Joint density function of $X$ and $X-Y$, where $X, Ysim U(-1,1)$












0












$begingroup$


Let $X$ and $Y$ be independent random variables following $U(-1,1)$. Find the joint CDF of $U=X-Y$ and $V=X$.



I found the Jacobian of the transformation to be equal to $1$, and $f_Xf_Y=frac{1}{4}$. I'm confused regarding the final solution.



Is $displaystyle P(Uleq s,Vleq t) = int_{-1}^{t}int_{-2}^{s}frac{1}{8}du dv$? Somehow this seems wrong to me.






probability density-function







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$endgroup$












  • $begingroup$
    I guess you meant $P()$ instead of $F()$, no? Also, you surely need to specify the range of $s,t$, no?
    $endgroup$
    – leonbloy
    Nov 24 '18 at 2:14










  • $begingroup$
    Yes. I'm trying to edit it. $-2<s<2$ and $-1<t<1$.
    $endgroup$
    – FreeSid91
    Nov 24 '18 at 2:26
















0












$begingroup$


Let $X$ and $Y$ be independent random variables following $U(-1,1)$. Find the joint CDF of $U=X-Y$ and $V=X$.



I found the Jacobian of the transformation to be equal to $1$, and $f_Xf_Y=frac{1}{4}$. I'm confused regarding the final solution.



Is $displaystyle P(Uleq s,Vleq t) = int_{-1}^{t}int_{-2}^{s}frac{1}{8}du dv$? Somehow this seems wrong to me.






probability density-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    I guess you meant $P()$ instead of $F()$, no? Also, you surely need to specify the range of $s,t$, no?
    $endgroup$
    – leonbloy
    Nov 24 '18 at 2:14










  • $begingroup$
    Yes. I'm trying to edit it. $-2<s<2$ and $-1<t<1$.
    $endgroup$
    – FreeSid91
    Nov 24 '18 at 2:26














0












0








0





$begingroup$


Let $X$ and $Y$ be independent random variables following $U(-1,1)$. Find the joint CDF of $U=X-Y$ and $V=X$.



I found the Jacobian of the transformation to be equal to $1$, and $f_Xf_Y=frac{1}{4}$. I'm confused regarding the final solution.



Is $displaystyle P(Uleq s,Vleq t) = int_{-1}^{t}int_{-2}^{s}frac{1}{8}du dv$? Somehow this seems wrong to me.






probability density-function







share|cite|improve this question











$endgroup$




Let $X$ and $Y$ be independent random variables following $U(-1,1)$. Find the joint CDF of $U=X-Y$ and $V=X$.



I found the Jacobian of the transformation to be equal to $1$, and $f_Xf_Y=frac{1}{4}$. I'm confused regarding the final solution.



Is $displaystyle P(Uleq s,Vleq t) = int_{-1}^{t}int_{-2}^{s}frac{1}{8}du dv$? Somehow this seems wrong to me.






probability density-function




probability density-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 '18 at 2:52









Rócherz

2,7762721




2,7762721










asked Nov 24 '18 at 2:07









FreeSid91FreeSid91

1




1












  • $begingroup$
    I guess you meant $P()$ instead of $F()$, no? Also, you surely need to specify the range of $s,t$, no?
    $endgroup$
    – leonbloy
    Nov 24 '18 at 2:14










  • $begingroup$
    Yes. I'm trying to edit it. $-2<s<2$ and $-1<t<1$.
    $endgroup$
    – FreeSid91
    Nov 24 '18 at 2:26


















  • $begingroup$
    I guess you meant $P()$ instead of $F()$, no? Also, you surely need to specify the range of $s,t$, no?
    $endgroup$
    – leonbloy
    Nov 24 '18 at 2:14










  • $begingroup$
    Yes. I'm trying to edit it. $-2<s<2$ and $-1<t<1$.
    $endgroup$
    – FreeSid91
    Nov 24 '18 at 2:26
















