Probability question about calculating a expected value of a continuous random variable (A alternative way...











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I'm a senior studying calculus based probability and also studying for exam P. I need help walking through every part of this problem please. I understand there is a different way to do it but I need to know how to do the double integration way. I can't seem to visually or conceptually understand the problem and for me that is necessary. Here is a link to the problem.



Expected value of a continuous random variable: interchanging the order of integration



I just don't get what is going on when we are integrating, why does the inner integral go from y to infinity then from 0 to x. I just don't understand the process, I just was wondering if someone could walk me through each step and explain what is going on. I understand and derived out the other method which is A + the integral from A to B of 1-F(x) dx, where F(x) is the cumulative distribution function or just the anti-derivative of f(x).










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  • 2




    You have given a link to a question about changing the order of integration. There are several answers under that question. What is your question?
    – David K
    Nov 15 at 21:48












  • I just don't get what is going on when we are integrating, why does the inner integral go from y to infinity then from 0 to x. I just don't understand the process, I just was wondering if someone could walk me through each step and explain what is going on. I understand and derived out the other method which is A + the integral from A to B of 1-F(x) dx, where F(x) is the cumulative distribution function or just the anti-derivative of f(x).
    – BeepBoop
    Nov 16 at 13:16












  • Do you mean to say you read all the answers to the other question (which were specifically about why you can change the double integral from one form to the other) and you did not understand any of them? When you say "visually," does that mean a picture might help?
    – David K
    Nov 16 at 14:00










  • By the way, it's generally advised to put the specific information about what you have done and what you didn't understand in the question itself rather than in the comments.
    – David K
    Nov 16 at 14:13










  • Hi David, I did read the other posts and still sort of lost on why the process works. I do understand the last comment though and how to do the alternative method but I need to know and want to know why and how the other method works. A picture would help allot honestly, if it is something that can be reasoned geometrically. Okay thanks I will fix my post!
    – BeepBoop
    Nov 16 at 14:28

















up vote
0
down vote

favorite
1












I'm a senior studying calculus based probability and also studying for exam P. I need help walking through every part of this problem please. I understand there is a different way to do it but I need to know how to do the double integration way. I can't seem to visually or conceptually understand the problem and for me that is necessary. Here is a link to the problem.



Expected value of a continuous random variable: interchanging the order of integration



I just don't get what is going on when we are integrating, why does the inner integral go from y to infinity then from 0 to x. I just don't understand the process, I just was wondering if someone could walk me through each step and explain what is going on. I understand and derived out the other method which is A + the integral from A to B of 1-F(x) dx, where F(x) is the cumulative distribution function or just the anti-derivative of f(x).










share|cite|improve this question




















  • 2




    You have given a link to a question about changing the order of integration. There are several answers under that question. What is your question?
    – David K
    Nov 15 at 21:48












  • I just don't get what is going on when we are integrating, why does the inner integral go from y to infinity then from 0 to x. I just don't understand the process, I just was wondering if someone could walk me through each step and explain what is going on. I understand and derived out the other method which is A + the integral from A to B of 1-F(x) dx, where F(x) is the cumulative distribution function or just the anti-derivative of f(x).
    – BeepBoop
    Nov 16 at 13:16












  • Do you mean to say you read all the answers to the other question (which were specifically about why you can change the double integral from one form to the other) and you did not understand any of them? When you say "visually," does that mean a picture might help?
    – David K
    Nov 16 at 14:00










  • By the way, it's generally advised to put the specific information about what you have done and what you didn't understand in the question itself rather than in the comments.
    – David K
    Nov 16 at 14:13










  • Hi David, I did read the other posts and still sort of lost on why the process works. I do understand the last comment though and how to do the alternative method but I need to know and want to know why and how the other method works. A picture would help allot honestly, if it is something that can be reasoned geometrically. Okay thanks I will fix my post!
    – BeepBoop
    Nov 16 at 14:28















up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





I'm a senior studying calculus based probability and also studying for exam P. I need help walking through every part of this problem please. I understand there is a different way to do it but I need to know how to do the double integration way. I can't seem to visually or conceptually understand the problem and for me that is necessary. Here is a link to the problem.



