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Calculate $f_X(x)$ and $f_Y(y)$ given a pair $(X,Y)$ of continuous random variables with a joint PDF of…

Joint PDF of random variablesJoint pdf of discrete and continuous random variablesGiven Joint PDF, find conditional probabilityFind continuous stochastic variable $X$ with PDF $f_X = frac1x^2$How to find the joint distribution and joint density functions of two random variables?Finding $pdf$ of $Y=frac1X$ given $pdf$ of $X$ is $f_X(x) = frac24x^4I_(2,∞)(x)$Sums of Continuous Random Variables using ConvolutionsExpected value given joint PDFFinding joint density function of two independent random variablesFinding the PDF of a function if X is exponentially distributed with a given parameter $lambda$ = 3

0

$begingroup$

Calculate $f_X(x)$ and $f_Y(y)$ given a pair (X,Y) of continuous random variables with a joint PDF of:

$f(x,y)=$
begincases
3 & 0leq x leq 1 & 0leq y leq x^2 \
0 & textotherwise
endcases

This problem was given to me as a review for an upcoming exam.

My current workings:

$f_X(x) = int_-infty^infty f(x,y) dy$

$f_Y(y) = int_-infty^infty f(x,y) dx$

I'm not exactly sure how to use the f(x,y) in the integral. For $f_X(x)$ do I plug in $x^2$ into the integral and 1 for $f_Y(y)$? If someone can point in the correct direction on what to integrate I should be able to continue from there.

Updated attempt:

$f_X(x) = int_0^x^2 3 dy = 3x^2$

$f_Y(y) = int_sqrty^1 3 dx = 3-3sqrty$

$f_X(x) =$
begincases
3x^2 & 0leq x leq 1 \
0 & textotherwise
endcases

$f_Y(y) =$
begincases
3-3sqrty & 0leq y leq 1 \
0 & textotherwise
endcases

probability statistics

share|cite|improve this question

edited Mar 21 at 15:01

YuiTo Cheng

2,1863937

asked Mar 20 at 17:55

JoeJoe

596

$endgroup$

add a comment | 

0

$begingroup$

Calculate $f_X(x)$ and $f_Y(y)$ given a pair (X,Y) of continuous random variables with a joint PDF of:

$f(x,y)=$
begincases
3 & 0leq x leq 1 & 0leq y leq x^2 \
0 & textotherwise
endcases

This problem was given to me as a review for an upcoming exam.

My current workings:

$f_X(x) = int_-infty^infty f(x,y) dy$

$f_Y(y) = int_-infty^infty f(x,y) dx$

I'm not exactly sure how to use the f(x,y) in the integral. For $f_X(x)$ do I plug in $x^2$ into the integral and 1 for $f_Y(y)$? If someone can point in the correct direction on what to integrate I should be able to continue from there.

Updated attempt:

$f_X(x) = int_0^x^2 3 dy = 3x^2$

$f_Y(y) = int_sqrty^1 3 dx = 3-3sqrty$

$f_X(x) =$
begincases
3x^2 & 0leq x leq 1 \
0 & textotherwise
endcases

$f_Y(y) =$
begincases
3-3sqrty & 0leq y leq 1 \
0 & textotherwise
endcases

probability statistics

share|cite|improve this question

edited Mar 21 at 15:01

YuiTo Cheng

2,1863937

asked Mar 20 at 17:55

JoeJoe

596

$endgroup$

add a comment | 

$begingroup$

Calculate $f_X(x)$ and $f_Y(y)$ given a pair (X,Y) of continuous random variables with a joint PDF of:

$f(x,y)=$
begincases
3 & 0leq x leq 1 & 0leq y leq x^2 \
0 & textotherwise
endcases

This problem was given to me as a review for an upcoming exam.

My current workings:

$f_X(x) = int_-infty^infty f(x,y) dy$

$f_Y(y) = int_-infty^infty f(x,y) dx$

I'm not exactly sure how to use the f(x,y) in the integral. For $f_X(x)$ do I plug in $x^2$ into the integral and 1 for $f_Y(y)$? If someone can point in the correct direction on what to integrate I should be able to continue from there.

Updated attempt:

$f_X(x) = int_0^x^2 3 dy = 3x^2$

$f_Y(y) = int_sqrty^1 3 dx = 3-3sqrty$

$f_X(x) =$
begincases
3x^2 & 0leq x leq 1 \
0 & textotherwise
endcases

$f_Y(y) =$
begincases
3-3sqrty & 0leq y leq 1 \
0 & textotherwise
endcases

probability statistics

share|cite|improve this question

edited Mar 21 at 15:01

YuiTo Cheng

2,1863937

asked Mar 20 at 17:55

JoeJoe

596

$endgroup$

share|cite|improve this question

edited Mar 21 at 15:01

YuiTo Cheng

2,1863937

asked Mar 20 at 17:55

JoeJoe

596

share|cite|improve this question

edited Mar 21 at 15:01

YuiTo Cheng

2,1863937

asked Mar 20 at 17:55

JoeJoe

596

edited Mar 21 at 15:01

YuiTo Cheng

2,1863937

edited Mar 21 at 15:01

YuiTo Cheng

2,1863937

asked Mar 20 at 17:55

JoeJoe

596

asked Mar 20 at 17:55

JoeJoe

596

2 Answers
2

active

oldest

votes

2

$begingroup$

Hint

according the the definition of $f_XY(x,y)$ we obtain $$f_X(x)=int_y=0^y=x^2f_XY(x,y)dy$$and$$f_Y(y)=int_x=sqrt y^x=1f_XY(x,y)dy$$

share|cite|improve this answer

answered Mar 20 at 18:27

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

add a comment | 

0

$begingroup$

For $x<0$ or $x>1$, it's clear that $f_X=0$ because for such $x$, $f(x,y)=0$ for all $y$. For $0leq xleq 1$, we have:
$$
f_X(x)=int_-infty^infty f(x,y)dy=int_0^x^2 3dy=3x^2.
$$
You can do something similar for $f_Y$.

