Mixing
Find PDF of Y from (constant) joint PDF
$begingroup$
Given the following joint PDF:
$$
f_{X,Y}(t,s) =
begin{cases}
frac{2}{3} & 0leq tleq 1, -1leq sleq t \
0 & otherwise
end{cases}
$$
I need to find $f_y(s)$
So according to defintion:
$f_Y(s)=int_{-infty}^{infty}f_{X,Y}(t,s) dt = frac{2}{3}int_{0}^{1} dt = frac{2}{3}$
Which is TOTALLY wrong since it's not even a valid density (doesn't integrate to 1). I believe I have some component missing here, but I can't see what. I guess it has something with the fact that $s$ is bounded by $t$ in some of the region, but I can't understand what to do with that.
Could anyone please enlighten me? I really really wish to understand what's wrong with my way of thinking here.
Thanks!
probability probability-distributions density-function
asked Jan 31 at 12:44
superuser123superuser123
48628
$endgroup$
add a comment |
$begingroup$
Given the following joint PDF:
$$
f_{X,Y}(t,s) =
begin{cases}
frac{2}{3} & 0leq tleq 1, -1leq sleq t \
0 & otherwise
end{cases}
$$
I need to find $f_y(s)$
So according to defintion:
$f_Y(s)=int_{-infty}^{infty}f_{X,Y}(t,s) dt = frac{2}{3}int_{0}^{1} dt = frac{2}{3}$
Which is TOTALLY wrong since it's not even a valid density (doesn't integrate to 1). I believe I have some component missing here, but I can't see what. I guess it has something with the fact that $s$ is bounded by $t$ in some of the region, but I can't understand what to do with that.
Could anyone please enlighten me? I really really wish to understand what's wrong with my way of thinking here.
Thanks!
probability probability-distributions density-function
asked Jan 31 at 12:44
superuser123superuser123
48628
$endgroup$
$begingroup$
Writing the joint density with full support in the integral would make this easy to understand: $$f_Y(s)=intfrac{2}{3}mathbf1_{0<t<1,-1<s<t},dt=frac{2}{3}int mathbf1_{max(0,s)<t<1}mathbf1_{-1<s<1},dt=cdots$$
$endgroup$
– StubbornAtom
Jan 31 at 13:48
$begingroup$
@StubbornAtom thanks, I am not sure though how to continue. I guess that $1_{0<t<1, -1<s<t}$ is and indicator, but Iv'e never used it in such cases. Can you please share more details?
$endgroup$
– superuser123
Jan 31 at 13:52
$begingroup$
The joint PDF is not constant (none is), for example $f_{X,Y}(0,frac12)ne f_{X,Y}(frac12,0)$.
$endgroup$
– Did
Jan 31 at 13:53
1
$begingroup$
@superuser123 You should use the indicator in these cases to help you keep track of the bounds of the integral. Just rewrite the condition $0<t<1,-1<s<t$ to get $max(0,s)<t<1,,,-1<s<1$. Now you can see that $t$ runs from $max(0,s)$ to $1$ whenever $s$ runs from $-1$ to $1$.
$endgroup$
– StubbornAtom
Jan 31 at 13:59
add a comment |
$begingroup$
Given the following joint PDF:
$$
f_{X,Y}(t,s) =
begin{cases}
frac{2}{3} & 0leq tleq 1, -1leq sleq t \
0 & otherwise
end{cases}
$$
I need to find $f_y(s)$
So according to defintion:
$f_Y(s)=int_{-infty}^{infty}f_{X,Y}(t,s) dt = frac{2}{3}int_{0}^{1} dt = frac{2}{3}$
Which is TOTALLY wrong since it's not even a valid density (doesn't integrate to 1). I believe I have some component missing here, but I can't see what. I guess it has something with the fact that $s$ is bounded by $t$ in some of the region, but I can't understand what to do with that.
Could anyone please enlighten me? I really really wish to understand what's wrong with my way of thinking here.
