variables are functions anyways
Figured it out it was proved in my probability book with a great proof!.
Question Picture
calculus probability transformation
asked Jan 6 at 15:24
BeepBoopBeepBoop
85
$endgroup$
add a comment |
$begingroup$
I'm an undergraduate student studying for the actuarial exams and was wondering if someone could please walk me through the proof and intuition of this please? I haven't taken an analysis course yet, I'm a senior and have taken calculus 1 through 3, linear algebra, differential equations, probability, theory of interest, and this year taking statistical inference, mathematics of finance, and number theory. I wanted to put my math background here so that someone can help me knowing the level of mathematics I have. I understand the FTC but for some reason this stumped me and can't seem to find a reference for proofing this. I linked the pictures. Thank you to anyone who has the time to help me with this, your time is much appreciated. The Y and X here are random variables but I don't think that matters since random variables are functions anyways
Figured it out it was proved in my probability book with a great proof!.
Question Picture
calculus probability transformation
asked Jan 6 at 15:24
BeepBoopBeepBoop
85
$endgroup$
$begingroup$
Are you asking about differentiation under the integral sign?
$endgroup$
– Henry
Jan 6 at 15:27
$begingroup$
Hi, thank you for your response. I'm not sure if that is what I'm looking for. I checked that out and that is something beyond my level of mathematics to be honest. Also that page doesn't link any intuition to the proofs which is what I'm seeking.
$endgroup$
– BeepBoop
Jan 6 at 15:35
add a comment |
$begingroup$
I'm an undergraduate student studying for the actuarial exams and was wondering if someone could please walk me through the proof and intuition of this please? I haven't taken an analysis course yet, I'm a senior and have taken calculus 1 through 3, linear algebra, differential equations, probability, theory of interest, and this year taking statistical inference, mathematics of finance, and number theory. I wanted to put my math background here so that someone can help me knowing the level of mathematics I have. I understand the FTC but for some reason this stumped me and can't seem to find a reference for proofing this. I linked the pictures. Thank you to anyone who has the time to help me with this, your time is much appreciated. The Y and X here are random variables but I don't think that matters since random variables are functions anyways
Figured it out it was proved in my probability book with a great proof!.
Question Picture
calculus probability transformation
asked Jan 6 at 15:24
BeepBoopBeepBoop
85
$endgroup$
I'm an undergraduate student studying for the actuarial exams and was wondering if someone could please walk me through the proof and intuition of this please? I haven't taken an analysis course yet, I'm a senior and have taken calculus 1 through 3, linear algebra, differential equations, probability, theory of interest, and this year taking statistical inference, mathematics of finance, and number theory. I wanted to put my math background here so that someone can help me knowing the level of mathematics I have. I understand the FTC but for some reason this stumped me and can't seem to find a reference for proofing this. I linked the pictures. Thank you to anyone who has the time to help me with this, your time is much appreciated. The Y and X here are random variables but I don't think that matters since random variables are functions anyways
Figured it out it was proved in my probability book with a great proof!.
Question Picture
calculus probability transformation
calculus probability transformation
asked Jan 6 at 15:24
BeepBoopBeepBoop
85
asked Jan 6 at 15:24
BeepBoopBeepBoop
85
edited Jan 6 at 20:52
BeepBoop
asked Jan 6 at 15:24
BeepBoopBeepBoop
85
asked Jan 6 at 15:24
BeepBoopBeepBoop
85
asked Jan 6 at 15:24
BeepBoopBeepBoop
85
85
$begingroup$
Are you asking about differentiation under the integral sign?
$endgroup$
– Henry
Jan 6 at 15:27
$begingroup$
Hi, thank you for your response. I'm not sure if that is what I'm looking for. I checked that out and that is something beyond my level of mathematics to be honest. Also that page doesn't link any intuition to the proofs which is what I'm seeking.
$endgroup$
– BeepBoop
Jan 6 at 15:35
add a comment |
$begingroup$
Are you asking about differentiation under the integral sign?