$begingroup$
I guess you meant $P()$ instead of $F()$, no? Also, you surely need to specify the range of $s,t$, no?
$endgroup$
– leonbloy
Nov 24 '18 at 2:14




$begingroup$
I guess you meant $P()$ instead of $F()$, no? Also, you surely need to specify the range of $s,t$, no?
$endgroup$
– leonbloy
Nov 24 '18 at 2:14












$begingroup$
Yes. I'm trying to edit it. $-2<s<2$ and $-1<t<1$.
$endgroup$
– FreeSid91
Nov 24 '18 at 2:26




$begingroup$
Yes. I'm trying to edit it. $-2<s<2$ and $-1<t<1$.
$endgroup$
– FreeSid91
Nov 24 '18 at 2:26










1 Answer
1






active

oldest

votes


















0












$begingroup$

If you are using the transformation formula, you should have got, for the joint density



$$f_{U,V}(u,v)=frac{f_{X,Y}(x,y)}{left |frac{partial(U,V) }{partial(X,Y)}right|}=frac{1}{4}$$



but this is not the end of the story, you need to get the support of the transformed variables. To write $-2<U<2$ and $-1<V<1$ is not totally right: both inequalities are true, but they don't give you the support, because (say) you cannot have simultaneously $V=0.9$ and $U=1.9$.



(You can also guess that something is wrong in that the integral of the density over the support must be one).



The correct way is to note that if we allow for $V=X$ its full range $-1<V<1$ then we must put that dependence into the other variable: $U=Y-X=Y-V$ , hence the range of $U$ is $(-1-V, 1-V)$



Then the support is $-1<V<1$ and $-1-V< U <1-V$ which corresponds to a parallelogram.



(Notice BTW that two variables with uniform joint density over a -straight- rectangular support are independent - which is the case for $X,Y$, but it's not -it should not- for $U,V$ )



Sanity check:



$$ int f_{U,V}= int_{-1}^1 int_{-1-V}^{1-V} frac{1}{4} dU dV= frac{1}{4} int_{-1}^1 2 dV= 1$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Doesn't the Jacobian have to multiply $f_{X,Y}(x,y)$ instead of divide it? (here it doesn't matter much because $|J| = 1$).
    $endgroup$
    – Thomas Bladt
    Nov 24 '18 at 3:05










  • $begingroup$
    @leonbloy Thanks! I was getting confused as I was only changing the lower bound of $U$ instead of both.
    $endgroup$
    – FreeSid91
    Nov 24 '18 at 3:50












  • $begingroup$
    @ThomasBladt It depens on which derivative we are taking. I clarified the notation
    $endgroup$
    – leonbloy
    Nov 24 '18 at 11:20











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









0












$begingroup$

If you are using the transformation formula, you should have got, for the joint density



$$f_{U,V}(u,v)=frac{f_{X,Y}(x,y)}{left |frac{partial(U,V) }{partial(X,Y)}right|}=frac{1}{4}$$



but this is not the end of the story, you need to get the support of the transformed variables. To write $-2<U<2$ and $-1<V<1$ is not totally right: both inequalities are true, but they don't give you the support, because (say) you cannot have simultaneously $V=0.9$ and $U=1.9$.



(You can also guess that something is wrong in that the integral of the density over the support must be one).



The correct way is to note that if we allow for $V=X$ its full range $-1<V<1$ then we must put that dependence into the other variable: $U=Y-X=Y-V$ , hence the range of $U$ is $(-1-V, 1-V)$



Then the support is $-1<V<1$ and $-1-V< U <1-V$ which corresponds to a parallelogram.