share|cite|improve this answer

answered Mar 20 at 18:06

yurneroyurnero

7,4291926

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Once I solve the integrals, is the domain for each of the functions basically given? So $0 leq xleq 1$ and $0 leq yleq 1$ (Because x can only be from 0 to 1, so the maximum value for y would be $1^2=1$) I've updated my attempt to demonstrate my comment.
  $endgroup$
  – Joe
  Mar 20 at 18:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Joe The domain is the real line. The support (where the density is non zero) is an interval.
  $endgroup$
  – yurnero
  Mar 20 at 19:24
  
  

  
  
  
  

add a comment | 

Your Answer

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2

$begingroup$

Hint

according the the definition of $f_XY(x,y)$ we obtain $$f_X(x)=int_y=0^y=x^2f_XY(x,y)dy$$and$$f_Y(y)=int_x=sqrt y^x=1f_XY(x,y)dy$$

share|cite|improve this answer

answered Mar 20 at 18:27

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

add a comment | 

2

$begingroup$

Hint

according the the definition of $f_XY(x,y)$ we obtain $$f_X(x)=int_y=0^y=x^2f_XY(x,y)dy$$and$$f_Y(y)=int_x=sqrt y^x=1f_XY(x,y)dy$$

share|cite|improve this answer

answered Mar 20 at 18:27

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

add a comment | 

$begingroup$

Hint

according the the definition of $f_XY(x,y)$ we obtain $$f_X(x)=int_y=0^y=x^2f_XY(x,y)dy$$and$$f_Y(y)=int_x=sqrt y^x=1f_XY(x,y)dy$$

share|cite|improve this answer

answered Mar 20 at 18:27

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

share|cite|improve this answer

answered Mar 20 at 18:27

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 20 at 18:27

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 20 at 18:27

Mostafa AyazMostafa Ayaz

18.2k31040

0

$begingroup$

For $x<0$ or $x>1$, it's clear that $f_X=0$ because for such $x$, $f(x,y)=0$ for all $y$. For $0leq xleq 1$, we have:
$$
f_X(x)=int_-infty^infty f(x,y)dy=int_0^x^2 3dy=3x^2.
$$
You can do something similar for $f_Y$.

share|cite|improve this answer

answered Mar 20 at 18:06

yurneroyurnero

7,4291926

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Once I solve the integrals, is the domain for each of the functions basically given? So $0 leq xleq 1$ and $0 leq yleq 1$ (Because x can only be from 0 to 1, so the maximum value for y would be $1^2=1$) I've updated my attempt to demonstrate my comment.
  $endgroup$
  – Joe
  Mar 20 at 18:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Joe The domain is the real line. The support (where the density is non zero) is an interval.
  $endgroup$
  – yurnero
  Mar 20 at 19:24
  
  

  
  
  
  

add a comment | 

0

$begingroup$

For $x<0$ or $x>1$, it's clear that $f_X=0$ because for such $x$, $f(x,y)=0$ for all $y$. For $0leq xleq 1$, we have:
$$
f_X(x)=int_-infty^infty f(x,y)dy=int_0^x^2 3dy=3x^2.
$$
You can do something similar for $f_Y$.

share|cite|improve this answer

answered Mar 20 at 18:06

yurneroyurnero

7,4291926

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Once I solve the integrals, is the domain for each of the functions basically given? So $0 leq xleq 1$ and $0 leq yleq 1$ (Because x can only be from 0 to 1, so the maximum value for y would be $1^2=1$) I've updated my attempt to demonstrate my comment.
  $endgroup$
  – Joe
  Mar 20 at 18:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Joe The domain is the real line. The support (where the density is non zero) is an interval.
  $endgroup$
  – yurnero
  Mar 20 at 19:24
  
  

  
  
  
  

add a comment | 

$begingroup$

For $x<0$ or $x>1$, it's clear that $f_X=0$ because for such $x$, $f(x,y)=0$ for all $y$. For $0leq xleq 1$, we have:
$$
f_X(x)=int_-infty^infty f(x,y)dy=int_0^x^2 3dy=3x^2.
$$
You can do something similar for $f_Y$.

share|cite|improve this answer

answered Mar 20 at 18:06

yurneroyurnero

7,4291926

$endgroup$

share|cite|improve this answer

answered Mar 20 at 18:06

yurneroyurnero

7,4291926

answered Mar 20 at 18:06

yurneroyurnero

7,4291926

answered Mar 20 at 18:06

yurneroyurnero

7,4291926