Find $E[Xmid Y]$ where X, Y are random variable

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Let $Y$ a uniform random variable in $[0,1]$, and let $X$ a uniform variable in $[1,e^Y]$

My work

By definition

$E(Xmid Y)=int_xin Axp(xmid y),dx$

Now, I need the function $f(xmid y)$.

By definition, $f(x|y)=fracf(x,y)f_y(y)$

I'm stuck trying to finding $f(x,y)$ and $f_y(y)$. Can someone help me?

probability

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edited Jul 14 at 17:34

Bernard

110k635103

asked Jul 14 at 16:18

Bvss12

1,609516

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Hint: $X|Y=y$ is uniformly distributed on $[1,e^y]$.
  – Stan Tendijck
  Jul 14 at 16:30
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

Let $Y$ a uniform random variable in $[0,1]$, and let $X$ a uniform variable in $[1,e^Y]$

My work

By definition

$E(Xmid Y)=int_xin Axp(xmid y),dx$

Now, I need the function $f(xmid y)$.

By definition, $f(x|y)=fracf(x,y)f_y(y)$

I'm stuck trying to finding $f(x,y)$ and $f_y(y)$. Can someone help me?

probability

share|cite|improve this question

edited Jul 14 at 17:34

Bernard

110k635103

asked Jul 14 at 16:18

Bvss12

1,609516

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Hint: $X|Y=y$ is uniformly distributed on $[1,e^y]$.
  – Stan Tendijck
  Jul 14 at 16:30
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $Y$ a uniform random variable in $[0,1]$, and let $X$ a uniform variable in $[1,e^Y]$

My work

By definition

$E(Xmid Y)=int_xin Axp(xmid y),dx$

Now, I need the function $f(xmid y)$.

By definition, $f(x|y)=fracf(x,y)f_y(y)$

I'm stuck trying to finding $f(x,y)$ and $f_y(y)$. Can someone help me?

probability

share|cite|improve this question

edited Jul 14 at 17:34

Bernard

110k635103

asked Jul 14 at 16:18

Bvss12

1,609516

Let $Y$ a uniform random variable in $[0,1]$, and let $X$ a uniform variable in $[1,e^Y]$

My work

By definition

$E(Xmid Y)=int_xin Axp(xmid y),dx$

Now, I need the function $f(xmid y)$.

By definition, $f(x|y)=fracf(x,y)f_y(y)$

I'm stuck trying to finding $f(x,y)$ and $f_y(y)$. Can someone help me?

probability

share|cite|improve this question

edited Jul 14 at 17:34

Bernard

110k635103

asked Jul 14 at 16:18

Bvss12

1,609516

share|cite|improve this question

share|cite|improve this question

edited Jul 14 at 17:34

Bernard

110k635103

edited Jul 14 at 17:34

Bernard

110k635103

edited Jul 14 at 17:34

Bernard

110k635103

110k635103

asked Jul 14 at 16:18

Bvss12

1,609516

asked Jul 14 at 16:18

Bvss12

1,609516

asked Jul 14 at 16:18

Bvss12

1,609516

1,609516

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Hint: $X|Y=y$ is uniformly distributed on $[1,e^y]$.
  – Stan Tendijck
  Jul 14 at 16:30
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Hint: $X|Y=y$ is uniformly distributed on $[1,e^y]$.
  – Stan Tendijck
  Jul 14 at 16:30
  

  
  
  
  

2

2

Hint: $X|Y=y$ is uniformly distributed on $[1,e^y]$.
– Stan Tendijck
Jul 14 at 16:30

Hint: $X|Y=y$ is uniformly distributed on $[1,e^y]$.
– Stan Tendijck
Jul 14 at 16:30

add a comment | 

2 Answers
2

active

oldest

votes

up vote
4
down vote

You don't need $f_Y$ nor $f_XY$ here because $f(x|y)$ is given straightfowardly.

You know that $Xsim mathcalU[1,e^Y]$, what (being $Y$ a r.v.) gives actually the conditional density of $X$ given $Y$. In particular, what we know is that if you are aware of the fact that $Y$ gets the value $y$, then $Xsim mathcalU[1,e^y]$.