(Notice BTW that two variables with uniform joint density over a -straight- rectangular support are independent - which is the case for $X,Y$, but it's not -it should not- for $U,V$ )



Sanity check:



$$ int f_{U,V}= int_{-1}^1 int_{-1-V}^{1-V} frac{1}{4} dU dV= frac{1}{4} int_{-1}^1 2 dV= 1$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Doesn't the Jacobian have to multiply $f_{X,Y}(x,y)$ instead of divide it? (here it doesn't matter much because $|J| = 1$).
    $endgroup$
    – Thomas Bladt
    Nov 24 '18 at 3:05










  • $begingroup$
    @leonbloy Thanks! I was getting confused as I was only changing the lower bound of $U$ instead of both.
    $endgroup$
    – FreeSid91
    Nov 24 '18 at 3:50












  • $begingroup$
    @ThomasBladt It depens on which derivative we are taking. I clarified the notation
    $endgroup$
    – leonbloy
    Nov 24 '18 at 11:20
















0












$begingroup$

If you are using the transformation formula, you should have got, for the joint density



$$f_{U,V}(u,v)=frac{f_{X,Y}(x,y)}{left |frac{partial(U,V) }{partial(X,Y)}right|}=frac{1}{4}$$



but this is not the end of the story, you need to get the support of the transformed variables. To write $-2<U<2$ and $-1<V<1$ is not totally right: both inequalities are true, but they don't give you the support, because (say) you cannot have simultaneously $V=0.9$ and $U=1.9$.



(You can also guess that something is wrong in that the integral of the density over the support must be one).



The correct way is to note that if we allow for $V=X$ its full range $-1<V<1$ then we must put that dependence into the other variable: $U=Y-X=Y-V$ , hence the range of $U$ is $(-1-V, 1-V)$



Then the support is $-1<V<1$ and $-1-V< U <1-V$ which corresponds to a parallelogram.



(Notice BTW that two variables with uniform joint density over a -straight- rectangular support are independent - which is the case for $X,Y$, but it's not -it should not- for $U,V$ )



Sanity check:



$$ int f_{U,V}= int_{-1}^1 int_{-1-V}^{1-V} frac{1}{4} dU dV= frac{1}{4} int_{-1}^1 2 dV= 1$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Doesn't the Jacobian have to multiply $f_{X,Y}(x,y)$ instead of divide it? (here it doesn't matter much because $|J| = 1$).
    $endgroup$
    – Thomas Bladt
    Nov 24 '18 at 3:05










  • $begingroup$
    @leonbloy Thanks! I was getting confused as I was only changing the lower bound of $U$ instead of both.
    $endgroup$
    – FreeSid91
    Nov 24 '18 at 3:50












  • $begingroup$
    @ThomasBladt It depens on which derivative we are taking. I clarified the notation
    $endgroup$
    – leonbloy
    Nov 24 '18 at 11:20














0












0








0





$begingroup$

If you are using the transformation formula, you should have got, for the joint density



$$f_{U,V}(u,v)=frac{f_{X,Y}(x,y)}{left |frac{partial(U,V) }{partial(X,Y)}right|}=frac{1}{4}$$



but this is not the end of the story, you need to get the support of the transformed variables. To write $-2<U<2$ and $-1<V<1$ is not totally right: both inequalities are true, but they don't give you the support, because (say) you cannot have simultaneously $V=0.9$ and $U=1.9$.



(You can also guess that something is wrong in that the integral of the density over the support must be one).



The correct way is to note that if we allow for $V=X$ its full range $-1<V<1$ then we must put that dependence into the other variable: $U=Y-X=Y-V$ , hence the range of $U$ is $(-1-V, 1-V)$



Then the support is $-1<V<1$ and $-1-V< U <1-V$ which corresponds to a parallelogram.