Thanks!
probability probability-distributions density-function
asked Jan 31 at 12:44
superuser123superuser123
48628
$endgroup$
Given the following joint PDF:
$$
f_{X,Y}(t,s) =
begin{cases}
frac{2}{3} & 0leq tleq 1, -1leq sleq t \
0 & otherwise
end{cases}
$$
I need to find $f_y(s)$
So according to defintion:
$f_Y(s)=int_{-infty}^{infty}f_{X,Y}(t,s) dt = frac{2}{3}int_{0}^{1} dt = frac{2}{3}$
Which is TOTALLY wrong since it's not even a valid density (doesn't integrate to 1). I believe I have some component missing here, but I can't see what. I guess it has something with the fact that $s$ is bounded by $t$ in some of the region, but I can't understand what to do with that.
Could anyone please enlighten me? I really really wish to understand what's wrong with my way of thinking here.
Thanks!
probability probability-distributions density-function
probability probability-distributions density-function
asked Jan 31 at 12:44
superuser123superuser123
48628
asked Jan 31 at 12:44
superuser123superuser123
48628
asked Jan 31 at 12:44
superuser123superuser123
48628
asked Jan 31 at 12:44
superuser123superuser123
48628
asked Jan 31 at 12:44
superuser123superuser123
48628
48628
$begingroup$
Writing the joint density with full support in the integral would make this easy to understand: $$f_Y(s)=intfrac{2}{3}mathbf1_{0<t<1,-1<s<t},dt=frac{2}{3}int mathbf1_{max(0,s)<t<1}mathbf1_{-1<s<1},dt=cdots$$
$endgroup$
– StubbornAtom
Jan 31 at 13:48
$begingroup$
@StubbornAtom thanks, I am not sure though how to continue. I guess that $1_{0<t<1, -1<s<t}$ is and indicator, but Iv'e never used it in such cases. Can you please share more details?
$endgroup$
– superuser123
Jan 31 at 13:52
$begingroup$
The joint PDF is not constant (none is), for example $f_{X,Y}(0,frac12)ne f_{X,Y}(frac12,0)$.
$endgroup$
– Did
Jan 31 at 13:53
1
$begingroup$
@superuser123 You should use the indicator in these cases to help you keep track of the bounds of the integral. Just rewrite the condition $0<t<1,-1<s<t$ to get $max(0,s)<t<1,,,-1<s<1$. Now you can see that $t$ runs from $max(0,s)$ to $1$ whenever $s$ runs from $-1$ to $1$.
$endgroup$
– StubbornAtom
Jan 31 at 13:59
add a comment |
$begingroup$
Writing the joint density with full support in the integral would make this easy to understand: $$f_Y(s)=intfrac{2}{3}mathbf1_{0<t<1,-1<s<t},dt=frac{2}{3}int mathbf1_{max(0,s)<t<1}mathbf1_{-1<s<1},dt=cdots$$
$endgroup$
– StubbornAtom
Jan 31 at 13:48
$begingroup$
@StubbornAtom thanks, I am not sure though how to continue. I guess that $1_{0<t<1, -1<s<t}$ is and indicator, but Iv'e never used it in such cases. Can you please share more details?
$endgroup$
– superuser123
Jan 31 at 13:52
$begingroup$
The joint PDF is not constant (none is), for example $f_{X,Y}(0,frac12)ne f_{X,Y}(frac12,0)$.
$endgroup$
– Did
Jan 31 at 13:53
1
$begingroup$
@superuser123 You should use the indicator in these cases to help you keep track of the bounds of the integral. Just rewrite the condition $0<t<1,-1<s<t$ to get $max(0,s)<t<1,,,-1<s<1$. Now you can see that $t$ runs from $max(0,s)$ to $1$ whenever $s$ runs from $-1$ to $1$.
$endgroup$
– StubbornAtom
Jan 31 at 13:59
$begingroup$
Writing the joint density with full support in the integral would make this easy to understand: $$f_Y(s)=intfrac{2}{3}mathbf1_{0<t<1,-1<s<t},dt=frac{2}{3}int mathbf1_{max(0,s)<t<1}mathbf1_{-1<s<1},dt=cdots$$
$endgroup$
– StubbornAtom
Jan 31 at 13:48
$begingroup$
Writing the joint density with full support in the integral would make this easy to understand: $$f_Y(s)=intfrac{2}{3}mathbf1_{0<t<1,-1<s<t},dt=frac{2}{3}int mathbf1_{max(0,s)<t<1}mathbf1_{-1<s<1},dt=cdots$$
$endgroup$
– StubbornAtom
Jan 31 at 13:48
$begingroup$
@StubbornAtom thanks, I am not sure though how to continue. I guess that $1_{0<t<1, -1<s<t}$ is and indicator, but Iv'e never used it in such cases. Can you please share more details?