On the other hand, if you want $f(y|x)$, then you should calculate both $f_X$ and $f_XY$.

share|cite|improve this answer

edited Jul 26 at 13:43

answered Jul 14 at 16:32

Alejandro Nasif Salum

3,07617

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, then $f(x|y)= begincases 0, & x le 1 \ x, & 1 < x le e^Y \ 1, & x > e^Y. endcases$ no? but i don't sure of this, can you review this? thanks!
  – Bvss12
  Jul 14 at 16:42
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  No, first of all I think you're confusing the Cumulative Distribution Function (CDF) with the Probability Density Function (PDF). That is, $F_X$ and $f_X$. So the density is $$f(x|y)= begincases k & 1 le x le e^y \0 & otherwise,\endcases$$ where $k$ is such that guarantee this integrates to $1$.
  – Alejandro Nasif Salum
  Jul 14 at 17:05
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Oh, thanks... sorry. thanks for your answer.
  – Bvss12
  Jul 14 at 17:07
  

  
  
  
  

add a comment | 

up vote
0
down vote

$F_y (y) = max(0, min(x, 1))$ by definition of uniform continuous distribution.

The random vector $(X, Y)$ is uniformly distributed on $ 0 leq y leq 1, 1 leq x leq e^y$, that means $$F(x, y) =
begincases
1 & quad textif 1 leq y text and e^y leq x\
e^y - y - 1 & quad textif 0 leq y leq 1 text and e^y leq x\
x(1 - ln x) + (x-1)min(1,y) - 1 & quad textif 1 leq x leq e^y text and 0 leq y\
0 & quad textif x leq 1 text or y leq 0\
endcases $$
This formula comes from the fact that $P(Xleq x, Y leq y) = fracmu( 0 leq z leq min(1, y), 1 leq t leq min(e^z, x))mu((z, t))$

share|cite|improve this answer

answered Jul 14 at 17:06

Yanior Weg

1,0761730

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

up vote
4
down vote

You don't need $f_Y$ nor $f_XY$ here because $f(x|y)$ is given straightfowardly.

You know that $Xsim mathcalU[1,e^Y]$, what (being $Y$ a r.v.) gives actually the conditional density of $X$ given $Y$. In particular, what we know is that if you are aware of the fact that $Y$ gets the value $y$, then $Xsim mathcalU[1,e^y]$.

On the other hand, if you want $f(y|x)$, then you should calculate both $f_X$ and $f_XY$.

share|cite|improve this answer

edited Jul 26 at 13:43

answered Jul 14 at 16:32

Alejandro Nasif Salum

3,07617

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, then $f(x|y)= begincases 0, & x le 1 \ x, & 1 < x le e^Y \ 1, & x > e^Y. endcases$ no? but i don't sure of this, can you review this? thanks!
  – Bvss12
  Jul 14 at 16:42
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  No, first of all I think you're confusing the Cumulative Distribution Function (CDF) with the Probability Density Function (PDF). That is, $F_X$ and $f_X$. So the density is $$f(x|y)= begincases k & 1 le x le e^y \0 & otherwise,\endcases$$ where $k$ is such that guarantee this integrates to $1$.
  – Alejandro Nasif Salum
  Jul 14 at 17:05
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Oh, thanks... sorry. thanks for your answer.
  – Bvss12
  Jul 14 at 17:07
  

  
  
  
  

add a comment | 

up vote
4
down vote

You don't need $f_Y$ nor $f_XY$ here because $f(x|y)$ is given straightfowardly.

You know that $Xsim mathcalU[1,e^Y]$, what (being $Y$ a r.v.) gives actually the conditional density of $X$ given $Y$. In particular, what we know is that if you are aware of the fact that $Y$ gets the value $y$, then $Xsim mathcalU[1,e^y]$.

On the other hand, if you want $f(y|x)$, then you should calculate both $f_X$ and $f_XY$.

share|cite|improve this answer

edited Jul 26 at 13:43

answered Jul 14 at 16:32

Alejandro Nasif Salum

3,07617

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, then $f(x|y)= begincases 0, & x le 1 \ x, & 1 < x le e^Y \ 1, & x > e^Y. endcases$ no? but i don't sure of this, can you review this? thanks!
  – Bvss12
  Jul 14 at 16:42
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  No, first of all I think you're confusing the Cumulative Distribution Function (CDF) with the Probability Density Function (PDF). That is, $F_X$ and $f_X$. So the density is $$f(x|y)= begincases k & 1 le x le e^y \0 & otherwise,\endcases$$ where $k$ is such that guarantee this integrates to $1$.
  – Alejandro Nasif Salum
  Jul 14 at 17:05
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Oh, thanks... sorry. thanks for your answer.
  – Bvss12
  Jul 14 at 17:07
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

You don't need $f_Y$ nor $f_XY$ here because $f(x|y)$ is given straightfowardly.