Expected value of a continuous random variable: interchanging the order of integration



I just don't get what is going on when we are integrating, why does the inner integral go from y to infinity then from 0 to x. I just don't understand the process, I just was wondering if someone could walk me through each step and explain what is going on. I understand and derived out the other method which is A + the integral from A to B of 1-F(x) dx, where F(x) is the cumulative distribution function or just the anti-derivative of f(x).










share|cite|improve this question















I'm a senior studying calculus based probability and also studying for exam P. I need help walking through every part of this problem please. I understand there is a different way to do it but I need to know how to do the double integration way. I can't seem to visually or conceptually understand the problem and for me that is necessary. Here is a link to the problem.



Expected value of a continuous random variable: interchanging the order of integration



I just don't get what is going on when we are integrating, why does the inner integral go from y to infinity then from 0 to x. I just don't understand the process, I just was wondering if someone could walk me through each step and explain what is going on. I understand and derived out the other method which is A + the integral from A to B of 1-F(x) dx, where F(x) is the cumulative distribution function or just the anti-derivative of f(x).







probability probability-theory probability-distributions






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edited Nov 16 at 14:29

























asked Nov 15 at 19:43









BeepBoop

34




34








  • 2




    You have given a link to a question about changing the order of integration. There are several answers under that question. What is your question?
    – David K
    Nov 15 at 21:48












  • I just don't get what is going on when we are integrating, why does the inner integral go from y to infinity then from 0 to x. I just don't understand the process, I just was wondering if someone could walk me through each step and explain what is going on. I understand and derived out the other method which is A + the integral from A to B of 1-F(x) dx, where F(x) is the cumulative distribution function or just the anti-derivative of f(x).
    – BeepBoop
    Nov 16 at 13:16












  • Do you mean to say you read all the answers to the other question (which were specifically about why you can change the double integral from one form to the other) and you did not understand any of them? When you say "visually," does that mean a picture might help?
    – David K
    Nov 16 at 14:00










  • By the way, it's generally advised to put the specific information about what you have done and what you didn't understand in the question itself rather than in the comments.
    – David K
    Nov 16 at 14:13










  • Hi David, I did read the other posts and still sort of lost on why the process works. I do understand the last comment though and how to do the alternative method but I need to know and want to know why and how the other method works. A picture would help allot honestly, if it is something that can be reasoned geometrically. Okay thanks I will fix my post!
    – BeepBoop
    Nov 16 at 14:28
















  • 2




    You have given a link to a question about changing the order of integration. There are several answers under that question. What is your question?
    – David K
    Nov 15 at 21:48












  • I just don't get what is going on when we are integrating, why does the inner integral go from y to infinity then from 0 to x. I just don't understand the process, I just was wondering if someone could walk me through each step and explain what is going on. I understand and derived out the other method which is A + the integral from A to B of 1-F(x) dx, where F(x) is the cumulative distribution function or just the anti-derivative of f(x).
    – BeepBoop
    Nov 16 at 13:16












  • Do you mean to say you read all the answers to the other question (which were specifically about why you can change the double integral from one form to the other) and you did not understand any of them? When you say "visually," does that mean a picture might help?
    – David K
    Nov 16 at 14:00










  • By the way, it's generally advised to put the specific information about what you have done and what you didn't understand in the question itself rather than in the comments.
    – David K
    Nov 16 at 14:13










  • Hi David, I did read the other posts and still sort of lost on why the process works. I do understand the last comment though and how to do the alternative method but I need to know and want to know why and how the other method works. A picture would help allot honestly, if it is something that can be reasoned geometrically. Okay thanks I will fix my post!
    – BeepBoop
    Nov 16 at 14:28