$endgroup$
– superuser123
Jan 31 at 13:52
$begingroup$
@StubbornAtom thanks, I am not sure though how to continue. I guess that $1_{0<t<1, -1<s<t}$ is and indicator, but Iv'e never used it in such cases. Can you please share more details?
$endgroup$
– superuser123
Jan 31 at 13:52
$begingroup$
The joint PDF is not constant (none is), for example $f_{X,Y}(0,frac12)ne f_{X,Y}(frac12,0)$.
$endgroup$
– Did
Jan 31 at 13:53
$begingroup$
The joint PDF is not constant (none is), for example $f_{X,Y}(0,frac12)ne f_{X,Y}(frac12,0)$.
$endgroup$
– Did
Jan 31 at 13:53
1
1
$begingroup$
@superuser123 You should use the indicator in these cases to help you keep track of the bounds of the integral. Just rewrite the condition $0<t<1,-1<s<t$ to get $max(0,s)<t<1,,,-1<s<1$. Now you can see that $t$ runs from $max(0,s)$ to $1$ whenever $s$ runs from $-1$ to $1$.
$endgroup$
– StubbornAtom
Jan 31 at 13:59
$begingroup$
@superuser123 You should use the indicator in these cases to help you keep track of the bounds of the integral. Just rewrite the condition $0<t<1,-1<s<t$ to get $max(0,s)<t<1,,,-1<s<1$. Now you can see that $t$ runs from $max(0,s)$ to $1$ whenever $s$ runs from $-1$ to $1$.
$endgroup$
– StubbornAtom
Jan 31 at 13:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The integration is not from $t= 0$ to $t= 1$. For $s< 0$, the integration is from $t= 0$ to $t= 1$ but for $sge 0$ the integration is from $t= s$ to $t= 1$.
edited Jan 31 at 13:21
pointguard0
1,56211121
answered Jan 31 at 13:01
user247327user247327
11.5k1516
$endgroup$
$begingroup$
Thanks, but, how did you see that?
$endgroup$
– superuser123
Jan 31 at 13:46
2
$begingroup$
Draw a graph of values for x and y. x goes from 0 to 1 while y goes from -1 to x. That region is the square with vertices (0, -1), (1, -1), (0,0), and (1,0), topped by the triangle with vertices (0, 0), (1, 0), and (1, 1). For each y, if y< 0, x goes from 0 to 1 but if $yge 0$, x goes from x= y, on the left, to x= 1, on the right.
$endgroup$
– user247327
Jan 31 at 13:52
add a comment |
$begingroup$
$$f_Y(s)=int_{-infty}^{infty}f_{X,Y}(t,s) dt = int_{s}^{1} frac{2}{3} dt = frac{2}{3}(1 - s),$$
which is indeed a proper density function for $s > 0$.
Note that for $s le 0$ we have $f_Y(s)$ as you've calculated, hence
$$f_Y(s) = begin{cases}frac{2}{3}(1 - s), s ge 0 \ frac{2}{3}, s < 0 end{cases}$$
answered Jan 31 at 13:02
pointguard0pointguard0
1,56211121
$endgroup$
$begingroup$
How did you come up with that?
$endgroup$
– superuser123
Jan 31 at 13:46
$begingroup$
check carefully the limits from your system. You have $t in [0, 1]$ but that's not it, you also have $-1 le s le t$, which means that the least value of $t$ is $s$ and the most is $1$, hence the limits are $s$ and $1$.
$endgroup$
– pointguard0
Jan 31 at 13:58
$begingroup$
@pointguard0 You have only considered the case $s>0$, so the pdf of $Y$ is incomplete.
$endgroup$
– StubbornAtom
Jan 31 at 14:03
$begingroup$
@StubbornAtom you are right, thanks, just fixed it.
$endgroup$
– pointguard0
Jan 31 at 14:14
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 integration is not from $t= 0$ to $t= 1$. For $s< 0$, the integration is from $t= 0$ to $t= 1$ but for $sge 0$ the integration is from $t= s$ to $t= 1$.
edited Jan 31 at 13:21
pointguard0
1,56211121
answered Jan 31 at 13:01
user247327user247327
11.5k1516
$endgroup$
$begingroup$
Thanks, but, how did you see that?