$endgroup$
– Henry
Jan 6 at 15:27
$begingroup$
Hi, thank you for your response. I'm not sure if that is what I'm looking for. I checked that out and that is something beyond my level of mathematics to be honest. Also that page doesn't link any intuition to the proofs which is what I'm seeking.
$endgroup$
– BeepBoop
Jan 6 at 15:35
$begingroup$
Are you asking about differentiation under the integral sign?
$endgroup$
– Henry
Jan 6 at 15:27
$begingroup$
Are you asking about differentiation under the integral sign?
$endgroup$
– Henry
Jan 6 at 15:27
$begingroup$
Hi, thank you for your response. I'm not sure if that is what I'm looking for. I checked that out and that is something beyond my level of mathematics to be honest. Also that page doesn't link any intuition to the proofs which is what I'm seeking.
$endgroup$
– BeepBoop
Jan 6 at 15:35
$begingroup$
Hi, thank you for your response. I'm not sure if that is what I'm looking for. I checked that out and that is something beyond my level of mathematics to be honest. Also that page doesn't link any intuition to the proofs which is what I'm seeking.
$endgroup$
– BeepBoop
Jan 6 at 15:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The substitution rule may be of use to you: https://en.wikipedia.org/wiki/Integration_by_substitution
In this case, (looking at the proof on Wikipedia) let $phi$ be $g^{-1}$ $-$ that should transform it into a more FTC-friendly form but, as you may have noticed, this can only work if you are allowed to assume $g(-infty)$ exists (so you can rewrite the lower bound as $g^{-1}(g(-infty))$).
As one of the commenters mentioned, knowing general differentiation under the integral sign may be more broadly useful, but this should work well enough if you're not looking for anything too fancy.
answered Jan 6 at 15:41
Cardioid_Ass_22Cardioid_Ass_22
31114
$endgroup$
$begingroup$
Thank you, I don't mind fancy proofs I actually think that is the heart of mathematics but was just wondering what this would look like visually also? I just don't totally understand the general differentiation under the integral sign and honestly never heard of that. At what level of mathematics would one study that, or what course covers that?
$endgroup$
– BeepBoop
Jan 6 at 15:52
$begingroup$
@BeepBoop The idea of the proof for the DUIS formula (as presented in the link Henry provided) may not be too hard to get even with just a background in introductory calculus. The easiest way to see this is by looking at the section "General Form With Variable Limits" Wikipedia The whole thing is kind of just a more general FTC if you think about it (the difference is that now the integrand also has an explicit dependence on $alpha$ and the bounds can vary as more general functions in $alpha$. All that's really used is the MVT .
$endgroup$
– Cardioid_Ass_22
Jan 6 at 16:22
$begingroup$
Okay thank you!
$endgroup$
– BeepBoop
Jan 6 at 16:40
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
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votes
$begingroup$
The substitution rule may be of use to you: https://en.wikipedia.org/wiki/Integration_by_substitution
In this case, (looking at the proof on Wikipedia) let $phi$ be $g^{-1}$ $-$ that should transform it into a more FTC-friendly form but, as you may have noticed, this can only work if you are allowed to assume $g(-infty)$ exists (so you can rewrite the lower bound as $g^{-1}(g(-infty))$).
As one of the commenters mentioned, knowing general differentiation under the integral sign may be more broadly useful, but this should work well enough if you're not looking for anything too fancy.
answered Jan 6 at 15:41
Cardioid_Ass_22Cardioid_Ass_22
31114
$endgroup$
$begingroup$
Thank you, I don't mind fancy proofs I actually think that is the heart of mathematics but was just wondering what this would look like visually also? I just don't totally understand the general differentiation under the integral sign and honestly never heard of that. At what level of mathematics would one study that, or what course covers that?
$endgroup$
– BeepBoop
Jan 6 at 15:52
$begingroup$
@BeepBoop The idea of the proof for the DUIS formula (as presented in the link Henry provided) may not be too hard to get even with just a background in introductory calculus. The easiest way to see this is by looking at the section "General Form With Variable Limits" Wikipedia The whole thing is kind of just a more general FTC if you think about it (the difference is that now the integrand also has an explicit dependence on $alpha$ and the bounds can vary as more general functions in $alpha$. All that's really used is the MVT .
