Mixing
Assume that $X$ and $Y$ have joint probability density function $f_{X,Y}$. Calculate the joint probability...
$begingroup$
Assume that $X$ and $Y$ have the following joint probability density function
$$f_{X,Y}(x,y) =
begin{cases}
displaystylefrac{1}{x^{2}y^{2}} & text{if},,xgeq 1,,text{and},,ygeq 1\\
,,,,,0 & text{otherwise}
end{cases}$$
(a) Calculate the joint probability density function of $U = XY$ and $V = X/Y$.
(b) What are the marginal density functions?
MY SOLUTION
(a) To begin with, notice that $u geq 1$ and $v > 0$. Moreover, we have $X = sqrt{UV}$ and $Y = sqrt{U/V}$. From whence we conclude that
begin{align*}
f_{U,V}(u,v) = f_{X,Y}(sqrt{uv},sqrt{uv^{-1}})|det J(u,v)|
end{align*}
where $J(u,v)$ is given by
begin{align*}
begin{vmatrix}
displaystylefrac{partial x}{partial u} & displaystylefrac{partial x}{partial v} \
displaystylefrac{partial y}{partial u} & displaystylefrac{partial y}{partial v}
end{vmatrix} =
begin{vmatrix}
displaystylefrac{v}{2sqrt{uv}} & displaystylefrac{u}{2sqrt{uv}} \
displaystylefrac{1}{2sqrt{uv}} & displaystyle-frac{sqrt{u}}{2sqrt{v^{3}}}
end{vmatrix} = -frac{1}{4v} - frac{1}{4v} = -frac{1}{2v}
end{align*}
Therefore we have
begin{align*}
f_{U,V}(u,v) = frac{1}{2u^{2}v}
end{align*}
(b) Once you have the joint probability density function, you can determine the marginal distributions through the formulas
begin{cases}
f_{U}(u) = displaystyleint_{-infty}^{+infty}f_{U,V}(u,v)mathrm{d}v\\
f_{V}(v) = displaystyleint_{-infty}^{+infty}f_{U,V}(u,v)mathrm{d}u
end{cases}
My question is: where things went wrong? I cannot find out where are the miscalculations of $f_{U,V}$ since its integral does not converge to $1$. Any help is appreciated. Thanks in advance!
probability probability-theory density-function change-of-variable
asked Feb 1 at 0:26
APC89APC89
2,371720
$endgroup$
add a comment |
$begingroup$
Assume that $X$ and $Y$ have the following joint probability density function
$$f_{X,Y}(x,y) =
begin{cases}
displaystylefrac{1}{x^{2}y^{2}} & text{if},,xgeq 1,,text{and},,ygeq 1\\
,,,,,0 & text{otherwise}
end{cases}$$
(a) Calculate the joint probability density function of $U = XY$ and $V = X/Y$.
(b) What are the marginal density functions?
MY SOLUTION
(a) To begin with, notice that $u geq 1$ and $v > 0$. Moreover, we have $X = sqrt{UV}$ and $Y = sqrt{U/V}$. From whence we conclude that
begin{align*}
f_{U,V}(u,v) = f_{X,Y}(sqrt{uv},sqrt{uv^{-1}})|det J(u,v)|
end{align*}
where $J(u,v)$ is given by
begin{align*}
begin{vmatrix}
displaystylefrac{partial x}{partial u} & displaystylefrac{partial x}{partial v} \
displaystylefrac{partial y}{partial u} & displaystylefrac{partial y}{partial v}
end{vmatrix} =
begin{vmatrix}
displaystylefrac{v}{2sqrt{uv}} & displaystylefrac{u}{2sqrt{uv}} \
displaystylefrac{1}{2sqrt{uv}} & displaystyle-frac{sqrt{u}}{2sqrt{v^{3}}}
end{vmatrix} = -frac{1}{4v} - frac{1}{4v} = -frac{1}{2v}
end{align*}
Therefore we have
begin{align*}
f_{U,V}(u,v) = frac{1}{2u^{2}v}
end{align*}
(b) Once you have the joint probability density function, you can determine the marginal distributions through the formulas
begin{cases}
f_{U}(u) = displaystyleint_{-infty}^{+infty}f_{U,V}(u,v)mathrm{d}v\\
f_{V}(v) = displaystyleint_{-infty}^{+infty}f_{U,V}(u,v)mathrm{d}u
end{cases}
My question is: where things went wrong? I cannot find out where are the miscalculations of $f_{U,V}$ since its integral does not converge to $1$. Any help is appreciated. Thanks in advance!
