Dtjytyk

How is the claim "I am in New York only if I am in America" the same as "If I am in New York, then I am in America?

What does CI-V stand for?

Do I have a twin with permutated remainders?

Today is the Center

Do VLANs within a subnet need to have their own subnet for router on a stick?

Why does Kotter return in Welcome Back Kotter?

LaTeX closing $ signs makes cursor jump

Dragon forelimb placement

Can a Warlock become Neutral Good?

Is it legal for company to use my work email to pretend I still work there?

Why do falling prices hurt debtors?

To string or not to string

What are the differences between the usage of 'it' and 'they'?

Is it unprofessional to ask if a job posting on GlassDoor is real?

What are these boxed doors outside store fronts in New York?

What does "Puller Prush Person" mean?

Why Is Death Allowed In the Matrix?

Can an x86 CPU running in real mode be considered to be basically an 8086 CPU?

Is this a crack on the carbon frame?

How to format long polynomial?

Can I ask the recruiters in my resume to put the reason why I am rejected?

Test whether all array elements are factors of a number

Python: next in for loop

Why did the Germans forbid the possession of pet pigeons in Rostov-on-Don in 1941?

How partial derivatives in this answer were calculated, and how to understand a notation in the answer

Differentiate a conditional expected value by its probabilityIntegral $int_{-1}^{1} frac{1}{x}sqrt{frac{1+x}{1-x}} log left( frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} right) , mathrm dx$Leibniz Notation Second Derivative Chain Rule?Partial derivatives and chain rule explanation.How does this partial differentiation work?Partial derivatives of the marginal likelihood of a Gaussian ProcessDifferentiating problem under integral signProbs. 10 (a), (b), and (c), Chap. 6, in Baby Rudin: Holder's Inequality for IntegralsWhy am I not getting the right answer for this integral?Leibniz integration rule applied twiceIf $g(x) = f(ax)$, then $tilde g(k) = frac{1}{ |a|} tilde f left(frac{k}{a}right)$ - why we need to take the absolute value of $a$?

1

$begingroup$

There is a question asking how to calculate $frac{partial tilde{x}}{partial a}$, where $tilde{x}$ is defined as follows.

$$mathbb{E} left( x | x > a right) = frac{ int_a^{infty} x fleft(xright) dx }{ int_a^{infty} fleft(xright) dx }equiv tilde{x}$$ and $$Pr left (x> aright) = int_a^{infty} fleft(xright) dxequiv alpha$$

This answer states that the answer is as follows

$$
defdd#1#2{frac{mathrm d#1}{mathrm d#2}}dd{tilde x}alpha=dd{tilde x}add aalpha=left(frac{af(a)}{alpha}-frac{tilde xf(a)}{alpha}right)f(a)^{-1}=frac{a-tilde x}{alpha};.
$$

My questions are

What is the meaning/interpretation of $frac{da}{dalpha}$. Isn't $a$ just a constant?

How was this partial derivative actually calculated? I believe it is just Leibniz rule, but do I combine Liebniz rule with the quotient rule? An example of What I have in Mind is below

---

I believe Leibniz rule on the numerator would be

$$
frac{dtilde{x}}{da} int_a^{infty} x fleft(xright) dx=-af(a)
$$

and on the denominator
$$
frac{dtilde{x}}{da}int_a^{infty} fleft(xright) dx= -f(a)
$$

and then just take these two derivatives and use quotient rule to get $frac{dtilde{x}}{da}$

But I'm not sure if I'm applying Leibniz rule correct, in particular because Wikipedia article says there should be a (sorry change of notation) $int_{b(a)}^{c(a)}frac{d}{da}f(x,a)dx$ term, but here $f$ is not a function of $a$. So is it just that $frac{d}{da}f(x) =0$?

integration derivatives

share|cite|improve this question

asked Mar 20 at 0:11

user106860user106860

380315

$endgroup$

add a comment | 

1

$begingroup$

There is a question asking how to calculate $frac{partial tilde{x}}{partial a}$, where $tilde{x}$ is defined as follows.

$$mathbb{E} left( x | x > a right) = frac{ int_a^{infty} x fleft(xright) dx }{ int_a^{infty} fleft(xright) dx }equiv tilde{x}$$ and $$Pr left (x> aright) = int_a^{infty} fleft(xright) dxequiv alpha$$

This answer states that the answer is as follows

$$
defdd#1#2{frac{mathrm d#1}{mathrm d#2}}dd{tilde x}alpha=dd{tilde x}add aalpha=left(frac{af(a)}{alpha}-frac{tilde xf(a)}{alpha}right)f(a)^{-1}=frac{a-tilde x}{alpha};.
$$

My questions are

What is the meaning/interpretation of $frac{da}{dalpha}$. Isn't $a$ just a constant?