2




2




You have given a link to a question about changing the order of integration. There are several answers under that question. What is your question?
– David K
Nov 15 at 21:48






You have given a link to a question about changing the order of integration. There are several answers under that question. What is your question?
– David K
Nov 15 at 21:48














I just don't get what is going on when we are integrating, why does the inner integral go from y to infinity then from 0 to x. I just don't understand the process, I just was wondering if someone could walk me through each step and explain what is going on. I understand and derived out the other method which is A + the integral from A to B of 1-F(x) dx, where F(x) is the cumulative distribution function or just the anti-derivative of f(x).
– BeepBoop
Nov 16 at 13:16






I just don't get what is going on when we are integrating, why does the inner integral go from y to infinity then from 0 to x. I just don't understand the process, I just was wondering if someone could walk me through each step and explain what is going on. I understand and derived out the other method which is A + the integral from A to B of 1-F(x) dx, where F(x) is the cumulative distribution function or just the anti-derivative of f(x).
– BeepBoop
Nov 16 at 13:16














Do you mean to say you read all the answers to the other question (which were specifically about why you can change the double integral from one form to the other) and you did not understand any of them? When you say "visually," does that mean a picture might help?
– David K
Nov 16 at 14:00




Do you mean to say you read all the answers to the other question (which were specifically about why you can change the double integral from one form to the other) and you did not understand any of them? When you say "visually," does that mean a picture might help?
– David K
Nov 16 at 14:00












By the way, it's generally advised to put the specific information about what you have done and what you didn't understand in the question itself rather than in the comments.
– David K
Nov 16 at 14:13




By the way, it's generally advised to put the specific information about what you have done and what you didn't understand in the question itself rather than in the comments.
– David K
Nov 16 at 14:13












Hi David, I did read the other posts and still sort of lost on why the process works. I do understand the last comment though and how to do the alternative method but I need to know and want to know why and how the other method works. A picture would help allot honestly, if it is something that can be reasoned geometrically. Okay thanks I will fix my post!
– BeepBoop
Nov 16 at 14:28






Hi David, I did read the other posts and still sort of lost on why the process works. I do understand the last comment though and how to do the alternative method but I need to know and want to know why and how the other method works. A picture would help allot honestly, if it is something that can be reasoned geometrically. Okay thanks I will fix my post!
– BeepBoop
Nov 16 at 14:28












1 Answer
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up vote
0
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accepted










Here is an equation similar to the one in the proof, in which the inner integral bounds are $y$ to infinity on the left of the equation and $0$ to $x$ on the right:
$$
int_0^infty int_y^infty f_Y(x),dx,dy
= int_0^infty int_0^x f_Y(x),dy,dx.
$$

In this form I think it's a little clearer that the only things that change are the order of the integration variables and the bounds of the inner integral.



Both sides of the equation are integrating over the same region of integration.
The region is the infinite area above the $x$-axis and below the line $y=x,$
shown as the shaded portion of the figure below.



enter image description here



Each integral integrates $f_Y(x)$ over the entire shaded region.
On the left side, to find any of the points in the region of integration,
you first choose $y$ within the bounds given by the outer integral
and then choose $x$ within the bounds given by the inner integral.
Observe that in the shaded region, $y$ is never greater than $x$;
instead, $x$ is always greater than or equal to $y.$
Hence the only $x$ values over which we can integrate are between $y$ and infinity;
but at any particular value of $y,$ all the $x$ values from $y$ to inifinity
give points in the integration region. So the bounds on the left are correct.



On the right, the outer integral is over $dx,$ so to find a point in the integration region of this integral you first choose $x$ within the bounds of the outer integral and then choose $y$ within the bounds of the inner integral.
On this side, for any given $x,$ the $y$ values in the shaded region are
$0 leq y leq x,$ so we integrate over $y$ from $0$ to $x.$



The rest of the proof is just further rewriting of the same integral to or from one of these two forms.
In particular, we can rewrite the right-hand side this way:
begin{align}
int_0^infty int_0^x f_Y(x),dy,dx
&= int_0^infty left( int_0^x f_Y(x),dyright)dx \
&= int_0^infty left( f_Y(x) int_0^x dyright)dx \
&= int_0^infty left(int_0^x dyright) f_Y(x),dx,
end{align}

and now we have the right-hand side in the form in which it was written in the proof.