$endgroup$
– Cardioid_Ass_22
Jan 6 at 16:22
$begingroup$
Okay thank you!
$endgroup$
– BeepBoop
Jan 6 at 16:40
add a comment |
$begingroup$
The substitution rule may be of use to you: https://en.wikipedia.org/wiki/Integration_by_substitution
In this case, (looking at the proof on Wikipedia) let $phi$ be $g^{-1}$ $-$ that should transform it into a more FTC-friendly form but, as you may have noticed, this can only work if you are allowed to assume $g(-infty)$ exists (so you can rewrite the lower bound as $g^{-1}(g(-infty))$).
As one of the commenters mentioned, knowing general differentiation under the integral sign may be more broadly useful, but this should work well enough if you're not looking for anything too fancy.
answered Jan 6 at 15:41
Cardioid_Ass_22Cardioid_Ass_22
31114
$endgroup$
$begingroup$
Thank you, I don't mind fancy proofs I actually think that is the heart of mathematics but was just wondering what this would look like visually also? I just don't totally understand the general differentiation under the integral sign and honestly never heard of that. At what level of mathematics would one study that, or what course covers that?
$endgroup$
– BeepBoop
Jan 6 at 15:52
$begingroup$
@BeepBoop The idea of the proof for the DUIS formula (as presented in the link Henry provided) may not be too hard to get even with just a background in introductory calculus. The easiest way to see this is by looking at the section "General Form With Variable Limits" Wikipedia The whole thing is kind of just a more general FTC if you think about it (the difference is that now the integrand also has an explicit dependence on $alpha$ and the bounds can vary as more general functions in $alpha$. All that's really used is the MVT .
$endgroup$
– Cardioid_Ass_22
Jan 6 at 16:22
$begingroup$
Okay thank you!
$endgroup$
– BeepBoop
Jan 6 at 16:40
add a comment |
$begingroup$
The substitution rule may be of use to you: https://en.wikipedia.org/wiki/Integration_by_substitution
In this case, (looking at the proof on Wikipedia) let $phi$ be $g^{-1}$ $-$ that should transform it into a more FTC-friendly form but, as you may have noticed, this can only work if you are allowed to assume $g(-infty)$ exists (so you can rewrite the lower bound as $g^{-1}(g(-infty))$).
As one of the commenters mentioned, knowing general differentiation under the integral sign may be more broadly useful, but this should work well enough if you're not looking for anything too fancy.
answered Jan 6 at 15:41
Cardioid_Ass_22Cardioid_Ass_22
31114
$endgroup$
The substitution rule may be of use to you: https://en.wikipedia.org/wiki/Integration_by_substitution
In this case, (looking at the proof on Wikipedia) let $phi$ be $g^{-1}$ $-$ that should transform it into a more FTC-friendly form but, as you may have noticed, this can only work if you are allowed to assume $g(-infty)$ exists (so you can rewrite the lower bound as $g^{-1}(g(-infty))$).
As one of the commenters mentioned, knowing general differentiation under the integral sign may be more broadly useful, but this should work well enough if you're not looking for anything too fancy.
answered Jan 6 at 15:41
Cardioid_Ass_22Cardioid_Ass_22
31114
answered Jan 6 at 15:41
Cardioid_Ass_22Cardioid_Ass_22
31114
answered Jan 6 at 15:41
Cardioid_Ass_22Cardioid_Ass_22
31114
answered Jan 6 at 15:41
Cardioid_Ass_22Cardioid_Ass_22
31114
31114
$begingroup$
Thank you, I don't mind fancy proofs I actually think that is the heart of mathematics but was just wondering what this would look like visually also? I just don't totally understand the general differentiation under the integral sign and honestly never heard of that. At what level of mathematics would one study that, or what course covers that?
