The problem for conditional expectation

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The joint probability density function for random variables $X$, $Y$ is given by
$$f(x, y)=begincases
2(x+y) & textif 0<x<y<1 \ 0 & textotherwise
&
endcases.$$
When the conditional expectation of $X$ is $E(X | Y=aX)=frac29$, what is the real number $a$?

I'm struggle with this problem. Any help please.

probability expectation conditional-expectation

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edited Jul 30 at 2:46

Math Lover

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up vote
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down vote

favorite

The joint probability density function for random variables $X$, $Y$ is given by
$$f(x, y)=begincases
2(x+y) & textif 0<x<y<1 \ 0 & textotherwise
&
endcases.$$
When the conditional expectation of $X$ is $E(X | Y=aX)=frac29$, what is the real number $a$?

I'm struggle with this problem. Any help please.

probability expectation conditional-expectation

share|cite|improve this question

edited Jul 30 at 2:46

Math Lover

12.3k21232

asked Jul 30 at 2:40

purecj

495

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

The joint probability density function for random variables $X$, $Y$ is given by
$$f(x, y)=begincases
2(x+y) & textif 0<x<y<1 \ 0 & textotherwise
&
endcases.$$
When the conditional expectation of $X$ is $E(X | Y=aX)=frac29$, what is the real number $a$?

I'm struggle with this problem. Any help please.

probability expectation conditional-expectation

share|cite|improve this question

edited Jul 30 at 2:46

Math Lover

12.3k21232

asked Jul 30 at 2:40

purecj

495

The joint probability density function for random variables $X$, $Y$ is given by
$$f(x, y)=begincases
2(x+y) & textif 0<x<y<1 \ 0 & textotherwise
&
endcases.$$
When the conditional expectation of $X$ is $E(X | Y=aX)=frac29$, what is the real number $a$?

I'm struggle with this problem. Any help please.

probability expectation conditional-expectation

share|cite|improve this question

edited Jul 30 at 2:46

Math Lover

12.3k21232

asked Jul 30 at 2:40

purecj

495

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 2:46

Math Lover

12.3k21232

edited Jul 30 at 2:46

Math Lover

12.3k21232

edited Jul 30 at 2:46

Math Lover

12.3k21232

12.3k21232

asked Jul 30 at 2:40

purecj

495

asked Jul 30 at 2:40

purecj

495

asked Jul 30 at 2:40

purecj

495

495

add a comment | 

add a comment | 

2 Answers
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Hint: For $a > 1$, the conditional density of $X$ given $Y = aX$ is $f_X mid Y=aX(x) := fracf_X,Y(x, ax)int_0^1/a f_X,Y(t, at) , dt$ for $x in (0,1/a)$.

Continuing from above, $$f_X mid Y=aX(x) := fracf_X,Y(x, ax)int_0^1/a f_X,Y(t, at) , dt= fracx+axint_0^1/a (t+at) , dt = frac(1+a)x(1+a)/(2a^2) = 2a^2 x.$$ Now that we have the conditional density, the desired conditional expectation is $$int_0^1/a x (2 a^2 x) , dx = frac23a.$$

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answered Jul 30 at 2:54

angryavian

34.5k12874

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you....^^
  – purecj
  Jul 30 at 3:22
  

  
  
  
  

add a comment | 

up vote
1
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By definition, $$beginalignmathsf E(Xmid Y=aX) ~&=~dfracdisplaystyleint_Bbb R x, f(x,ax),mathsf d xdisplaystyleint_Bbb R f(x,ax),mathsf d x\[2ex]~&=~dfracdisplaystyleint_0^1/a 2(1+a)x^2,mathsf d xdisplaystyleint_0^1/a 2(1+a)x,mathsf d x&&textif ageq 1endalign$$

share|cite|improve this answer

edited Jul 30 at 2:50

Math Lover

12.3k21232

answered Jul 30 at 2:47

Graham Kemp

80k43275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you....^^
  – purecj
  Jul 30 at 3:22
  

  
  
  
  

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
2
down vote

Hint: For $a > 1$, the conditional density of $X$ given $Y = aX$ is $f_X mid Y=aX(x) := fracf_X,Y(x, ax)int_0^1/a f_X,Y(t, at) , dt$ for $x in (0,1/a)$.

Continuing from above, $$f_X mid Y=aX(x) := fracf_X,Y(x, ax)int_0^1/a f_X,Y(t, at) , dt= fracx+axint_0^1/a (t+at) , dt = frac(1+a)x(1+a)/(2a^2) = 2a^2 x.$$ Now that we have the conditional density, the desired conditional expectation is $$int_0^1/a x (2 a^2 x) , dx = frac23a.$$

share|cite|improve this answer

answered Jul 30 at 2:54

angryavian

34.5k12874

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you....^^
  – purecj
  Jul 30 at 3:22
  

  
  
  
  

add a comment | 

up vote
2
down vote

Hint: For $a > 1$, the conditional density of $X$ given $Y = aX$ is $f_X mid Y=aX(x) := fracf_X,Y(x, ax)int_0^1/a f_X,Y(t, at) , dt$ for $x in (0,1/a)$.