As an aside (to relate this to the answers to the other question),
the formal proof of the integral equation uses things called "indicator functions" to identify the region of integration in the figure above.
In the proof, each indicator function takes the value $1$ whenever the pair of variables $(x,y)$ is in the shaded region of the figure, and the indicator function takes the value $0$ everywhere else.
That way the proof can set the integral up as an integral over the entire first quadrant, for $0 leq x < infty$ and $0 leq y < infty,$
and because the integrand is multiplied by the indicator function, which is zero everywhere in the unshaded region, the integration over the unshaded region is just an integral of zero, which comes out to zero. As a result, the integral over the entire first quadrant comes out to just the positive value contributed by the shaded region.






share|cite|improve this answer





















  • Thanks David! I really appreciate your answer and your time to help me out. I understand why it is valid to change the order of integration now. I didn't realize we were integrating under the line y=x. I was just wondering how did you know that was the region of integration?
    – BeepBoop
    Nov 17 at 18:29










  • When you've done enough integrals like this you might start to recognize patterns like that. The real insight here--the thing that's really not obvious--is the fact that this integral will compute the expected value. But I didn't think of that, I just managed to (eventually) understand this proof that someone else came up with.
    – David K
    Nov 18 at 1:03











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1 Answer
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1 Answer
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active

oldest

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oldest

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active

oldest

votes








up vote
0
down vote



accepted










Here is an equation similar to the one in the proof, in which the inner integral bounds are $y$ to infinity on the left of the equation and $0$ to $x$ on the right:
$$
int_0^infty int_y^infty f_Y(x),dx,dy
= int_0^infty int_0^x f_Y(x),dy,dx.
$$

In this form I think it's a little clearer that the only things that change are the order of the integration variables and the bounds of the inner integral.



Both sides of the equation are integrating over the same region of integration.
The region is the infinite area above the $x$-axis and below the line $y=x,$
shown as the shaded portion of the figure below.



enter image description here



Each integral integrates $f_Y(x)$ over the entire shaded region.
On the left side, to find any of the points in the region of integration,
you first choose $y$ within the bounds given by the outer integral
and then choose $x$ within the bounds given by the inner integral.
Observe that in the shaded region, $y$ is never greater than $x$;
instead, $x$ is always greater than or equal to $y.$
Hence the only $x$ values over which we can integrate are between $y$ and infinity;
but at any particular value of $y,$ all the $x$ values from $y$ to inifinity
give points in the integration region. So the bounds on the left are correct.



On the right, the outer integral is over $dx,$ so to find a point in the integration region of this integral you first choose $x$ within the bounds of the outer integral and then choose $y$ within the bounds of the inner integral.
On this side, for any given $x,$ the $y$ values in the shaded region are
$0 leq y leq x,$ so we integrate over $y$ from $0$ to $x.$



The rest of the proof is just further rewriting of the same integral to or from one of these two forms.
In particular, we can rewrite the right-hand side this way:
begin{align}
int_0^infty int_0^x f_Y(x),dy,dx
&= int_0^infty left( int_0^x f_Y(x),dyright)dx \
&= int_0^infty left( f_Y(x) int_0^x dyright)dx \
&= int_0^infty left(int_0^x dyright) f_Y(x),dx,
end{align}

and now we have the right-hand side in the form in which it was written in the proof.





