Mixing
Calculating $Y|X$ density with distribution $U[X,X+1]$
$begingroup$
I have two random variables $X$ and $Y$ where $X$ has distribution $U [0; 1]$ and $Y | X$ has distribution $U [X; X +1]$ (that is, $f (x) = 1$ for $x$ and $[0, 1]$ and $0$ cc, and $fY|X(y|x) = 1$ if $y in[x; x + 1]$ and $0$ cc, for all $x$.)
How can I calculate the density of Y?
Also, can you clarify how to work with $U [X; X +1]$?
Any hint?
probability probability-theory
asked Jan 11 at 2:06
LauraLaura
2358
$endgroup$
add a comment |
$begingroup$
I have two random variables $X$ and $Y$ where $X$ has distribution $U [0; 1]$ and $Y | X$ has distribution $U [X; X +1]$ (that is, $f (x) = 1$ for $x$ and $[0, 1]$ and $0$ cc, and $fY|X(y|x) = 1$ if $y in[x; x + 1]$ and $0$ cc, for all $x$.)
How can I calculate the density of Y?
Also, can you clarify how to work with $U [X; X +1]$?
Any hint?
probability probability-theory
asked Jan 11 at 2:06
LauraLaura
2358
$endgroup$
add a comment |
$begingroup$
I have two random variables $X$ and $Y$ where $X$ has distribution $U [0; 1]$ and $Y | X$ has distribution $U [X; X +1]$ (that is, $f (x) = 1$ for $x$ and $[0, 1]$ and $0$ cc, and $fY|X(y|x) = 1$ if $y in[x; x + 1]$ and $0$ cc, for all $x$.)
How can I calculate the density of Y?
Also, can you clarify how to work with $U [X; X +1]$?
Any hint?
probability probability-theory
asked Jan 11 at 2:06
LauraLaura
2358
$endgroup$
I have two random variables $X$ and $Y$ where $X$ has distribution $U [0; 1]$ and $Y | X$ has distribution $U [X; X +1]$ (that is, $f (x) = 1$ for $x$ and $[0, 1]$ and $0$ cc, and $fY|X(y|x) = 1$ if $y in[x; x + 1]$ and $0$ cc, for all $x$.)
How can I calculate the density of Y?
Also, can you clarify how to work with $U [X; X +1]$?
Any hint?
probability probability-theory
probability probability-theory
asked Jan 11 at 2:06
LauraLaura
2358
asked Jan 11 at 2:06
LauraLaura
2358
asked Jan 11 at 2:06
LauraLaura
2358
asked Jan 11 at 2:06
LauraLaura
2358
asked Jan 11 at 2:06
LauraLaura
2358
2358
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Given the marginal distribution for $X$, and the conditional distribution $Y|X$, you can recover the distribution for $Y$ from a continuous version of the law of total probability:
$$
f_Y(y) = int f_{Y|X=x}(y)f_X(x),dx
$$
The marginal pdf $f_X(x)$ is $1$ on $[0,1]$ and $0$ elsewhere, so this becomes
$$
f_Y(y)=int_0^1 f_{Y|X=x}(y),dx
$$
The conditional pdf $f_{Y|X=x}(y)$ is equal to $1$ when $x<y<x+1$. Otherwise, it is $0$. Effectively, in addition to the bounds $0<x<1$ of integration, we also have the bounds $y-1<x<y$. These two imply we are integrating the function $1$ over the region where $max(0,y-1)<x<min(1,y)$. Whenever $0<y<2$, this region is nonempty, and we get
$$
f_Y(y)=int_{max(0,y-1)}^{min(1,y)}1,dx=min(1,y)-max(0,y-1).
$$
When $0<y<1$, $f_Y(y)=y$. When $1<y<2$, $f_Y(y)=2-y$. Otherwise, $f_Y(y)=0$.
edited Jan 13 at 20:45
Did
247k23223460
answered Jan 11 at 3:27
Mike EarnestMike Earnest
22.4k12051
$endgroup$
$begingroup$
thank you very much. Its really difficult this question for me. Thanks!
$endgroup$
– Laura
Jan 14 at 0:20
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given the marginal distribution for $X$, and the conditional distribution $Y|X$, you can recover the distribution for $Y$ from a continuous version of the law of total probability:
$$
f_Y(y) = int f_{Y|X=x}(y)f_X(x),dx
$$
The marginal pdf $f_X(x)$ is $1$ on $[0,1]$ and $0$ elsewhere, so this becomes
$$
f_Y(y)=int_0^1 f_{Y|X=x}(y),dx
$$
The conditional pdf $f_{Y|X=x}(y)$ is equal to $1$ when $x<y<x+1$. Otherwise, it is $0$. Effectively, in addition to the bounds $0<x<1$ of integration, we also have the bounds $y-1<x<y$. These two imply we are integrating the function $1$ over the region where $max(0,y-1)<x<min(1,y)$. Whenever $0<y<2$, this region is nonempty, and we get
$$
f_Y(y)=int_{max(0,y-1)}^{min(1,y)}1,dx=min(1,y)-max(0,y-1).
$$
When $0<y<1$, $f_Y(y)=y$. When $1<y<2$, $f_Y(y)=2-y$. Otherwise, $f_Y(y)=0$.
edited Jan 13 at 20:45
Did
247k23223460
answered Jan 11 at 3:27
Mike EarnestMike Earnest
22.4k12051
$endgroup$
$begingroup$
thank you very much. Its really difficult this question for me. Thanks!