Continuing from above, $$f_X mid Y=aX(x) := fracf_X,Y(x, ax)int_0^1/a f_X,Y(t, at) , dt= fracx+axint_0^1/a (t+at) , dt = frac(1+a)x(1+a)/(2a^2) = 2a^2 x.$$ Now that we have the conditional density, the desired conditional expectation is $$int_0^1/a x (2 a^2 x) , dx = frac23a.$$

share|cite|improve this answer

answered Jul 30 at 2:54

angryavian

34.5k12874

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you....^^
  – purecj
  Jul 30 at 3:22
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

Hint: For $a > 1$, the conditional density of $X$ given $Y = aX$ is $f_X mid Y=aX(x) := fracf_X,Y(x, ax)int_0^1/a f_X,Y(t, at) , dt$ for $x in (0,1/a)$.

Continuing from above, $$f_X mid Y=aX(x) := fracf_X,Y(x, ax)int_0^1/a f_X,Y(t, at) , dt= fracx+axint_0^1/a (t+at) , dt = frac(1+a)x(1+a)/(2a^2) = 2a^2 x.$$ Now that we have the conditional density, the desired conditional expectation is $$int_0^1/a x (2 a^2 x) , dx = frac23a.$$

share|cite|improve this answer

answered Jul 30 at 2:54

angryavian

34.5k12874

Hint: For $a > 1$, the conditional density of $X$ given $Y = aX$ is $f_X mid Y=aX(x) := fracf_X,Y(x, ax)int_0^1/a f_X,Y(t, at) , dt$ for $x in (0,1/a)$.

Continuing from above, $$f_X mid Y=aX(x) := fracf_X,Y(x, ax)int_0^1/a f_X,Y(t, at) , dt= fracx+axint_0^1/a (t+at) , dt = frac(1+a)x(1+a)/(2a^2) = 2a^2 x.$$ Now that we have the conditional density, the desired conditional expectation is $$int_0^1/a x (2 a^2 x) , dx = frac23a.$$

share|cite|improve this answer

answered Jul 30 at 2:54

angryavian

34.5k12874

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 2:54

angryavian

34.5k12874

answered Jul 30 at 2:54

angryavian

34.5k12874

answered Jul 30 at 2:54

angryavian

34.5k12874

34.5k12874

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you....^^
  – purecj
  Jul 30 at 3:22
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you....^^
  – purecj
  Jul 30 at 3:22
  

  
  
  
  

Thank you....^^
– purecj
Jul 30 at 3:22

Thank you....^^
– purecj
Jul 30 at 3:22

add a comment | 

up vote
1
down vote

By definition, $$beginalignmathsf E(Xmid Y=aX) ~&=~dfracdisplaystyleint_Bbb R x, f(x,ax),mathsf d xdisplaystyleint_Bbb R f(x,ax),mathsf d x\[2ex]~&=~dfracdisplaystyleint_0^1/a 2(1+a)x^2,mathsf d xdisplaystyleint_0^1/a 2(1+a)x,mathsf d x&&textif ageq 1endalign$$

share|cite|improve this answer

edited Jul 30 at 2:50

Math Lover

12.3k21232

answered Jul 30 at 2:47

Graham Kemp

80k43275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you....^^
  – purecj
  Jul 30 at 3:22
  

  
  
  
  

add a comment | 

up vote
1
down vote

By definition, $$beginalignmathsf E(Xmid Y=aX) ~&=~dfracdisplaystyleint_Bbb R x, f(x,ax),mathsf d xdisplaystyleint_Bbb R f(x,ax),mathsf d x\[2ex]~&=~dfracdisplaystyleint_0^1/a 2(1+a)x^2,mathsf d xdisplaystyleint_0^1/a 2(1+a)x,mathsf d x&&textif ageq 1endalign$$

share|cite|improve this answer

edited Jul 30 at 2:50

Math Lover

12.3k21232

answered Jul 30 at 2:47

Graham Kemp

80k43275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you....^^
  – purecj
  Jul 30 at 3:22
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

By definition, $$beginalignmathsf E(Xmid Y=aX) ~&=~dfracdisplaystyleint_Bbb R x, f(x,ax),mathsf d xdisplaystyleint_Bbb R f(x,ax),mathsf d x\[2ex]~&=~dfracdisplaystyleint_0^1/a 2(1+a)x^2,mathsf d xdisplaystyleint_0^1/a 2(1+a)x,mathsf d x&&textif ageq 1endalign$$

share|cite|improve this answer

edited Jul 30 at 2:50

Math Lover

12.3k21232

answered Jul 30 at 2:47

Graham Kemp

80k43275

By definition, $$beginalignmathsf E(Xmid Y=aX) ~&=~dfracdisplaystyleint_Bbb R x, f(x,ax),mathsf d xdisplaystyleint_Bbb R f(x,ax),mathsf d x\[2ex]~&=~dfracdisplaystyleint_0^1/a 2(1+a)x^2,mathsf d xdisplaystyleint_0^1/a 2(1+a)x,mathsf d x&&textif ageq 1endalign$$

share|cite|improve this answer

edited Jul 30 at 2:50

Math Lover

12.3k21232

answered Jul 30 at 2:47

Graham Kemp

80k43275

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 2:50

Math Lover

12.3k21232

edited Jul 30 at 2:50

Math Lover

12.3k21232

edited Jul 30 at 2:50

Math Lover

12.3k21232

12.3k21232

answered Jul 30 at 2:47

Graham Kemp

80k43275

answered Jul 30 at 2:47

Graham Kemp

80k43275

answered Jul 30 at 2:47

Graham Kemp

80k43275

80k43275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you....^^
  – purecj
  Jul 30 at 3:22
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you....^^
  – purecj
  Jul 30 at 3:22
  

  
  
  
  

Thank you....^^
– purecj
Jul 30 at 3:22

Thank you....^^
– purecj
Jul 30 at 3:22

add a comment | 

 

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