probability probability-theory density-function change-of-variable
asked Feb 1 at 0:26
APC89APC89
2,371720
$endgroup$
$begingroup$
Part (a) does not include the support for the function, therefore how can you know where to integrate in part (b).
$endgroup$
– Graham Kemp
Feb 1 at 0:53
$begingroup$
Perhaps I have misunderstood your question, but I have pointed out that $(u,v)in[1,+infty)times(0,+infty)$. Is this it you were talking about?
$endgroup$
– APC89
Feb 1 at 0:55
$begingroup$
That's not the support. As hypermova notes, you have to be very careful with the transformation.
$endgroup$
– Graham Kemp
Feb 1 at 1:12
add a comment |
$begingroup$
Assume that $X$ and $Y$ have the following joint probability density function
$$f_{X,Y}(x,y) =
begin{cases}
displaystylefrac{1}{x^{2}y^{2}} & text{if},,xgeq 1,,text{and},,ygeq 1\\
,,,,,0 & text{otherwise}
end{cases}$$
(a) Calculate the joint probability density function of $U = XY$ and $V = X/Y$.
(b) What are the marginal density functions?
MY SOLUTION
(a) To begin with, notice that $u geq 1$ and $v > 0$. Moreover, we have $X = sqrt{UV}$ and $Y = sqrt{U/V}$. From whence we conclude that
begin{align*}
f_{U,V}(u,v) = f_{X,Y}(sqrt{uv},sqrt{uv^{-1}})|det J(u,v)|
end{align*}
where $J(u,v)$ is given by
begin{align*}
begin{vmatrix}
displaystylefrac{partial x}{partial u} & displaystylefrac{partial x}{partial v} \
displaystylefrac{partial y}{partial u} & displaystylefrac{partial y}{partial v}
end{vmatrix} =
begin{vmatrix}
displaystylefrac{v}{2sqrt{uv}} & displaystylefrac{u}{2sqrt{uv}} \
displaystylefrac{1}{2sqrt{uv}} & displaystyle-frac{sqrt{u}}{2sqrt{v^{3}}}
end{vmatrix} = -frac{1}{4v} - frac{1}{4v} = -frac{1}{2v}
end{align*}
Therefore we have
begin{align*}
f_{U,V}(u,v) = frac{1}{2u^{2}v}
end{align*}
(b) Once you have the joint probability density function, you can determine the marginal distributions through the formulas
begin{cases}
f_{U}(u) = displaystyleint_{-infty}^{+infty}f_{U,V}(u,v)mathrm{d}v\\
f_{V}(v) = displaystyleint_{-infty}^{+infty}f_{U,V}(u,v)mathrm{d}u
end{cases}
My question is: where things went wrong? I cannot find out where are the miscalculations of $f_{U,V}$ since its integral does not converge to $1$. Any help is appreciated. Thanks in advance!
probability probability-theory density-function change-of-variable
asked Feb 1 at 0:26
APC89APC89
2,371720
$endgroup$
Assume that $X$ and $Y$ have the following joint probability density function
$$f_{X,Y}(x,y) =
begin{cases}
displaystylefrac{1}{x^{2}y^{2}} & text{if},,xgeq 1,,text{and},,ygeq 1\\
,,,,,0 & text{otherwise}
end{cases}$$
(a) Calculate the joint probability density function of $U = XY$ and $V = X/Y$.