You know that $Xsim mathcalU[1,e^Y]$, what (being $Y$ a r.v.) gives actually the conditional density of $X$ given $Y$. In particular, what we know is that if you are aware of the fact that $Y$ gets the value $y$, then $Xsim mathcalU[1,e^y]$.

On the other hand, if you want $f(y|x)$, then you should calculate both $f_X$ and $f_XY$.

share|cite|improve this answer

edited Jul 26 at 13:43

answered Jul 14 at 16:32

Alejandro Nasif Salum

3,07617

You don't need $f_Y$ nor $f_XY$ here because $f(x|y)$ is given straightfowardly.

You know that $Xsim mathcalU[1,e^Y]$, what (being $Y$ a r.v.) gives actually the conditional density of $X$ given $Y$. In particular, what we know is that if you are aware of the fact that $Y$ gets the value $y$, then $Xsim mathcalU[1,e^y]$.

On the other hand, if you want $f(y|x)$, then you should calculate both $f_X$ and $f_XY$.

share|cite|improve this answer

edited Jul 26 at 13:43

answered Jul 14 at 16:32

Alejandro Nasif Salum

3,07617

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 13:43

edited Jul 26 at 13:43

edited Jul 26 at 13:43

answered Jul 14 at 16:32

Alejandro Nasif Salum

3,07617

answered Jul 14 at 16:32

Alejandro Nasif Salum

3,07617

answered Jul 14 at 16:32

Alejandro Nasif Salum

3,07617

3,07617

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, then $f(x|y)= begincases 0, & x le 1 \ x, & 1 < x le e^Y \ 1, & x > e^Y. endcases$ no? but i don't sure of this, can you review this? thanks!
  – Bvss12
  Jul 14 at 16:42
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  No, first of all I think you're confusing the Cumulative Distribution Function (CDF) with the Probability Density Function (PDF). That is, $F_X$ and $f_X$. So the density is $$f(x|y)= begincases k & 1 le x le e^y \0 & otherwise,\endcases$$ where $k$ is such that guarantee this integrates to $1$.
  – Alejandro Nasif Salum
  Jul 14 at 17:05
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Oh, thanks... sorry. thanks for your answer.
  – Bvss12
  Jul 14 at 17:07
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, then $f(x|y)= begincases 0, & x le 1 \ x, & 1 < x le e^Y \ 1, & x > e^Y. endcases$ no? but i don't sure of this, can you review this? thanks!
  – Bvss12
  Jul 14 at 16:42
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  No, first of all I think you're confusing the Cumulative Distribution Function (CDF) with the Probability Density Function (PDF). That is, $F_X$ and $f_X$. So the density is $$f(x|y)= begincases k & 1 le x le e^y \0 & otherwise,\endcases$$ where $k$ is such that guarantee this integrates to $1$.
  – Alejandro Nasif Salum
  Jul 14 at 17:05
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Oh, thanks... sorry. thanks for your answer.
  – Bvss12
  Jul 14 at 17:07
  

  
  
  
  

Okay, then $f(x|y)= begincases 0, & x le 1 \ x, & 1 < x le e^Y \ 1, & x > e^Y. endcases$ no? but i don't sure of this, can you review this? thanks!
– Bvss12
Jul 14 at 16:42

Okay, then $f(x|y)= begincases 0, & x le 1 \ x, & 1 < x le e^Y \ 1, & x > e^Y. endcases$ no? but i don't sure of this, can you review this? thanks!
– Bvss12
Jul 14 at 16:42

1

1

No, first of all I think you're confusing the Cumulative Distribution Function (CDF) with the Probability Density Function (PDF). That is, $F_X$ and $f_X$. So the density is $$f(x|y)= begincases k & 1 le x le e^y \0 & otherwise,\endcases$$ where $k$ is such that guarantee this integrates to $1$.
– Alejandro Nasif Salum
Jul 14 at 17:05

No, first of all I think you're confusing the Cumulative Distribution Function (CDF) with the Probability Density Function (PDF). That is, $F_X$ and $f_X$. So the density is $$f(x|y)= begincases k & 1 le x le e^y \0 & otherwise,\endcases$$ where $k$ is such that guarantee this integrates to $1$.
– Alejandro Nasif Salum
Jul 14 at 17:05

2

2

Oh, thanks... sorry. thanks for your answer.
– Bvss12
Jul 14 at 17:07

Oh, thanks... sorry. thanks for your answer.
– Bvss12
Jul 14 at 17:07

add a comment | 

up vote
0
down vote

$F_y (y) = max(0, min(x, 1))$ by definition of uniform continuous distribution.

