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












0














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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  • I guess you meant $P()$ instead of $F()$, no? Also, you surely need to specify the range of $s,t$, no?
    – leonbloy
    Nov 24 at 2:14










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
















0














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
























  • I guess you meant $P()$ instead of $F()$, no? Also, you surely need to specify the range of $s,t$, no?
    – leonbloy
    Nov 24 at 2:14










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














0












0








0







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















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






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share|cite|improve this question













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edited Nov 24 at 2:52









Rócherz

2,7262721




2,7262721










asked Nov 24 at 2:07









FreeSid91

1




1












  • I guess you meant $P()$ instead of $F()$, no? Also, you surely need to specify the range of $s,t$, no?
    – leonbloy
    Nov 24 at 2:14










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


















  • I guess you meant $P()$ instead of $F()$, no? Also, you surely need to specify the range of $s,t$, no?
    – leonbloy
    Nov 24 at 2:14










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
















I guess you meant $P()$ instead of $F()$, no? Also, you surely need to specify the range of $s,t$, no?
– leonbloy
Nov 24 at 2:14




I guess you meant $P()$ instead of $F()$, no? Also, you surely need to specify the range of $s,t$, no?
– leonbloy
Nov 24 at 2:14












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




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










1 Answer
1






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oldest

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0














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























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










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












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











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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























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










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












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
















0














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























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










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












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














0












0








0






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














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 at 11:19

























answered Nov 24 at 2:47









leonbloy

40.2k645107




40.2k645107












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










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












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


















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










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












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
















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




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












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






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














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




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


















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