As an aside (to relate this to the answers to the other question),
the formal proof of the integral equation uses things called "indicator functions" to identify the region of integration in the figure above.
In the proof, each indicator function takes the value $1$ whenever the pair of variables $(x,y)$ is in the shaded region of the figure, and the indicator function takes the value $0$ everywhere else.
That way the proof can set the integral up as an integral over the entire first quadrant, for $0 leq x < infty$ and $0 leq y < infty,$
and because the integrand is multiplied by the indicator function, which is zero everywhere in the unshaded region, the integration over the unshaded region is just an integral of zero, which comes out to zero. As a result, the integral over the entire first quadrant comes out to just the positive value contributed by the shaded region.






share|cite|improve this answer





















  • Thanks David! I really appreciate your answer and your time to help me out. I understand why it is valid to change the order of integration now. I didn't realize we were integrating under the line y=x. I was just wondering how did you know that was the region of integration?
    – BeepBoop
    Nov 17 at 18:29










  • When you've done enough integrals like this you might start to recognize patterns like that. The real insight here--the thing that's really not obvious--is the fact that this integral will compute the expected value. But I didn't think of that, I just managed to (eventually) understand this proof that someone else came up with.
    – David K
    Nov 18 at 1:03















up vote
0
down vote



accepted










Here is an equation similar to the one in the proof, in which the inner integral bounds are $y$ to infinity on the left of the equation and $0$ to $x$ on the right:
$$
int_0^infty int_y^infty f_Y(x),dx,dy
= int_0^infty int_0^x f_Y(x),dy,dx.
$$

In this form I think it's a little clearer that the only things that change are the order of the integration variables and the bounds of the inner integral.



Both sides of the equation are integrating over the same region of integration.
The region is the infinite area above the $x$-axis and below the line $y=x,$
shown as the shaded portion of the figure below.



enter image description here



Each integral integrates $f_Y(x)$ over the entire shaded region.
On the left side, to find any of the points in the region of integration,
you first choose $y$ within the bounds given by the outer integral
and then choose $x$ within the bounds given by the inner integral.
Observe that in the shaded region, $y$ is never greater than $x$;
instead, $x$ is always greater than or equal to $y.$
Hence the only $x$ values over which we can integrate are between $y$ and infinity;
but at any particular value of $y,$ all the $x$ values from $y$ to inifinity
give points in the integration region. So the bounds on the left are correct.



On the right, the outer integral is over $dx,$ so to find a point in the integration region of this integral you first choose $x$ within the bounds of the outer integral and then choose $y$ within the bounds of the inner integral.
On this side, for any given $x,$ the $y$ values in the shaded region are
$0 leq y leq x,$ so we integrate over $y$ from $0$ to $x.$



The rest of the proof is just further rewriting of the same integral to or from one of these two forms.
In particular, we can rewrite the right-hand side this way:
begin{align}
int_0^infty int_0^x f_Y(x),dy,dx
&= int_0^infty left( int_0^x f_Y(x),dyright)dx \
&= int_0^infty left( f_Y(x) int_0^x dyright)dx \
&= int_0^infty left(int_0^x dyright) f_Y(x),dx,
end{align}

and now we have the right-hand side in the form in which it was written in the proof.





As an aside (to relate this to the answers to the other question),
the formal proof of the integral equation uses things called "indicator functions" to identify the region of integration in the figure above.
In the proof, each indicator function takes the value $1$ whenever the pair of variables $(x,y)$ is in the shaded region of the figure, and the indicator function takes the value $0$ everywhere else.
That way the proof can set the integral up as an integral over the entire first quadrant, for $0 leq x < infty$ and $0 leq y < infty,$
and because the integrand is multiplied by the indicator function, which is zero everywhere in the unshaded region, the integration over the unshaded region is just an integral of zero, which comes out to zero. As a result, the integral over the entire first quadrant comes out to just the positive value contributed by the shaded region.






share|cite|improve this answer





















  • Thanks David! I really appreciate your answer and your time to help me out. I understand why it is valid to change the order of integration now. I didn't realize we were integrating under the line y=x. I was just wondering how did you know that was the region of integration?
    – BeepBoop
    Nov 17 at 18:29










