I cannot understand why the range of integral of $x, y$ are from $0$ to $1$ when $x > 0 $ and $y < 1$.












1












$begingroup$


First of all, I am not native speak so sorry for my poor English.



I am studying the text book named "the probability and statistics for engineering and scientists" by myself.



But I am not good at this subject.



So I need help from experts.(Please help me)




Q) Let X and Y be random variables with joint density function.



$$ f(x, y) = 4xy (0<x, y<1) $$



Find the expected value of $$ Z = {sqrt{(X^2 + Y^2)}} $$.




The problem is that I cannot understand why the range of integral of x, y are from 0 to 1.



Please explain that above..






probability expected-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    ...because $f(x,y)$ is defined $0 < x<1$ and $0<y<1)$.
    $endgroup$
    – David G. Stork
    Dec 18 '18 at 1:30












  • $begingroup$
    thank you for your answer. does that means x and y are same domain(or axis)
    $endgroup$
    – ty.kim
    Dec 18 '18 at 1:32












  • $begingroup$
    You may want to start from $f(x,y) = 4xy, 0 < x,y < alpha$ where $alpha$ is some positive real number. Then if $intint f(x,y)dxdy = 1$ (as we must have for a probability density) then we must set $alpha = 1$.
    $endgroup$
    – user113988
    Dec 18 '18 at 1:33












  • $begingroup$
    @ty.kim Your English is excellent!
    $endgroup$
    – Kavi Rama Murthy
    Dec 18 '18 at 6:22










  • $begingroup$
    After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    $endgroup$
    – Clement C.
    Dec 21 '18 at 1:12
















1












$begingroup$


First of all, I am not native speak so sorry for my poor English.



I am studying the text book named "the probability and statistics for engineering and scientists" by myself.



But I am not good at this subject.



So I need help from experts.(Please help me)




Q) Let X and Y be random variables with joint density function.



$$ f(x, y) = 4xy (0<x, y<1) $$



Find the expected value of $$ Z = {sqrt{(X^2 + Y^2)}} $$.




The problem is that I cannot understand why the range of integral of x, y are from 0 to 1.



Please explain that above..






probability expected-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    ...because $f(x,y)$ is defined $0 < x<1$ and $0<y<1)$.
    $endgroup$
    – David G. Stork
    Dec 18 '18 at 1:30












  • $begingroup$
    thank you for your answer. does that means x and y are same domain(or axis)
    $endgroup$
    – ty.kim
    Dec 18 '18 at 1:32












  • $begingroup$
    You may want to start from $f(x,y) = 4xy, 0 < x,y < alpha$ where $alpha$ is some positive real number. Then if $intint f(x,y)dxdy = 1$ (as we must have for a probability density) then we must set $alpha = 1$.
    $endgroup$
    – user113988
    Dec 18 '18 at 1:33












  • $begingroup$
    @ty.kim Your English is excellent!
    $endgroup$
    – Kavi Rama Murthy
    Dec 18 '18 at 6:22










  • $begingroup$
    After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    $endgroup$
    – Clement C.
    Dec 21 '18 at 1:12














1












1








1





$begingroup$


First of all, I am not native speak so sorry for my poor English.



I am studying the text book named "the probability and statistics for engineering and scientists" by myself.



But I am not good at this subject.



So I need help from experts.(Please help me)




Q) Let X and Y be random variables with joint density function.



$$ f(x, y) = 4xy (0<x, y<1) $$



Find the expected value of $$ Z = {sqrt{(X^2 + Y^2)}} $$.




The problem is that I cannot understand why the range of integral of x, y are from 0 to 1.



Please explain that above..






probability expected-value







share|cite|improve this question











$endgroup$




First of all, I am not native speak so sorry for my poor English.



I am studying the text book named "the probability and statistics for engineering and scientists" by myself.



But I am not good at this subject.