How was this partial derivative actually calculated? I believe it is just Leibniz rule, but do I combine Liebniz rule with the quotient rule? An example of What I have in Mind is below

---

I believe Leibniz rule on the numerator would be

$$
frac{dtilde{x}}{da} int_a^{infty} x fleft(xright) dx=-af(a)
$$

and on the denominator
$$
frac{dtilde{x}}{da}int_a^{infty} fleft(xright) dx= -f(a)
$$

and then just take these two derivatives and use quotient rule to get $frac{dtilde{x}}{da}$

But I'm not sure if I'm applying Leibniz rule correct, in particular because Wikipedia article says there should be a (sorry change of notation) $int_{b(a)}^{c(a)}frac{d}{da}f(x,a)dx$ term, but here $f$ is not a function of $a$. So is it just that $frac{d}{da}f(x) =0$?

integration derivatives

share|cite|improve this question

asked Mar 20 at 0:11

user106860user106860

380315

$endgroup$

add a comment | 

$begingroup$

There is a question asking how to calculate $frac{partial tilde{x}}{partial a}$, where $tilde{x}$ is defined as follows.

$$mathbb{E} left( x | x > a right) = frac{ int_a^{infty} x fleft(xright) dx }{ int_a^{infty} fleft(xright) dx }equiv tilde{x}$$ and $$Pr left (x> aright) = int_a^{infty} fleft(xright) dxequiv alpha$$

This answer states that the answer is as follows

$$
defdd#1#2{frac{mathrm d#1}{mathrm d#2}}dd{tilde x}alpha=dd{tilde x}add aalpha=left(frac{af(a)}{alpha}-frac{tilde xf(a)}{alpha}right)f(a)^{-1}=frac{a-tilde x}{alpha};.
$$

My questions are

What is the meaning/interpretation of $frac{da}{dalpha}$. Isn't $a$ just a constant?

How was this partial derivative actually calculated? I believe it is just Leibniz rule, but do I combine Liebniz rule with the quotient rule? An example of What I have in Mind is below

---

I believe Leibniz rule on the numerator would be

$$
frac{dtilde{x}}{da} int_a^{infty} x fleft(xright) dx=-af(a)
$$

and on the denominator
$$
frac{dtilde{x}}{da}int_a^{infty} fleft(xright) dx= -f(a)
$$

and then just take these two derivatives and use quotient rule to get $frac{dtilde{x}}{da}$

But I'm not sure if I'm applying Leibniz rule correct, in particular because Wikipedia article says there should be a (sorry change of notation) $int_{b(a)}^{c(a)}frac{d}{da}f(x,a)dx$ term, but here $f$ is not a function of $a$. So is it just that $frac{d}{da}f(x) =0$?

integration derivatives

share|cite|improve this question

asked Mar 20 at 0:11

user106860user106860

380315

$endgroup$

share|cite|improve this question

asked Mar 20 at 0:11

user106860user106860

380315

share|cite|improve this question

asked Mar 20 at 0:11

user106860user106860

380315

asked Mar 20 at 0:11

user106860user106860

380315

asked Mar 20 at 0:11

user106860user106860

380315

1 Answer
1

active

oldest

votes

0

$begingroup$

1. You can consider $a$ as a function of $alpha$, under some regularity condition on $f$. Then, you can apply the inverse function theorem on $a$ to compute $da / dalpha$.




2. You have applied the Leibniz rule correctly, assuming you mean $1/ da$ instead of $dtilde x / da$. Again, unter some regularity condition on $f$. In fact, you don't need the Leibniz rule. It is just the fundamental theorem of calculus.