(b) What are the marginal density functions?
MY SOLUTION
(a) To begin with, notice that $u geq 1$ and $v > 0$. Moreover, we have $X = sqrt{UV}$ and $Y = sqrt{U/V}$. From whence we conclude that
begin{align*}
f_{U,V}(u,v) = f_{X,Y}(sqrt{uv},sqrt{uv^{-1}})|det J(u,v)|
end{align*}
where $J(u,v)$ is given by
begin{align*}
begin{vmatrix}
displaystylefrac{partial x}{partial u} & displaystylefrac{partial x}{partial v} \
displaystylefrac{partial y}{partial u} & displaystylefrac{partial y}{partial v}
end{vmatrix} =
begin{vmatrix}
displaystylefrac{v}{2sqrt{uv}} & displaystylefrac{u}{2sqrt{uv}} \
displaystylefrac{1}{2sqrt{uv}} & displaystyle-frac{sqrt{u}}{2sqrt{v^{3}}}
end{vmatrix} = -frac{1}{4v} - frac{1}{4v} = -frac{1}{2v}
end{align*}
Therefore we have
begin{align*}
f_{U,V}(u,v) = frac{1}{2u^{2}v}
end{align*}
(b) Once you have the joint probability density function, you can determine the marginal distributions through the formulas
begin{cases}
f_{U}(u) = displaystyleint_{-infty}^{+infty}f_{U,V}(u,v)mathrm{d}v\\
f_{V}(v) = displaystyleint_{-infty}^{+infty}f_{U,V}(u,v)mathrm{d}u
end{cases}
My question is: where things went wrong? I cannot find out where are the miscalculations of $f_{U,V}$ since its integral does not converge to $1$. Any help is appreciated. Thanks in advance!
probability probability-theory density-function change-of-variable
probability probability-theory density-function change-of-variable
asked Feb 1 at 0:26
APC89APC89
2,371720
asked Feb 1 at 0:26
APC89APC89
2,371720
asked Feb 1 at 0:26
APC89APC89
2,371720
asked Feb 1 at 0:26
APC89APC89
2,371720
asked Feb 1 at 0:26
APC89APC89
2,371720
2,371720
$begingroup$
Part (a) does not include the support for the function, therefore how can you know where to integrate in part (b).
$endgroup$
– Graham Kemp
Feb 1 at 0:53
$begingroup$
Perhaps I have misunderstood your question, but I have pointed out that $(u,v)in[1,+infty)times(0,+infty)$. Is this it you were talking about?
$endgroup$
– APC89
Feb 1 at 0:55
$begingroup$
That's not the support. As hypermova notes, you have to be very careful with the transformation.
$endgroup$
– Graham Kemp
Feb 1 at 1:12
add a comment |
$begingroup$
Part (a) does not include the support for the function, therefore how can you know where to integrate in part (b).
$endgroup$
– Graham Kemp
Feb 1 at 0:53
$begingroup$
Perhaps I have misunderstood your question, but I have pointed out that $(u,v)in[1,+infty)times(0,+infty)$. Is this it you were talking about?
$endgroup$
– APC89
Feb 1 at 0:55
$begingroup$
That's not the support. As hypermova notes, you have to be very careful with the transformation.
$endgroup$
– Graham Kemp
Feb 1 at 1:12
$begingroup$
Part (a) does not include the support for the function, therefore how can you know where to integrate in part (b).
$endgroup$
– Graham Kemp
Feb 1 at 0:53
$begingroup$
Part (a) does not include the support for the function, therefore how can you know where to integrate in part (b).
$endgroup$
– Graham Kemp
Feb 1 at 0:53
$begingroup$
Perhaps I have misunderstood your question, but I have pointed out that $(u,v)in[1,+infty)times(0,+infty)$. Is this it you were talking about?
$endgroup$
– APC89
Feb 1 at 0:55
$begingroup$
Perhaps I have misunderstood your question, but I have pointed out that $(u,v)in[1,+infty)times(0,+infty)$. Is this it you were talking about?