$endgroup$
– superuser123
Jan 31 at 13:46
2
$begingroup$
Draw a graph of values for x and y. x goes from 0 to 1 while y goes from -1 to x. That region is the square with vertices (0, -1), (1, -1), (0,0), and (1,0), topped by the triangle with vertices (0, 0), (1, 0), and (1, 1). For each y, if y< 0, x goes from 0 to 1 but if $yge 0$, x goes from x= y, on the left, to x= 1, on the right.
$endgroup$
– user247327
Jan 31 at 13:52
add a comment |
$begingroup$
The integration is not from $t= 0$ to $t= 1$. For $s< 0$, the integration is from $t= 0$ to $t= 1$ but for $sge 0$ the integration is from $t= s$ to $t= 1$.
edited Jan 31 at 13:21
pointguard0
1,56211121
answered Jan 31 at 13:01
user247327user247327
11.5k1516
$endgroup$
$begingroup$
Thanks, but, how did you see that?
$endgroup$
– superuser123
Jan 31 at 13:46
2
$begingroup$
Draw a graph of values for x and y. x goes from 0 to 1 while y goes from -1 to x. That region is the square with vertices (0, -1), (1, -1), (0,0), and (1,0), topped by the triangle with vertices (0, 0), (1, 0), and (1, 1). For each y, if y< 0, x goes from 0 to 1 but if $yge 0$, x goes from x= y, on the left, to x= 1, on the right.
$endgroup$
– user247327
Jan 31 at 13:52
add a comment |
$begingroup$
The integration is not from $t= 0$ to $t= 1$. For $s< 0$, the integration is from $t= 0$ to $t= 1$ but for $sge 0$ the integration is from $t= s$ to $t= 1$.
edited Jan 31 at 13:21
pointguard0
1,56211121
answered Jan 31 at 13:01
user247327user247327
11.5k1516
$endgroup$
The integration is not from $t= 0$ to $t= 1$. For $s< 0$, the integration is from $t= 0$ to $t= 1$ but for $sge 0$ the integration is from $t= s$ to $t= 1$.
edited Jan 31 at 13:21
pointguard0
1,56211121
answered Jan 31 at 13:01
user247327user247327
11.5k1516
edited Jan 31 at 13:21
pointguard0
1,56211121
edited Jan 31 at 13:21
pointguard0
1,56211121
edited Jan 31 at 13:21
pointguard0
1,56211121
1,56211121
answered Jan 31 at 13:01
user247327user247327
11.5k1516
answered Jan 31 at 13:01
user247327user247327
11.5k1516
answered Jan 31 at 13:01
user247327user247327
11.5k1516
11.5k1516
$begingroup$
Thanks, but, how did you see that?
$endgroup$
– superuser123
Jan 31 at 13:46
2
$begingroup$
Draw a graph of values for x and y. x goes from 0 to 1 while y goes from -1 to x. That region is the square with vertices (0, -1), (1, -1), (0,0), and (1,0), topped by the triangle with vertices (0, 0), (1, 0), and (1, 1). For each y, if y< 0, x goes from 0 to 1 but if $yge 0$, x goes from x= y, on the left, to x= 1, on the right.
$endgroup$
– user247327
Jan 31 at 13:52
add a comment |
$begingroup$
Thanks, but, how did you see that?
$endgroup$
– superuser123
Jan 31 at 13:46
2
$begingroup$
Draw a graph of values for x and y. x goes from 0 to 1 while y goes from -1 to x. That region is the square with vertices (0, -1), (1, -1), (0,0), and (1,0), topped by the triangle with vertices (0, 0), (1, 0), and (1, 1). For each y, if y< 0, x goes from 0 to 1 but if $yge 0$, x goes from x= y, on the left, to x= 1, on the right.
$endgroup$
– user247327
Jan 31 at 13:52
$begingroup$
Thanks, but, how did you see that?
$endgroup$
– superuser123
Jan 31 at 13:46
$begingroup$
Thanks, but, how did you see that?