(Notice BTW that two variables with uniform joint density over a -straight- rectangular support are independent - which is the case for $X,Y$, but it's not -it should not- for $U,V$ )



Sanity check:



$$ int f_{U,V}= int_{-1}^1 int_{-1-V}^{1-V} frac{1}{4} dU dV= frac{1}{4} int_{-1}^1 2 dV= 1$$






share|cite|improve this answer











$endgroup$



If you are using the transformation formula, you should have got, for the joint density



$$f_{U,V}(u,v)=frac{f_{X,Y}(x,y)}{left |frac{partial(U,V) }{partial(X,Y)}right|}=frac{1}{4}$$



but this is not the end of the story, you need to get the support of the transformed variables. To write $-2<U<2$ and $-1<V<1$ is not totally right: both inequalities are true, but they don't give you the support, because (say) you cannot have simultaneously $V=0.9$ and $U=1.9$.



(You can also guess that something is wrong in that the integral of the density over the support must be one).



The correct way is to note that if we allow for $V=X$ its full range $-1<V<1$ then we must put that dependence into the other variable: $U=Y-X=Y-V$ , hence the range of $U$ is $(-1-V, 1-V)$



Then the support is $-1<V<1$ and $-1-V< U <1-V$ which corresponds to a parallelogram.



(Notice BTW that two variables with uniform joint density over a -straight- rectangular support are independent - which is the case for $X,Y$, but it's not -it should not- for $U,V$ )



Sanity check:



$$ int f_{U,V}= int_{-1}^1 int_{-1-V}^{1-V} frac{1}{4} dU dV= frac{1}{4} int_{-1}^1 2 dV= 1$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 24 '18 at 11:19

























answered Nov 24 '18 at 2:47









leonbloyleonbloy

40.4k645107




40.4k645107












  • $begingroup$
    Doesn't the Jacobian have to multiply $f_{X,Y}(x,y)$ instead of divide it? (here it doesn't matter much because $|J| = 1$).
    $endgroup$
    – Thomas Bladt
    Nov 24 '18 at 3:05










  • $begingroup$
    @leonbloy Thanks! I was getting confused as I was only changing the lower bound of $U$ instead of both.
    $endgroup$
    – FreeSid91
    Nov 24 '18 at 3:50












  • $begingroup$
    @ThomasBladt It depens on which derivative we are taking. I clarified the notation
    $endgroup$
    – leonbloy
    Nov 24 '18 at 11:20


















  • $begingroup$
    Doesn't the Jacobian have to multiply $f_{X,Y}(x,y)$ instead of divide it? (here it doesn't matter much because $|J| = 1$).
    $endgroup$
    – Thomas Bladt
    Nov 24 '18 at 3:05










  • $begingroup$
    @leonbloy Thanks! I was getting confused as I was only changing the lower bound of $U$ instead of both.
    $endgroup$
    – FreeSid91
    Nov 24 '18 at 3:50












  • $begingroup$
    @ThomasBladt It depens on which derivative we are taking. I clarified the notation
    $endgroup$
    – leonbloy
    Nov 24 '18 at 11:20
















$begingroup$
Doesn't the Jacobian have to multiply $f_{X,Y}(x,y)$ instead of divide it? (here it doesn't matter much because $|J| = 1$).
$endgroup$
– Thomas Bladt
Nov 24 '18 at 3:05




$begingroup$
Doesn't the Jacobian have to multiply $f_{X,Y}(x,y)$ instead of divide it? (here it doesn't matter much because $|J| = 1$).
$endgroup$
– Thomas Bladt
Nov 24 '18 at 3:05












$begingroup$
@leonbloy Thanks! I was getting confused as I was only changing the lower bound of $U$ instead of both.
$endgroup$
– FreeSid91
Nov 24 '18 at 3:50






$begingroup$
@leonbloy Thanks! I was getting confused as I was only changing the lower bound of $U$ instead of both.
$endgroup$
– FreeSid91
Nov 24 '18 at 3:50














$begingroup$
@ThomasBladt It depens on which derivative we are taking. I clarified the notation
$endgroup$
– leonbloy
Nov 24 '18 at 11:20




$begingroup$
@ThomasBladt It depens on which derivative we are taking. I clarified the notation
$endgroup$
– leonbloy
Nov 24 '18 at 11:20


















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