The random vector $(X, Y)$ is uniformly distributed on $ 0 leq y leq 1, 1 leq x leq e^y$, that means $$F(x, y) =
begincases
1 & quad textif 1 leq y text and e^y leq x\
e^y - y - 1 & quad textif 0 leq y leq 1 text and e^y leq x\
x(1 - ln x) + (x-1)min(1,y) - 1 & quad textif 1 leq x leq e^y text and 0 leq y\
0 & quad textif x leq 1 text or y leq 0\
endcases $$
This formula comes from the fact that $P(Xleq x, Y leq y) = fracmu( 0 leq z leq min(1, y), 1 leq t leq min(e^z, x))mu((z, t))$

share|cite|improve this answer

answered Jul 14 at 17:06

Yanior Weg

1,0761730

add a comment | 

up vote
0
down vote

$F_y (y) = max(0, min(x, 1))$ by definition of uniform continuous distribution.

The random vector $(X, Y)$ is uniformly distributed on $ 0 leq y leq 1, 1 leq x leq e^y$, that means $$F(x, y) =
begincases
1 & quad textif 1 leq y text and e^y leq x\
e^y - y - 1 & quad textif 0 leq y leq 1 text and e^y leq x\
x(1 - ln x) + (x-1)min(1,y) - 1 & quad textif 1 leq x leq e^y text and 0 leq y\
0 & quad textif x leq 1 text or y leq 0\
endcases $$
This formula comes from the fact that $P(Xleq x, Y leq y) = fracmu( 0 leq z leq min(1, y), 1 leq t leq min(e^z, x))mu((z, t))$

share|cite|improve this answer

answered Jul 14 at 17:06

Yanior Weg

1,0761730

add a comment | 

up vote
0
down vote

up vote
0
down vote

$F_y (y) = max(0, min(x, 1))$ by definition of uniform continuous distribution.

The random vector $(X, Y)$ is uniformly distributed on $ 0 leq y leq 1, 1 leq x leq e^y$, that means $$F(x, y) =
begincases
1 & quad textif 1 leq y text and e^y leq x\
e^y - y - 1 & quad textif 0 leq y leq 1 text and e^y leq x\
x(1 - ln x) + (x-1)min(1,y) - 1 & quad textif 1 leq x leq e^y text and 0 leq y\
0 & quad textif x leq 1 text or y leq 0\
endcases $$
This formula comes from the fact that $P(Xleq x, Y leq y) = fracmu( 0 leq z leq min(1, y), 1 leq t leq min(e^z, x))mu((z, t))$

share|cite|improve this answer

answered Jul 14 at 17:06

Yanior Weg

1,0761730

$F_y (y) = max(0, min(x, 1))$ by definition of uniform continuous distribution.

The random vector $(X, Y)$ is uniformly distributed on $ 0 leq y leq 1, 1 leq x leq e^y$, that means $$F(x, y) =
begincases
1 & quad textif 1 leq y text and e^y leq x\
e^y - y - 1 & quad textif 0 leq y leq 1 text and e^y leq x\
x(1 - ln x) + (x-1)min(1,y) - 1 & quad textif 1 leq x leq e^y text and 0 leq y\
0 & quad textif x leq 1 text or y leq 0\
endcases $$
This formula comes from the fact that $P(Xleq x, Y leq y) = fracmu( 0 leq z leq min(1, y), 1 leq t leq min(e^z, x))mu((z, t))$

share|cite|improve this answer

answered Jul 14 at 17:06

Yanior Weg

1,0761730

share|cite|improve this answer

share|cite|improve this answer

answered Jul 14 at 17:06

Yanior Weg

1,0761730

answered Jul 14 at 17:06

Yanior Weg

1,0761730

answered Jul 14 at 17:06

Yanior Weg

1,0761730

1,0761730

add a comment | 

add a comment | 

 

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