I cannot understand why the range of integral of $x, y$ are from $0$ to $1$ when $x > 0 $ and $y < 1$.
$begingroup$
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
edited Dec 18 '18 at 1:32
usermath
2,5601227
asked Dec 18 '18 at 1:26
ty.kimty.kim
62
$endgroup$
add a comment |
$begingroup$
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
edited Dec 18 '18 at 1:32
usermath
2,5601227
asked Dec 18 '18 at 1:26
ty.kimty.kim
62
$endgroup$
$begingroup$
...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
add a comment |
$begingroup$
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
edited Dec 18 '18 at 1:32
usermath
2,5601227
asked Dec 18 '18 at 1:26
ty.kimty.kim
62
$endgroup$
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
edited Dec 18 '18 at 1:32
usermath
2,5601227
asked Dec 18 '18 at 1:26
ty.kimty.kim
62
edited Dec 18 '18 at 1:32
usermath
2,5601227
asked Dec 18 '18 at 1:26
ty.kimty.kim
62
edited Dec 18 '18 at 1:32
usermath
2,5601227
edited Dec 18 '18 at 1:32
usermath
2,5601227
edited Dec 18 '18 at 1:32
usermath
2,5601227
2,5601227
asked Dec 18 '18 at 1:26
ty.kimty.kim
62
asked Dec 18 '18 at 1:26
ty.kimty.kim
62
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
add a comment |
$begingroup$
...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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $(0,1)times(0,1)$).
answered Dec 18 '18 at 1:32
Clement C.Clement C.
50.9k33992
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 18 '18 at 7:00
Hans LundmarkHans Lundmark
35.9k564115
$endgroup$
$begingroup$
"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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $(0,1)times(0,1)$).
answered Dec 18 '18 at 1:32
Clement C.Clement C.
50.9k33992
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $(0,1)times(0,1)$).
answered Dec 18 '18 at 1:32
Clement C.Clement C.
50.9k33992
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $(0,1)times(0,1)$).
answered Dec 18 '18 at 1:32
Clement C.Clement C.
50.9k33992
$endgroup$
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 $(0,1)times(0,1)$).
answered Dec 18 '18 at 1:32
Clement C.Clement C.
50.9k33992
answered Dec 18 '18 at 1:32
Clement C.Clement C.
50.9k33992
answered Dec 18 '18 at 1:32
Clement C.Clement C.
50.9k33992
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 18 '18 at 7:00
Hans LundmarkHans Lundmark
35.9k564115
$endgroup$
$begingroup$
"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
add a comment |
$begingroup$
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.
answered Dec 18 '18 at 7:00
Hans LundmarkHans Lundmark
35.9k564115
$endgroup$
$begingroup$
"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
add a comment |
$begingroup$
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.
answered Dec 18 '18 at 7:00
Hans LundmarkHans Lundmark
35.9k564115
$endgroup$
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.
answered Dec 18 '18 at 7:00
Hans LundmarkHans Lundmark
35.9k564115
answered Dec 18 '18 at 7:00
Hans LundmarkHans Lundmark
35.9k564115
answered Dec 18 '18 at 7:00
Hans LundmarkHans Lundmark
35.9k564115
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
add a comment |
$begingroup$
"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.
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– Clement C.
Dec 18 '18 at 17:33
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@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.
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– Hans Lundmark
Dec 18 '18 at 18:38
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That's fair. ${}{}$
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– Clement C.
Dec 18 '18 at 19:35
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@HansLundmark Thanks a lot!
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– ty.kim
Dec 18 '18 at 23:27
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"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.
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– 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
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@HansLundmark Thanks a lot!
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– ty.kim
Dec 18 '18 at 23:27
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@HansLundmark Thanks a lot!
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– ty.kim
Dec 18 '18 at 23:27
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...because $f(x,y)$ is defined $0 < x<1$ and $0<y<1)$.
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– David G. Stork
Dec 18 '18 at 1:30
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thank you for your answer. does that means x and y are same domain(or axis)
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– ty.kim
Dec 18 '18 at 1:32
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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 $alpha = 1$.
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– user113988
Dec 18 '18 at 1:33
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@ty.kim Your English is excellent!
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Dec 18 '18 at 6:22
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– Clement C.
Dec 21 '18 at 1:12