$endgroup$
– superuser123
Jan 31 at 13:46
2
2
$begingroup$
Draw a graph of values for x and y. x goes from 0 to 1 while y goes from -1 to x. That region is the square with vertices (0, -1), (1, -1), (0,0), and (1,0), topped by the triangle with vertices (0, 0), (1, 0), and (1, 1). For each y, if y< 0, x goes from 0 to 1 but if $yge 0$, x goes from x= y, on the left, to x= 1, on the right.
$endgroup$
– user247327
Jan 31 at 13:52
$begingroup$
Draw a graph of values for x and y. x goes from 0 to 1 while y goes from -1 to x. That region is the square with vertices (0, -1), (1, -1), (0,0), and (1,0), topped by the triangle with vertices (0, 0), (1, 0), and (1, 1). For each y, if y< 0, x goes from 0 to 1 but if $yge 0$, x goes from x= y, on the left, to x= 1, on the right.
$endgroup$
– user247327
Jan 31 at 13:52
add a comment |
$begingroup$
$$f_Y(s)=int_{-infty}^{infty}f_{X,Y}(t,s) dt = int_{s}^{1} frac{2}{3} dt = frac{2}{3}(1 - s),$$
which is indeed a proper density function for $s > 0$.
Note that for $s le 0$ we have $f_Y(s)$ as you've calculated, hence
$$f_Y(s) = begin{cases}frac{2}{3}(1 - s), s ge 0 \ frac{2}{3}, s < 0 end{cases}$$
answered Jan 31 at 13:02
pointguard0pointguard0
1,56211121
$endgroup$
$begingroup$
How did you come up with that?
$endgroup$
– superuser123
Jan 31 at 13:46
$begingroup$
check carefully the limits from your system. You have $t in [0, 1]$ but that's not it, you also have $-1 le s le t$, which means that the least value of $t$ is $s$ and the most is $1$, hence the limits are $s$ and $1$.
$endgroup$
– pointguard0
Jan 31 at 13:58
$begingroup$
@pointguard0 You have only considered the case $s>0$, so the pdf of $Y$ is incomplete.
$endgroup$
– StubbornAtom
Jan 31 at 14:03
$begingroup$
@StubbornAtom you are right, thanks, just fixed it.
$endgroup$
– pointguard0
Jan 31 at 14:14
add a comment |
$begingroup$
$$f_Y(s)=int_{-infty}^{infty}f_{X,Y}(t,s) dt = int_{s}^{1} frac{2}{3} dt = frac{2}{3}(1 - s),$$
which is indeed a proper density function for $s > 0$.
Note that for $s le 0$ we have $f_Y(s)$ as you've calculated, hence
$$f_Y(s) = begin{cases}frac{2}{3}(1 - s), s ge 0 \ frac{2}{3}, s < 0 end{cases}$$
answered Jan 31 at 13:02
pointguard0pointguard0
1,56211121
$endgroup$
$begingroup$
How did you come up with that?
$endgroup$
– superuser123
Jan 31 at 13:46
$begingroup$
check carefully the limits from your system. You have $t in [0, 1]$ but that's not it, you also have $-1 le s le t$, which means that the least value of $t$ is $s$ and the most is $1$, hence the limits are $s$ and $1$.
$endgroup$
– pointguard0
Jan 31 at 13:58
$begingroup$
@pointguard0 You have only considered the case $s>0$, so the pdf of $Y$ is incomplete.
$endgroup$
– StubbornAtom
Jan 31 at 14:03
$begingroup$
@StubbornAtom you are right, thanks, just fixed it.
$endgroup$
– pointguard0
Jan 31 at 14:14
add a comment |
$begingroup$
$$f_Y(s)=int_{-infty}^{infty}f_{X,Y}(t,s) dt = int_{s}^{1} frac{2}{3} dt = frac{2}{3}(1 - s),$$
which is indeed a proper density function for $s > 0$.
Note that for $s le 0$ we have $f_Y(s)$ as you've calculated, hence
$$f_Y(s) = begin{cases}frac{2}{3}(1 - s), s ge 0 \ frac{2}{3}, s < 0 end{cases}$$
answered Jan 31 at 13:02
pointguard0pointguard0
1,56211121
$endgroup$
$$f_Y(s)=int_{-infty}^{infty}f_{X,Y}(t,s) dt = int_{s}^{1} frac{2}{3} dt = frac{2}{3}(1 - s),$$
which is indeed a proper density function for $s > 0$.