So I need help from experts.(Please help me)




Q) Let X and Y be random variables with joint density function.



$$ f(x, y) = 4xy (0<x, y<1) $$



Find the expected value of $$ Z = {sqrt{(X^2 + Y^2)}} $$.




The problem is that I cannot understand why the range of integral of x, y are from 0 to 1.



Please explain that above..






probability expected-value




probability expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 1:32









usermath

2,5601227




2,5601227










asked Dec 18 '18 at 1:26









ty.kimty.kim

62




62












  • $begingroup$
    ...because $f(x,y)$ is defined $0 < x<1$ and $0<y<1)$.
    $endgroup$
    – David G. Stork
    Dec 18 '18 at 1:30












  • $begingroup$
    thank you for your answer. does that means x and y are same domain(or axis)
    $endgroup$
    – ty.kim
    Dec 18 '18 at 1:32












  • $begingroup$
    You may want to start from $f(x,y) = 4xy, 0 < x,y < alpha$ where $alpha$ is some positive real number. Then if $intint f(x,y)dxdy = 1$ (as we must have for a probability density) then we must set $alpha = 1$.
    $endgroup$
    – user113988
    Dec 18 '18 at 1:33












  • $begingroup$
    @ty.kim Your English is excellent!
    $endgroup$
    – Kavi Rama Murthy
    Dec 18 '18 at 6:22










  • $begingroup$
    After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    $endgroup$
    – Clement C.
    Dec 21 '18 at 1:12


















  • $begingroup$
    ...because $f(x,y)$ is defined $0 < x<1$ and $0<y<1)$.
    $endgroup$
    – David G. Stork
    Dec 18 '18 at 1:30












  • $begingroup$
    thank you for your answer. does that means x and y are same domain(or axis)
    $endgroup$
    – ty.kim
    Dec 18 '18 at 1:32












  • $begingroup$
    You may want to start from $f(x,y) = 4xy, 0 < x,y < alpha$ where $alpha$ is some positive real number. Then if $intint f(x,y)dxdy = 1$ (as we must have for a probability density) then we must set $alpha = 1$.
    $endgroup$
    – user113988
    Dec 18 '18 at 1:33












  • $begingroup$
    @ty.kim Your English is excellent!
    $endgroup$
    – Kavi Rama Murthy
    Dec 18 '18 at 6:22










  • $begingroup$
    After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    $endgroup$
    – Clement C.
    Dec 21 '18 at 1:12
















$begingroup$
...because $f(x,y)$ is defined $0 < x<1$ and $0<y<1)$.
$endgroup$
– David G. Stork
Dec 18 '18 at 1:30






$begingroup$
...because $f(x,y)$ is defined $0 < x<1$ and $0<y<1)$.
$endgroup$
– David G. Stork
Dec 18 '18 at 1:30














$begingroup$
thank you for your answer. does that means x and y are same domain(or axis)
$endgroup$
– ty.kim
Dec 18 '18 at 1:32






$begingroup$
thank you for your answer. does that means x and y are same domain(or axis)
$endgroup$
– ty.kim
Dec 18 '18 at 1:32














$begingroup$
You may want to start from $f(x,y) = 4xy, 0 < x,y < alpha$ where $alpha$ is some positive real number. Then if $intint f(x,y)dxdy = 1$ (as we must have for a probability density) then we must set $alpha = 1$.
$endgroup$
– user113988
Dec 18 '18 at 1:33






$begingroup$
You may want to start from $f(x,y) = 4xy, 0 < x,y < alpha$ where $alpha$ is some positive real number. Then if $intint f(x,y)dxdy = 1$ (as we must have for a probability density) then we must set $alpha = 1$.
$endgroup$
– user113988
Dec 18 '18 at 1:33














$begingroup$
@ty.kim Your English is excellent!
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 6:22




$begingroup$
@ty.kim Your English is excellent!
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 6:22












$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Clement C.
Dec 21 '18 at 1:12




$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Clement C.
Dec 21 '18 at 1:12