$endgroup$
– APC89
Feb 1 at 0:55
$begingroup$
That's not the support. As hypermova notes, you have to be very careful with the transformation.
$endgroup$
– Graham Kemp
Feb 1 at 1:12
$begingroup$
That's not the support. As hypermova notes, you have to be very careful with the transformation.
$endgroup$
– Graham Kemp
Feb 1 at 1:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider $U=XY, V=X/Y$, gives us that $X^2/U=V$ and $Y^2V=U$
Then as $1leq X, 1leq Y$, we have ${(U,V):1leq U ~,~ 1/Uleq Vleq U}$ as the support.
answered Feb 1 at 2:07
Graham KempGraham Kemp
87.8k43578
$endgroup$
$begingroup$
Thank you very much! Could you please include the marginal density functions in your answer? I'd be greatly thankful for that.
$endgroup$
– APC89
Feb 1 at 2:17
1
$begingroup$
Just integrate $$begin{align}f_U(u) &= mathbf 1_{1leq u}dfrac 1{2u^2} int_{1/u}^u dfrac 1vmathsf d v\ f_V(v) &=mathbf 1_{0<v< 1}dfrac 1{2v}int_{1/v}^inftydfrac1{u^2}mathsf du+mathbf 1_{1leq v}dfrac 1{2v}int_{v}^inftydfrac1{u^2}mathsf du end{align}$$
$endgroup$
– Graham Kemp
Feb 1 at 3:34
$begingroup$
Thanks again :)
$endgroup$
– APC89
Feb 1 at 4:17
add a comment |
$begingroup$
The problem lies in that the domain
$$
left(x,yright)inleft[1,inftyright)^2
$$
and
$$
left(u,vright)inleft[1,inftyright)timesleft(0,inftyright)
$$
are not equivalent under your transformation. Instead, you may check that
$$
left{left(xy,x/yright):left(x,yright)inleft[1,inftyright)^2right}=left{yle xright}capleft{yge 1/xright}.
$$
answered Feb 1 at 1:06
hypernovahypernova
4,979514
$endgroup$
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3095674%2fassume-that-x-and-y-have-joint-probability-density-function-f-x-y-calcu%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
$begingroup$
Consider $U=XY, V=X/Y$, gives us that $X^2/U=V$ and $Y^2V=U$
Then as $1leq X, 1leq Y$, we have ${(U,V):1leq U ~,~ 1/Uleq Vleq U}$ as the support.
answered Feb 1 at 2:07
Graham KempGraham Kemp
87.8k43578
$endgroup$
$begingroup$
Thank you very much! Could you please include the marginal density functions in your answer? I'd be greatly thankful for that.
$endgroup$
– APC89
Feb 1 at 2:17
1
$begingroup$
Just integrate $$begin{align}f_U(u) &= mathbf 1_{1leq u}dfrac 1{2u^2} int_{1/u}^u dfrac 1vmathsf d v\ f_V(v) &=mathbf 1_{0<v< 1}dfrac 1{2v}int_{1/v}^inftydfrac1{u^2}mathsf du+mathbf 1_{1leq v}dfrac 1{2v}int_{v}^inftydfrac1{u^2}mathsf du end{align}$$
$endgroup$
– Graham Kemp
Feb 1 at 3:34
$begingroup$
Thanks again :)
$endgroup$
– APC89
Feb 1 at 4:17
add a comment |
$begingroup$
Consider $U=XY, V=X/Y$, gives us that $X^2/U=V$ and $Y^2V=U$
Then as $1leq X, 1leq Y$, we have ${(U,V):1leq U ~,~ 1/Uleq Vleq U}$ as the support.
answered Feb 1 at 2:07
Graham KempGraham Kemp
87.8k43578
$endgroup$
$begingroup$
Thank you very much! Could you please include the marginal density functions in your answer? I'd be greatly thankful for that.