  • When you've done enough integrals like this you might start to recognize patterns like that. The real insight here--the thing that's really not obvious--is the fact that this integral will compute the expected value. But I didn't think of that, I just managed to (eventually) understand this proof that someone else came up with.
    – David K
    Nov 18 at 1:03













up vote
0
down vote



accepted







up vote
0
down vote



accepted






Here is an equation similar to the one in the proof, in which the inner integral bounds are $y$ to infinity on the left of the equation and $0$ to $x$ on the right:
$$
int_0^infty int_y^infty f_Y(x),dx,dy
= int_0^infty int_0^x f_Y(x),dy,dx.
$$

In this form I think it's a little clearer that the only things that change are the order of the integration variables and the bounds of the inner integral.



Both sides of the equation are integrating over the same region of integration.
The region is the infinite area above the $x$-axis and below the line $y=x,$
shown as the shaded portion of the figure below.



enter image description here



Each integral integrates $f_Y(x)$ over the entire shaded region.
On the left side, to find any of the points in the region of integration,
you first choose $y$ within the bounds given by the outer integral
and then choose $x$ within the bounds given by the inner integral.
Observe that in the shaded region, $y$ is never greater than $x$;
instead, $x$ is always greater than or equal to $y.$
Hence the only $x$ values over which we can integrate are between $y$ and infinity;
but at any particular value of $y,$ all the $x$ values from $y$ to inifinity
give points in the integration region. So the bounds on the left are correct.



On the right, the outer integral is over $dx,$ so to find a point in the integration region of this integral you first choose $x$ within the bounds of the outer integral and then choose $y$ within the bounds of the inner integral.
On this side, for any given $x,$ the $y$ values in the shaded region are
$0 leq y leq x,$ so we integrate over $y$ from $0$ to $x.$



The rest of the proof is just further rewriting of the same integral to or from one of these two forms.
In particular, we can rewrite the right-hand side this way:
begin{align}
int_0^infty int_0^x f_Y(x),dy,dx
&= int_0^infty left( int_0^x f_Y(x),dyright)dx \
&= int_0^infty left( f_Y(x) int_0^x dyright)dx \
&= int_0^infty left(int_0^x dyright) f_Y(x),dx,
end{align}

and now we have the right-hand side in the form in which it was written in the proof.





As an aside (to relate this to the answers to the other question),
the formal proof of the integral equation uses things called "indicator functions" to identify the region of integration in the figure above.
In the proof, each indicator function takes the value $1$ whenever the pair of variables $(x,y)$ is in the shaded region of the figure, and the indicator function takes the value $0$ everywhere else.
That way the proof can set the integral up as an integral over the entire first quadrant, for $0 leq x < infty$ and $0 leq y < infty,$
and because the integrand is multiplied by the indicator function, which is zero everywhere in the unshaded region, the integration over the unshaded region is just an integral of zero, which comes out to zero. As a result, the integral over the entire first quadrant comes out to just the positive value contributed by the shaded region.






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Here is an equation similar to the one in the proof, in which the inner integral bounds are $y$ to infinity on the left of the equation and $0$ to $x$ on the right:
$$
int_0^infty int_y^infty f_Y(x),dx,dy
= int_0^infty int_0^x f_Y(x),dy,dx.
$$

In this form I think it's a little clearer that the only things that change are the order of the integration variables and the bounds of the inner integral.



Both sides of the equation are integrating over the same region of integration.
The region is the infinite area above the $x$-axis and below the line $y=x,$
shown as the shaded portion of the figure below.



enter image description here



Each integral integrates $f_Y(x)$ over the entire shaded region.
On the left side, to find any of the points in the region of integration,
you first choose $y$ within the bounds given by the outer integral
and then choose $x$ within the bounds given by the inner integral.
Observe that in the shaded region, $y$ is never greater than $x$;
instead, $x$ is always greater than or equal to $y.$
Hence the only $x$ values over which we can integrate are between $y$ and infinity;
but at any particular value of $y,$ all the $x$ values from $y$ to inifinity
give points in the integration region. So the bounds on the left are correct.