3. Correct again. As both integrands do not depend on $a$, their derivatives are zero.

Concerning regularity condition on $f$: For example, it is sufficient if $f$ is continuous and positive.

share|cite|improve this answer

edited Mar 20 at 2:14

answered Mar 20 at 1:58

user251257user251257

7,64721128

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hello, and thank you. This is helpful. For 1) though, what you mean is that I can think of $alpha$ as $alpha(a)$ and then I know the derivative of $a(alpha(a)) = alpha(a)^{-1} = frac{1}{a'(alpha)}$? (pardon my notation at the end may not be correct)
  $endgroup$
  – user106860
  Mar 20 at 2:09
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user106860 i meant the inverse function theorem in my answer (i corrected it). I hope it is more clear now.
  $endgroup$
  – user251257
  Mar 20 at 2:15
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3154832%2fhow-partial-derivatives-in-this-answer-were-calculated-and-how-to-understand-a%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

1. You can consider $a$ as a function of $alpha$, under some regularity condition on $f$. Then, you can apply the inverse function theorem on $a$ to compute $da / dalpha$.




2. You have applied the Leibniz rule correctly, assuming you mean $1/ da$ instead of $dtilde x / da$. Again, unter some regularity condition on $f$. In fact, you don't need the Leibniz rule. It is just the fundamental theorem of calculus.




3. Correct again. As both integrands do not depend on $a$, their derivatives are zero.

Concerning regularity condition on $f$: For example, it is sufficient if $f$ is continuous and positive.

share|cite|improve this answer

edited Mar 20 at 2:14

answered Mar 20 at 1:58

user251257user251257

7,64721128

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hello, and thank you. This is helpful. For 1) though, what you mean is that I can think of $alpha$ as $alpha(a)$ and then I know the derivative of $a(alpha(a)) = alpha(a)^{-1} = frac{1}{a'(alpha)}$? (pardon my notation at the end may not be correct)
  $endgroup$
  – user106860
  Mar 20 at 2:09
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user106860 i meant the inverse function theorem in my answer (i corrected it). I hope it is more clear now.
  $endgroup$
  – user251257
  Mar 20 at 2:15
  

  
  
  
  

add a comment | 

0

$begingroup$

1. You can consider $a$ as a function of $alpha$, under some regularity condition on $f$. Then, you can apply the inverse function theorem on $a$ to compute $da / dalpha$.




2. You have applied the Leibniz rule correctly, assuming you mean $1/ da$ instead of $dtilde x / da$. Again, unter some regularity condition on $f$. In fact, you don't need the Leibniz rule. It is just the fundamental theorem of calculus.




3. Correct again. As both integrands do not depend on $a$, their derivatives are zero.

Concerning regularity condition on $f$: For example, it is sufficient if $f$ is continuous and positive.

share|cite|improve this answer

edited Mar 20 at 2:14

answered Mar 20 at 1:58

user251257user251257

7,64721128

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hello, and thank you. This is helpful. For 1) though, what you mean is that I can think of $alpha$ as $alpha(a)$ and then I know the derivative of $a(alpha(a)) = alpha(a)^{-1} = frac{1}{a'(alpha)}$? (pardon my notation at the end may not be correct)
  $endgroup$
  – user106860
  Mar 20 at 2:09
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user106860 i meant the inverse function theorem in my answer (i corrected it). I hope it is more clear now.
  $endgroup$
  – user251257
  Mar 20 at 2:15
  

  
  
  
  

add a comment | 

$begingroup$

1. You can consider $a$ as a function of $alpha$, under some regularity condition on $f$. Then, you can apply the inverse function theorem on $a$ to compute $da / dalpha$.




2. You have applied the Leibniz rule correctly, assuming you mean $1/ da$ instead of $dtilde x / da$. Again, unter some regularity condition on $f$. In fact, you don't need the Leibniz rule. It is just the fundamental theorem of calculus.




3. Correct again. As both integrands do not depend on $a$, their derivatives are zero.

Concerning regularity condition on $f$: For example, it is sufficient if $f$ is continuous and positive.

share|cite|improve this answer

edited Mar 20 at 2:14

answered Mar 20 at 1:58

user251257user251257

7,64721128

$endgroup$

share|cite|improve this answer

edited Mar 20 at 2:14

answered Mar 20 at 1:58

user251257user251257

7,64721128

answered Mar 20 at 1:58

user251257user251257

7,64721128

answered Mar 20 at 1:58

user251257user251257

7,64721128