$endgroup$
– APC89
Feb 1 at 2:17
1
$begingroup$
Just integrate $$begin{align}f_U(u) &= mathbf 1_{1leq u}dfrac 1{2u^2} int_{1/u}^u dfrac 1vmathsf d v\ f_V(v) &=mathbf 1_{0<v< 1}dfrac 1{2v}int_{1/v}^inftydfrac1{u^2}mathsf du+mathbf 1_{1leq v}dfrac 1{2v}int_{v}^inftydfrac1{u^2}mathsf du end{align}$$
$endgroup$
– Graham Kemp
Feb 1 at 3:34
$begingroup$
Thanks again :)
$endgroup$
– APC89
Feb 1 at 4:17
add a comment |
$begingroup$
Consider $U=XY, V=X/Y$, gives us that $X^2/U=V$ and $Y^2V=U$
Then as $1leq X, 1leq Y$, we have ${(U,V):1leq U ~,~ 1/Uleq Vleq U}$ as the support.
answered Feb 1 at 2:07
Graham KempGraham Kemp
87.8k43578
$endgroup$
Consider $U=XY, V=X/Y$, gives us that $X^2/U=V$ and $Y^2V=U$
Then as $1leq X, 1leq Y$, we have ${(U,V):1leq U ~,~ 1/Uleq Vleq U}$ as the support.
answered Feb 1 at 2:07
Graham KempGraham Kemp
87.8k43578
answered Feb 1 at 2:07
Graham KempGraham Kemp
87.8k43578
answered Feb 1 at 2:07
Graham KempGraham Kemp
87.8k43578
answered Feb 1 at 2:07
Graham KempGraham Kemp
87.8k43578
87.8k43578
$begingroup$
Thank you very much! Could you please include the marginal density functions in your answer? I'd be greatly thankful for that.
$endgroup$
– APC89
Feb 1 at 2:17
1
$begingroup$
Just integrate $$begin{align}f_U(u) &= mathbf 1_{1leq u}dfrac 1{2u^2} int_{1/u}^u dfrac 1vmathsf d v\ f_V(v) &=mathbf 1_{0<v< 1}dfrac 1{2v}int_{1/v}^inftydfrac1{u^2}mathsf du+mathbf 1_{1leq v}dfrac 1{2v}int_{v}^inftydfrac1{u^2}mathsf du end{align}$$
$endgroup$
– Graham Kemp
Feb 1 at 3:34
$begingroup$
Thanks again :)
$endgroup$
– APC89
Feb 1 at 4:17
add a comment |
$begingroup$
Thank you very much! Could you please include the marginal density functions in your answer? I'd be greatly thankful for that.
$endgroup$
– APC89
Feb 1 at 2:17
1
$begingroup$
Just integrate $$begin{align}f_U(u) &= mathbf 1_{1leq u}dfrac 1{2u^2} int_{1/u}^u dfrac 1vmathsf d v\ f_V(v) &=mathbf 1_{0<v< 1}dfrac 1{2v}int_{1/v}^inftydfrac1{u^2}mathsf du+mathbf 1_{1leq v}dfrac 1{2v}int_{v}^inftydfrac1{u^2}mathsf du end{align}$$
$endgroup$
– Graham Kemp
Feb 1 at 3:34
$begingroup$
Thanks again :)
$endgroup$
– APC89
Feb 1 at 4:17
$begingroup$
Thank you very much! Could you please include the marginal density functions in your answer? I'd be greatly thankful for that.
$endgroup$
– APC89
Feb 1 at 2:17
$begingroup$
Thank you very much! Could you please include the marginal density functions in your answer? I'd be greatly thankful for that.