2 Answers
2






active

oldest

votes


















1












$begingroup$

The integrals are "technically" not on $(0,1)times(0,1)$ only, but on $mathbb{R}^2$: you do have
$$
mathbb{E}[Z] = int_{mathbb{R}^2} dxdy f(x,y) sqrt{x^2+y^2}
$$
as... expected. However, by definition, $f(x,y)$ is non-zero only if $(x,y)in (0,1)times(0,1)$, so we end up having
$$
mathbb{E}[Z] = int_{mathbb{R}^2} dxdy 4xy mathbf{1}_{(0,1)times(0,1)}(x,y) sqrt{x^2+y^2}
= int_{(0,1)times(0,1)} dxdy 4xy sqrt{x^2+y^2}
$$



This is because $mathbf{1}_{(0,1)times(0,1)}(x,y)$ is by definition zero when $(x,y)in(0,1)times(0,1)$, and $1$ otherwise (it is the indicator function of $(0,1)times(0,1)$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you.. I will try to understand this question based on your answer
    $endgroup$
    – ty.kim
    Dec 18 '18 at 1:49










  • $begingroup$
    @ty.kim You're welcome. If you have any doubt or further question, let me know.
    $endgroup$
    – Clement C.
    Dec 18 '18 at 1:52



















1












$begingroup$

Sometimes people write “$0<x,y<1$” to mean “$0<x<1$ and $0<y<1$”,
similar to how one might write “$x,y in (0,1)$” as an abbreviation for “$x in (0,1)$ and $y in (0,1)$”. And I think this is what's meant here. But this notation is a bit dangerous, since it can be interpreted (as you seem to have done) as “$0<x$ and $y<1$”, so I would recommend avoiding it.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    "as you seem to have done" Based on the last sentence, "The problem is that I cannot understand why the range of integral of x, y are from 0 to 1.", that does not look like what happened.
    $endgroup$
    – Clement C.
    Dec 18 '18 at 17:33










  • $begingroup$
    @ClementC.: My conclusion is based on the phrasing “when $x>0$ and $y<1$” in the title of the question. And I'm thinking that with this misinterpretation, it would indeed be hard to understand why the book integrates both $x$ and $y$ from 0 to 1 when computing the expected value.
    $endgroup$
    – Hans Lundmark
    Dec 18 '18 at 18:38












  • $begingroup$
    That's fair. ${}{}$
    $endgroup$
    – Clement C.
    Dec 18 '18 at 19:35










  • $begingroup$
    @HansLundmark Thanks a lot!
    $endgroup$
    – ty.kim
    Dec 18 '18 at 23:27











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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044679%2fi-cannot-understand-why-the-range-of-integral-of-x-y-are-from-0-to-1-when%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The integrals are "technically" not on $(0,1)times(0,1)$ only, but on $mathbb{R}^2$: you do have
$$
mathbb{E}[Z] = int_{mathbb{R}^2} dxdy f(x,y) sqrt{x^2+y^2}
$$
as... expected. However, by definition, $f(x,y)$ is non-zero only if $(x,y)in (0,1)times(0,1)$, so we end up having
$$
mathbb{E}[Z] = int_{mathbb{R}^2} dxdy 4xy mathbf{1}_{(0,1)times(0,1)}(x,y) sqrt{x^2+y^2}
= int_{(0,1)times(0,1)} dxdy 4xy sqrt{x^2+y^2}
$$



This is because $mathbf{1}_{(0,1)times(0,1)}(x,y)$ is by definition zero when $(x,y)in(0,1)times(0,1)$, and $1$ otherwise (it is the indicator function of $(0,1)times(0,1)$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you.. I will try to understand this question based on your answer
    $endgroup$
    – ty.kim
    Dec 18 '18 at 1:49










  • $begingroup$
    @ty.kim You're welcome. If you have any doubt or further question, let me know.
    $endgroup$
    – Clement C.
    Dec 18 '18 at 1:52
