Note that for $s le 0$ we have $f_Y(s)$ as you've calculated, hence
$$f_Y(s) = begin{cases}frac{2}{3}(1 - s), s ge 0 \ frac{2}{3}, s < 0 end{cases}$$
answered Jan 31 at 13:02
pointguard0pointguard0
1,56211121
edited Jan 31 at 14:14
answered Jan 31 at 13:02
pointguard0pointguard0
1,56211121
answered Jan 31 at 13:02
pointguard0pointguard0
1,56211121
answered Jan 31 at 13:02
pointguard0pointguard0
1,56211121
1,56211121
$begingroup$
How did you come up with that?
$endgroup$
– superuser123
Jan 31 at 13:46
$begingroup$
check carefully the limits from your system. You have $t in [0, 1]$ but that's not it, you also have $-1 le s le t$, which means that the least value of $t$ is $s$ and the most is $1$, hence the limits are $s$ and $1$.
$endgroup$
– pointguard0
Jan 31 at 13:58
$begingroup$
@pointguard0 You have only considered the case $s>0$, so the pdf of $Y$ is incomplete.
$endgroup$
– StubbornAtom
Jan 31 at 14:03
$begingroup$
@StubbornAtom you are right, thanks, just fixed it.
$endgroup$
– pointguard0
Jan 31 at 14:14
add a comment |
$begingroup$
How did you come up with that?
$endgroup$
– superuser123
Jan 31 at 13:46
$begingroup$
check carefully the limits from your system. You have $t in [0, 1]$ but that's not it, you also have $-1 le s le t$, which means that the least value of $t$ is $s$ and the most is $1$, hence the limits are $s$ and $1$.
$endgroup$
– pointguard0
Jan 31 at 13:58
$begingroup$
@pointguard0 You have only considered the case $s>0$, so the pdf of $Y$ is incomplete.
$endgroup$
– StubbornAtom
Jan 31 at 14:03
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@StubbornAtom you are right, thanks, just fixed it.
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– pointguard0
Jan 31 at 14:14
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How did you come up with that?
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– superuser123
Jan 31 at 13:46
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How did you come up with that?
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– superuser123
Jan 31 at 13:46
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check carefully the limits from your system. You have $t in [0, 1]$ but that's not it, you also have $-1 le s le t$, which means that the least value of $t$ is $s$ and the most is $1$, hence the limits are $s$ and $1$.
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– pointguard0
Jan 31 at 13:58
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check carefully the limits from your system. You have $t in [0, 1]$ but that's not it, you also have $-1 le s le t$, which means that the least value of $t$ is $s$ and the most is $1$, hence the limits are $s$ and $1$.
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– pointguard0
Jan 31 at 13:58
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@pointguard0 You have only considered the case $s>0$, so the pdf of $Y$ is incomplete.
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– StubbornAtom
Jan 31 at 14:03
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@pointguard0 You have only considered the case $s>0$, so the pdf of $Y$ is incomplete.
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– StubbornAtom
Jan 31 at 14:03
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@StubbornAtom you are right, thanks, just fixed it.
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– pointguard0
Jan 31 at 14:14
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@StubbornAtom you are right, thanks, just fixed it.
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– pointguard0
Jan 31 at 14:14
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Writing the joint density with full support in the integral would make this easy to understand: $$f_Y(s)=intfrac{2}{3}mathbf1_{0<t<1,-1<s<t},dt=frac{2}{3}int mathbf1_{max(0,s)<t<1}mathbf1_{-1<s<1},dt=cdots$$
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– StubbornAtom
Jan 31 at 13:48
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@StubbornAtom thanks, I am not sure though how to continue. I guess that $1_{0<t<1, -1<s<t}$ is and indicator, but Iv'e never used it in such cases. Can you please share more details?
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– superuser123
Jan 31 at 13:52
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The joint PDF is not constant (none is), for example $f_{X,Y}(0,frac12)ne f_{X,Y}(frac12,0)$.
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– Did
Jan 31 at 13:53
1
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@superuser123 You should use the indicator in these cases to help you keep track of the bounds of the integral. Just rewrite the condition $0<t<1,-1<s<t$ to get $max(0,s)<t<1,,,-1<s<1$. Now you can see that $t$ runs from $max(0,s)$ to $1$ whenever $s$ runs from $-1$ to $1$.
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– StubbornAtom
Jan 31 at 13:59