$endgroup$
– APC89
Feb 1 at 2:17
1
1
$begingroup$
Just integrate $$begin{align}f_U(u) &= mathbf 1_{1leq u}dfrac 1{2u^2} int_{1/u}^u dfrac 1vmathsf d v\ f_V(v) &=mathbf 1_{0<v< 1}dfrac 1{2v}int_{1/v}^inftydfrac1{u^2}mathsf du+mathbf 1_{1leq v}dfrac 1{2v}int_{v}^inftydfrac1{u^2}mathsf du end{align}$$
$endgroup$
– Graham Kemp
Feb 1 at 3:34
$begingroup$
Just integrate $$begin{align}f_U(u) &= mathbf 1_{1leq u}dfrac 1{2u^2} int_{1/u}^u dfrac 1vmathsf d v\ f_V(v) &=mathbf 1_{0<v< 1}dfrac 1{2v}int_{1/v}^inftydfrac1{u^2}mathsf du+mathbf 1_{1leq v}dfrac 1{2v}int_{v}^inftydfrac1{u^2}mathsf du end{align}$$
$endgroup$
– Graham Kemp
Feb 1 at 3:34
$begingroup$
Thanks again :)
$endgroup$
– APC89
Feb 1 at 4:17
$begingroup$
Thanks again :)
$endgroup$
– APC89
Feb 1 at 4:17
add a comment |
$begingroup$
The problem lies in that the domain
$$
left(x,yright)inleft[1,inftyright)^2
$$
and
$$
left(u,vright)inleft[1,inftyright)timesleft(0,inftyright)
$$
are not equivalent under your transformation. Instead, you may check that
$$
left{left(xy,x/yright):left(x,yright)inleft[1,inftyright)^2right}=left{yle xright}capleft{yge 1/xright}.
$$
answered Feb 1 at 1:06
hypernovahypernova
4,979514
$endgroup$
add a comment |
$begingroup$
The problem lies in that the domain
$$
left(x,yright)inleft[1,inftyright)^2
$$
and
$$
left(u,vright)inleft[1,inftyright)timesleft(0,inftyright)
$$
are not equivalent under your transformation. Instead, you may check that
$$
left{left(xy,x/yright):left(x,yright)inleft[1,inftyright)^2right}=left{yle xright}capleft{yge 1/xright}.
$$
answered Feb 1 at 1:06
hypernovahypernova
4,979514
$endgroup$
add a comment |
$begingroup$
The problem lies in that the domain
$$
left(x,yright)inleft[1,inftyright)^2
$$
and
$$
left(u,vright)inleft[1,inftyright)timesleft(0,inftyright)
$$
are not equivalent under your transformation. Instead, you may check that
$$
left{left(xy,x/yright):left(x,yright)inleft[1,inftyright)^2right}=left{yle xright}capleft{yge 1/xright}.
$$
answered Feb 1 at 1:06
hypernovahypernova
4,979514
$endgroup$
The problem lies in that the domain
$$
left(x,yright)inleft[1,inftyright)^2
$$
and
$$
left(u,vright)inleft[1,inftyright)timesleft(0,inftyright)
$$
are not equivalent under your transformation. Instead, you may check that
$$
left{left(xy,x/yright):left(x,yright)inleft[1,inftyright)^2right}=left{yle xright}capleft{yge 1/xright}.
$$
answered Feb 1 at 1:06
hypernovahypernova
4,979514
answered Feb 1 at 1:06
hypernovahypernova
4,979514
answered Feb 1 at 1:06
hypernovahypernova
4,979514
answered Feb 1 at 1:06
hypernovahypernova
4,979514
4,979514
add a comment |
add a comment |
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.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3095674%2fassume-that-x-and-y-have-joint-probability-density-function-f-x-y-calcu%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
Part (a) does not include the support for the function, therefore how can you know where to integrate in part (b).
$endgroup$
– Graham Kemp
Feb 1 at 0:53
$begingroup$
Perhaps I have misunderstood your question, but I have pointed out that $(u,v)in[1,+infty)times(0,+infty)$. Is this it you were talking about?
$endgroup$
– APC89
Feb 1 at 0:55
$begingroup$
That's not the support. As hypermova notes, you have to be very careful with the transformation.
$endgroup$
– Graham Kemp
Feb 1 at 1:12