1












$begingroup$

The integrals are "technically" not on $(0,1)times(0,1)$ only, but on $mathbb{R}^2$: you do have
$$
mathbb{E}[Z] = int_{mathbb{R}^2} dxdy f(x,y) sqrt{x^2+y^2}
$$
as... expected. However, by definition, $f(x,y)$ is non-zero only if $(x,y)in (0,1)times(0,1)$, so we end up having
$$
mathbb{E}[Z] = int_{mathbb{R}^2} dxdy 4xy mathbf{1}_{(0,1)times(0,1)}(x,y) sqrt{x^2+y^2}
= int_{(0,1)times(0,1)} dxdy 4xy sqrt{x^2+y^2}
$$



This is because $mathbf{1}_{(0,1)times(0,1)}(x,y)$ is by definition zero when $(x,y)in(0,1)times(0,1)$, and $1$ otherwise (it is the indicator function of $(0,1)times(0,1)$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you.. I will try to understand this question based on your answer
    $endgroup$
    – ty.kim
    Dec 18 '18 at 1:49










  • $begingroup$
    @ty.kim You're welcome. If you have any doubt or further question, let me know.
    $endgroup$
    – Clement C.
    Dec 18 '18 at 1:52














1












1








1





$begingroup$

The integrals are "technically" not on $(0,1)times(0,1)$ only, but on $mathbb{R}^2$: you do have
$$
mathbb{E}[Z] = int_{mathbb{R}^2} dxdy f(x,y) sqrt{x^2+y^2}
$$
as... expected. However, by definition, $f(x,y)$ is non-zero only if $(x,y)in (0,1)times(0,1)$, so we end up having
$$
mathbb{E}[Z] = int_{mathbb{R}^2} dxdy 4xy mathbf{1}_{(0,1)times(0,1)}(x,y) sqrt{x^2+y^2}
= int_{(0,1)times(0,1)} dxdy 4xy sqrt{x^2+y^2}
$$



This is because $mathbf{1}_{(0,1)times(0,1)}(x,y)$ is by definition zero when $(x,y)in(0,1)times(0,1)$, and $1$ otherwise (it is the indicator function of $(0,1)times(0,1)$).






share|cite|improve this answer









$endgroup$



The integrals are "technically" not on $(0,1)times(0,1)$ only, but on $mathbb{R}^2$: you do have
$$
mathbb{E}[Z] = int_{mathbb{R}^2} dxdy f(x,y) sqrt{x^2+y^2}
$$
as... expected. However, by definition, $f(x,y)$ is non-zero only if $(x,y)in (0,1)times(0,1)$, so we end up having
$$
mathbb{E}[Z] = int_{mathbb{R}^2} dxdy 4xy mathbf{1}_{(0,1)times(0,1)}(x,y) sqrt{x^2+y^2}
= int_{(0,1)times(0,1)} dxdy 4xy sqrt{x^2+y^2}
$$



This is because $mathbf{1}_{(0,1)times(0,1)}(x,y)$ is by definition zero when $(x,y)in(0,1)times(0,1)$, and $1$ otherwise (it is the indicator function of $(0,1)times(0,1)$).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 18 '18 at 1:32









Clement C.Clement C.

50.9k33992




50.9k33992












  • $begingroup$
    thank you.. I will try to understand this question based on your answer
    $endgroup$
    – ty.kim
    Dec 18 '18 at 1:49










  • $begingroup$
    @ty.kim You're welcome. If you have any doubt or further question, let me know.
    $endgroup$
    – Clement C.
    Dec 18 '18 at 1:52


















  • $begingroup$
    thank you.. I will try to understand this question based on your answer
    $endgroup$
    – ty.kim
    Dec 18 '18 at 1:49










  • $begingroup$
    @ty.kim You're welcome. If you have any doubt or further question, let me know.
    $endgroup$
    – Clement C.
    Dec 18 '18 at 1:52
