$endgroup$
– Laura
Jan 14 at 0:20
add a comment |
$begingroup$
Given the marginal distribution for $X$, and the conditional distribution $Y|X$, you can recover the distribution for $Y$ from a continuous version of the law of total probability:
$$
f_Y(y) = int f_{Y|X=x}(y)f_X(x),dx
$$
The marginal pdf $f_X(x)$ is $1$ on $[0,1]$ and $0$ elsewhere, so this becomes
$$
f_Y(y)=int_0^1 f_{Y|X=x}(y),dx
$$
The conditional pdf $f_{Y|X=x}(y)$ is equal to $1$ when $x<y<x+1$. Otherwise, it is $0$. Effectively, in addition to the bounds $0<x<1$ of integration, we also have the bounds $y-1<x<y$. These two imply we are integrating the function $1$ over the region where $max(0,y-1)<x<min(1,y)$. Whenever $0<y<2$, this region is nonempty, and we get
$$
f_Y(y)=int_{max(0,y-1)}^{min(1,y)}1,dx=min(1,y)-max(0,y-1).
$$
When $0<y<1$, $f_Y(y)=y$. When $1<y<2$, $f_Y(y)=2-y$. Otherwise, $f_Y(y)=0$.
edited Jan 13 at 20:45
Did
247k23223460
answered Jan 11 at 3:27
Mike EarnestMike Earnest
22.4k12051
$endgroup$
$begingroup$
thank you very much. Its really difficult this question for me. Thanks!
$endgroup$
– Laura
Jan 14 at 0:20
add a comment |
$begingroup$
Given the marginal distribution for $X$, and the conditional distribution $Y|X$, you can recover the distribution for $Y$ from a continuous version of the law of total probability:
$$
f_Y(y) = int f_{Y|X=x}(y)f_X(x),dx
$$
The marginal pdf $f_X(x)$ is $1$ on $[0,1]$ and $0$ elsewhere, so this becomes
$$
f_Y(y)=int_0^1 f_{Y|X=x}(y),dx
$$
The conditional pdf $f_{Y|X=x}(y)$ is equal to $1$ when $x<y<x+1$. Otherwise, it is $0$. Effectively, in addition to the bounds $0<x<1$ of integration, we also have the bounds $y-1<x<y$. These two imply we are integrating the function $1$ over the region where $max(0,y-1)<x<min(1,y)$. Whenever $0<y<2$, this region is nonempty, and we get
$$
f_Y(y)=int_{max(0,y-1)}^{min(1,y)}1,dx=min(1,y)-max(0,y-1).
$$
When $0<y<1$, $f_Y(y)=y$. When $1<y<2$, $f_Y(y)=2-y$. Otherwise, $f_Y(y)=0$.
edited Jan 13 at 20:45
Did
247k23223460
answered Jan 11 at 3:27
Mike EarnestMike Earnest
22.4k12051
$endgroup$
Given the marginal distribution for $X$, and the conditional distribution $Y|X$, you can recover the distribution for $Y$ from a continuous version of the law of total probability:
$$
f_Y(y) = int f_{Y|X=x}(y)f_X(x),dx
$$
The marginal pdf $f_X(x)$ is $1$ on $[0,1]$ and $0$ elsewhere, so this becomes
$$
f_Y(y)=int_0^1 f_{Y|X=x}(y),dx
$$
The conditional pdf $f_{Y|X=x}(y)$ is equal to $1$ when $x<y<x+1$. Otherwise, it is $0$. Effectively, in addition to the bounds $0<x<1$ of integration, we also have the bounds $y-1<x<y$. These two imply we are integrating the function $1$ over the region where $max(0,y-1)<x<min(1,y)$. Whenever $0<y<2$, this region is nonempty, and we get
$$
f_Y(y)=int_{max(0,y-1)}^{min(1,y)}1,dx=min(1,y)-max(0,y-1).
$$
When $0<y<1$, $f_Y(y)=y$. When $1<y<2$, $f_Y(y)=2-y$. Otherwise, $f_Y(y)=0$.
edited Jan 13 at 20:45
Did
247k23223460
answered Jan 11 at 3:27
Mike EarnestMike Earnest
22.4k12051
edited Jan 13 at 20:45
Did
247k23223460
edited Jan 13 at 20:45
Did
247k23223460
edited Jan 13 at 20:45
Did
247k23223460
247k23223460
answered Jan 11 at 3:27
Mike EarnestMike Earnest
22.4k12051
answered Jan 11 at 3:27
Mike EarnestMike Earnest
22.4k12051
answered Jan 11 at 3:27
Mike EarnestMike Earnest
22.4k12051
22.4k12051
$begingroup$
thank you very much. Its really difficult this question for me. Thanks!
$endgroup$
– Laura
Jan 14 at 0:20
add a comment |
$begingroup$
thank you very much. Its really difficult this question for me. Thanks!
$endgroup$
– Laura
Jan 14 at 0:20
$begingroup$
thank you very much. Its really difficult this question for me. Thanks!
$endgroup$
– Laura
Jan 14 at 0:20
$begingroup$
thank you very much. Its really difficult this question for me. Thanks!
$endgroup$
– Laura
Jan 14 at 0:20
add a comment |
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