$endgroup$
– BeepBoop
Jan 6 at 15:52
$begingroup$
@BeepBoop The idea of the proof for the DUIS formula (as presented in the link Henry provided) may not be too hard to get even with just a background in introductory calculus. The easiest way to see this is by looking at the section "General Form With Variable Limits" Wikipedia The whole thing is kind of just a more general FTC if you think about it (the difference is that now the integrand also has an explicit dependence on $alpha$ and the bounds can vary as more general functions in $alpha$. All that's really used is the MVT .
$endgroup$
– Cardioid_Ass_22
Jan 6 at 16:22
$begingroup$
Okay thank you!
$endgroup$
– BeepBoop
Jan 6 at 16:40
add a comment |
$begingroup$
Thank you, I don't mind fancy proofs I actually think that is the heart of mathematics but was just wondering what this would look like visually also? I just don't totally understand the general differentiation under the integral sign and honestly never heard of that. At what level of mathematics would one study that, or what course covers that?
$endgroup$
– BeepBoop
Jan 6 at 15:52
$begingroup$
@BeepBoop The idea of the proof for the DUIS formula (as presented in the link Henry provided) may not be too hard to get even with just a background in introductory calculus. The easiest way to see this is by looking at the section "General Form With Variable Limits" Wikipedia The whole thing is kind of just a more general FTC if you think about it (the difference is that now the integrand also has an explicit dependence on $alpha$ and the bounds can vary as more general functions in $alpha$. All that's really used is the MVT .
$endgroup$
– Cardioid_Ass_22
Jan 6 at 16:22
$begingroup$
Okay thank you!
$endgroup$
– BeepBoop
Jan 6 at 16:40
$begingroup$
Thank you, I don't mind fancy proofs I actually think that is the heart of mathematics but was just wondering what this would look like visually also? I just don't totally understand the general differentiation under the integral sign and honestly never heard of that. At what level of mathematics would one study that, or what course covers that?
$endgroup$
– BeepBoop
Jan 6 at 15:52
$begingroup$
Thank you, I don't mind fancy proofs I actually think that is the heart of mathematics but was just wondering what this would look like visually also? I just don't totally understand the general differentiation under the integral sign and honestly never heard of that. At what level of mathematics would one study that, or what course covers that?
$endgroup$
– BeepBoop
Jan 6 at 15:52
$begingroup$
@BeepBoop The idea of the proof for the DUIS formula (as presented in the link Henry provided) may not be too hard to get even with just a background in introductory calculus. The easiest way to see this is by looking at the section "General Form With Variable Limits" Wikipedia The whole thing is kind of just a more general FTC if you think about it (the difference is that now the integrand also has an explicit dependence on $alpha$ and the bounds can vary as more general functions in $alpha$. All that's really used is the MVT .
$endgroup$
– Cardioid_Ass_22
Jan 6 at 16:22
$begingroup$
@BeepBoop The idea of the proof for the DUIS formula (as presented in the link Henry provided) may not be too hard to get even with just a background in introductory calculus. The easiest way to see this is by looking at the section "General Form With Variable Limits" Wikipedia The whole thing is kind of just a more general FTC if you think about it (the difference is that now the integrand also has an explicit dependence on $alpha$ and the bounds can vary as more general functions in $alpha$. All that's really used is the MVT .
$endgroup$
– Cardioid_Ass_22
Jan 6 at 16:22
$begingroup$
Okay thank you!
$endgroup$
– BeepBoop
Jan 6 at 16:40
$begingroup$
Okay thank you!
$endgroup$
– BeepBoop
Jan 6 at 16:40
add a comment |
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$begingroup$
Are you asking about differentiation under the integral sign?
$endgroup$
– Henry
Jan 6 at 15:27
$begingroup$
Hi, thank you for your response. I'm not sure if that is what I'm looking for. I checked that out and that is something beyond my level of mathematics to be honest. Also that page doesn't link any intuition to the proofs which is what I'm seeking.
$endgroup$
– BeepBoop
Jan 6 at 15:35