$begingroup$
thank you.. I will try to understand this question based on your answer
$endgroup$
– ty.kim
Dec 18 '18 at 1:49




$begingroup$
thank you.. I will try to understand this question based on your answer
$endgroup$
– ty.kim
Dec 18 '18 at 1:49












$begingroup$
@ty.kim You're welcome. If you have any doubt or further question, let me know.
$endgroup$
– Clement C.
Dec 18 '18 at 1:52




$begingroup$
@ty.kim You're welcome. If you have any doubt or further question, let me know.
$endgroup$
– Clement C.
Dec 18 '18 at 1:52











1












$begingroup$

Sometimes people write “$0<x,y<1$” to mean “$0<x<1$ and $0<y<1$”,
similar to how one might write “$x,y in (0,1)$” as an abbreviation for “$x in (0,1)$ and $y in (0,1)$”. And I think this is what's meant here. But this notation is a bit dangerous, since it can be interpreted (as you seem to have done) as “$0<x$ and $y<1$”, so I would recommend avoiding it.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    "as you seem to have done" Based on the last sentence, "The problem is that I cannot understand why the range of integral of x, y are from 0 to 1.", that does not look like what happened.
    $endgroup$
    – Clement C.
    Dec 18 '18 at 17:33










  • $begingroup$
    @ClementC.: My conclusion is based on the phrasing “when $x>0$ and $y<1$” in the title of the question. And I'm thinking that with this misinterpretation, it would indeed be hard to understand why the book integrates both $x$ and $y$ from 0 to 1 when computing the expected value.
    $endgroup$
    – Hans Lundmark
    Dec 18 '18 at 18:38












  • $begingroup$
    That's fair. ${}{}$
    $endgroup$
    – Clement C.
    Dec 18 '18 at 19:35










  • $begingroup$
    @HansLundmark Thanks a lot!
    $endgroup$
    – ty.kim
    Dec 18 '18 at 23:27
















1












$begingroup$

Sometimes people write “$0<x,y<1$” to mean “$0<x<1$ and $0<y<1$”,
similar to how one might write “$x,y in (0,1)$” as an abbreviation for “$x in (0,1)$ and $y in (0,1)$”. And I think this is what's meant here. But this notation is a bit dangerous, since it can be interpreted (as you seem to have done) as “$0<x$ and $y<1$”, so I would recommend avoiding it.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    "as you seem to have done" Based on the last sentence, "The problem is that I cannot understand why the range of integral of x, y are from 0 to 1.", that does not look like what happened.
    $endgroup$
    – Clement C.
    Dec 18 '18 at 17:33










  • $begingroup$
    @ClementC.: My conclusion is based on the phrasing “when $x>0$ and $y<1$” in the title of the question. And I'm thinking that with this misinterpretation, it would indeed be hard to understand why the book integrates both $x$ and $y$ from 0 to 1 when computing the expected value.
    $endgroup$
    – Hans Lundmark
    Dec 18 '18 at 18:38












  • $begingroup$
    That's fair. ${}{}$
    $endgroup$
    – Clement C.
    Dec 18 '18 at 19:35










  • $begingroup$
    @HansLundmark Thanks a lot!
    $endgroup$
    – ty.kim
    Dec 18 '18 at 23:27














1












1








1





$begingroup$

Sometimes people write “$0<x,y<1$” to mean “$0<x<1$ and $0<y<1$”,
similar to how one might write “$x,y in (0,1)$” as an abbreviation for “$x in (0,1)$ and $y in (0,1)$”. And I think this is what's meant here. But this notation is a bit dangerous, since it can be interpreted (as you seem to have done) as “$0<x$ and $y<1$”, so I would recommend avoiding it.






share|cite|improve this answer









$endgroup$



Sometimes people write “$0<x,y<1$” to mean “$0<x<1$ and $0<y<1$”,
similar to how one might write “$x,y in (0,1)$” as an abbreviation for “$x in (0,1)$ and $y in (0,1)$”. And I think this is what's meant here. But this notation is a bit dangerous, since it can be interpreted (as you seem to have done) as “$0<x$ and $y<1$”, so I would recommend avoiding it.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 18 '18 at 7:00