On the right, the outer integral is over $dx,$ so to find a point in the integration region of this integral you first choose $x$ within the bounds of the outer integral and then choose $y$ within the bounds of the inner integral.
On this side, for any given $x,$ the $y$ values in the shaded region are
$0 leq y leq x,$ so we integrate over $y$ from $0$ to $x.$



The rest of the proof is just further rewriting of the same integral to or from one of these two forms.
In particular, we can rewrite the right-hand side this way:
begin{align}
int_0^infty int_0^x f_Y(x),dy,dx
&= int_0^infty left( int_0^x f_Y(x),dyright)dx \
&= int_0^infty left( f_Y(x) int_0^x dyright)dx \
&= int_0^infty left(int_0^x dyright) f_Y(x),dx,
end{align}

and now we have the right-hand side in the form in which it was written in the proof.





As an aside (to relate this to the answers to the other question),
the formal proof of the integral equation uses things called "indicator functions" to identify the region of integration in the figure above.
In the proof, each indicator function takes the value $1$ whenever the pair of variables $(x,y)$ is in the shaded region of the figure, and the indicator function takes the value $0$ everywhere else.
That way the proof can set the integral up as an integral over the entire first quadrant, for $0 leq x < infty$ and $0 leq y < infty,$
and because the integrand is multiplied by the indicator function, which is zero everywhere in the unshaded region, the integration over the unshaded region is just an integral of zero, which comes out to zero. As a result, the integral over the entire first quadrant comes out to just the positive value contributed by the shaded region.







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answered Nov 17 at 2:09









David K

51.5k340113




51.5k340113












  • Thanks David! I really appreciate your answer and your time to help me out. I understand why it is valid to change the order of integration now. I didn't realize we were integrating under the line y=x. I was just wondering how did you know that was the region of integration?
    – BeepBoop
    Nov 17 at 18:29










  • When you've done enough integrals like this you might start to recognize patterns like that. The real insight here--the thing that's really not obvious--is the fact that this integral will compute the expected value. But I didn't think of that, I just managed to (eventually) understand this proof that someone else came up with.
    – David K
    Nov 18 at 1:03


















  • Thanks David! I really appreciate your answer and your time to help me out. I understand why it is valid to change the order of integration now. I didn't realize we were integrating under the line y=x. I was just wondering how did you know that was the region of integration?
    – BeepBoop
    Nov 17 at 18:29










  • When you've done enough integrals like this you might start to recognize patterns like that. The real insight here--the thing that's really not obvious--is the fact that this integral will compute the expected value. But I didn't think of that, I just managed to (eventually) understand this proof that someone else came up with.
    – David K
    Nov 18 at 1:03
















Thanks David! I really appreciate your answer and your time to help me out. I understand why it is valid to change the order of integration now. I didn't realize we were integrating under the line y=x. I was just wondering how did you know that was the region of integration?
– BeepBoop
Nov 17 at 18:29




Thanks David! I really appreciate your answer and your time to help me out. I understand why it is valid to change the order of integration now. I didn't realize we were integrating under the line y=x. I was just wondering how did you know that was the region of integration?
– BeepBoop
Nov 17 at 18:29












When you've done enough integrals like this you might start to recognize patterns like that. The real insight here--the thing that's really not obvious--is the fact that this integral will compute the expected value. But I didn't think of that, I just managed to (eventually) understand this proof that someone else came up with.
– David K
Nov 18 at 1:03




When you've done enough integrals like this you might start to recognize patterns like that. The real insight here--the thing that's really not obvious--is the fact that this integral will compute the expected value. But I didn't think of that, I just managed to (eventually) understand this proof that someone else came up with.
– David K
Nov 18 at 1:03


















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