Hans LundmarkHans Lundmark

35.9k564115




35.9k564115












  • $begingroup$
    "as you seem to have done" Based on the last sentence, "The problem is that I cannot understand why the range of integral of x, y are from 0 to 1.", that does not look like what happened.
    $endgroup$
    – Clement C.
    Dec 18 '18 at 17:33










  • $begingroup$
    @ClementC.: My conclusion is based on the phrasing “when $x>0$ and $y<1$” in the title of the question. And I'm thinking that with this misinterpretation, it would indeed be hard to understand why the book integrates both $x$ and $y$ from 0 to 1 when computing the expected value.
    $endgroup$
    – Hans Lundmark
    Dec 18 '18 at 18:38












  • $begingroup$
    That's fair. ${}{}$
    $endgroup$
    – Clement C.
    Dec 18 '18 at 19:35










  • $begingroup$
    @HansLundmark Thanks a lot!
    $endgroup$
    – ty.kim
    Dec 18 '18 at 23:27


















  • $begingroup$
    "as you seem to have done" Based on the last sentence, "The problem is that I cannot understand why the range of integral of x, y are from 0 to 1.", that does not look like what happened.
    $endgroup$
    – Clement C.
    Dec 18 '18 at 17:33










  • $begingroup$
    @ClementC.: My conclusion is based on the phrasing “when $x>0$ and $y<1$” in the title of the question. And I'm thinking that with this misinterpretation, it would indeed be hard to understand why the book integrates both $x$ and $y$ from 0 to 1 when computing the expected value.
    $endgroup$
    – Hans Lundmark
    Dec 18 '18 at 18:38












  • $begingroup$
    That's fair. ${}{}$
    $endgroup$
    – Clement C.
    Dec 18 '18 at 19:35










  • $begingroup$
    @HansLundmark Thanks a lot!
    $endgroup$
    – ty.kim
    Dec 18 '18 at 23:27
















$begingroup$
"as you seem to have done" Based on the last sentence, "The problem is that I cannot understand why the range of integral of x, y are from 0 to 1.", that does not look like what happened.
$endgroup$
– Clement C.
Dec 18 '18 at 17:33




$begingroup$
"as you seem to have done" Based on the last sentence, "The problem is that I cannot understand why the range of integral of x, y are from 0 to 1.", that does not look like what happened.
$endgroup$
– Clement C.
Dec 18 '18 at 17:33












$begingroup$
@ClementC.: My conclusion is based on the phrasing “when $x>0$ and $y<1$” in the title of the question. And I'm thinking that with this misinterpretation, it would indeed be hard to understand why the book integrates both $x$ and $y$ from 0 to 1 when computing the expected value.
$endgroup$
– Hans Lundmark
Dec 18 '18 at 18:38






$begingroup$
@ClementC.: My conclusion is based on the phrasing “when $x>0$ and $y<1$” in the title of the question. And I'm thinking that with this misinterpretation, it would indeed be hard to understand why the book integrates both $x$ and $y$ from 0 to 1 when computing the expected value.
$endgroup$
– Hans Lundmark
Dec 18 '18 at 18:38














$begingroup$
That's fair. ${}{}$
$endgroup$
– Clement C.
Dec 18 '18 at 19:35




$begingroup$
That's fair. ${}{}$
$endgroup$
– Clement C.
Dec 18 '18 at 19:35












$begingroup$
@HansLundmark Thanks a lot!
$endgroup$
– ty.kim
Dec 18 '18 at 23:27




$begingroup$
@HansLundmark Thanks a lot!
$endgroup$
– ty.kim
Dec 18 '18 at 23:27


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044679%2fi-cannot-understand-why-the-range-of-integral-of-x-y-are-from-0-to-1-when%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
Read more