Random Variable X and Y has a joint probability density function.

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Random Variable X and Y has a joint probability density function.

$$f_X, Y (x, y) =begincases
cxy& 0 leq x leq y, 0 leq y leq5\
0 & textotherwise \
endcases
$$

(a) Find $f_ X(y | x)$

(b) $P(Y leq 4 | X = 3)$

---

$(a)$

$$f_X(x) = int_0^5 cxy dy = cx/2(y^2)bigg|_0^5 = fraccx252$$ with support $0 leq x leq y$

$$f_ X(y | x) = fraccxycx25/2 = frac2y25$$

$(b)$

$$f_Y, X(y | X=3) = frac2y25$$

$$int_0^4 f_Y, X(y | X=3) dy = int_0^4 frac2y25 dy = frac250left(y^2 right)bigg|_0^4 = frac1625$$

Apparently, this is wrong, why so?

probability probability-distributions

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asked Jul 19 at 22:34

Bas bas

39611

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up vote
0
down vote

favorite

Random Variable X and Y has a joint probability density function.

$$f_X, Y (x, y) =begincases
cxy& 0 leq x leq y, 0 leq y leq5\
0 & textotherwise \
endcases
$$

(a) Find $f_ X(y | x)$

(b) $P(Y leq 4 | X = 3)$

---

$(a)$

$$f_X(x) = int_0^5 cxy dy = cx/2(y^2)bigg|_0^5 = fraccx252$$ with support $0 leq x leq y$

$$f_ X(y | x) = fraccxycx25/2 = frac2y25$$

$(b)$

$$f_Y, X(y | X=3) = frac2y25$$

$$int_0^4 f_Y, X(y | X=3) dy = int_0^4 frac2y25 dy = frac250left(y^2 right)bigg|_0^4 = frac1625$$

Apparently, this is wrong, why so?

probability probability-distributions

share|cite|improve this question

asked Jul 19 at 22:34

Bas bas

39611

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Random Variable X and Y has a joint probability density function.

$$f_X, Y (x, y) =begincases
cxy& 0 leq x leq y, 0 leq y leq5\
0 & textotherwise \
endcases
$$

(a) Find $f_ X(y | x)$

(b) $P(Y leq 4 | X = 3)$

---

$(a)$

$$f_X(x) = int_0^5 cxy dy = cx/2(y^2)bigg|_0^5 = fraccx252$$ with support $0 leq x leq y$

$$f_ X(y | x) = fraccxycx25/2 = frac2y25$$

$(b)$

$$f_Y, X(y | X=3) = frac2y25$$

$$int_0^4 f_Y, X(y | X=3) dy = int_0^4 frac2y25 dy = frac250left(y^2 right)bigg|_0^4 = frac1625$$

Apparently, this is wrong, why so?

probability probability-distributions

share|cite|improve this question

asked Jul 19 at 22:34

Bas bas

39611

Random Variable X and Y has a joint probability density function.

$$f_X, Y (x, y) =begincases
cxy& 0 leq x leq y, 0 leq y leq5\
0 & textotherwise \
endcases
$$

(a) Find $f_ X(y | x)$

(b) $P(Y leq 4 | X = 3)$

---

$(a)$

$$f_X(x) = int_0^5 cxy dy = cx/2(y^2)bigg|_0^5 = fraccx252$$ with support $0 leq x leq y$

$$f_ X(y | x) = fraccxycx25/2 = frac2y25$$

$(b)$

$$f_Y, X(y | X=3) = frac2y25$$

$$int_0^4 f_Y, X(y | X=3) dy = int_0^4 frac2y25 dy = frac250left(y^2 right)bigg|_0^4 = frac1625$$

Apparently, this is wrong, why so?

probability probability-distributions

share|cite|improve this question

asked Jul 19 at 22:34

Bas bas

39611

share|cite|improve this question

share|cite|improve this question

asked Jul 19 at 22:34

Bas bas

39611

asked Jul 19 at 22:34

Bas bas

39611

asked Jul 19 at 22:34

Bas bas

39611

39611

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

The pdf only has support for $y>x,$ so actually $$ f_X(x) = int_x^5cxy;dy.$$

share|cite|improve this answer

answered Jul 19 at 22:46

spaceisdarkgreen

27.5k21547

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also do the same for (b). @Basbas
  – Graham Kemp
  Jul 19 at 22:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so, x to 4 for b?
  – Bas bas
  Jul 19 at 22:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Basbas No. What is the support of $f(ymid X=3)?$
  – spaceisdarkgreen
  Jul 19 at 22:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh. So, 3 to 4.
  – Bas bas
  Jul 19 at 23:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Basbas yes, the support of $f(ymid x)$ is $(x,5).$
  – spaceisdarkgreen
  Jul 19 at 23:09
  
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

The pdf only has support for $y>x,$ so actually $$ f_X(x) = int_x^5cxy;dy.$$

share|cite|improve this answer

answered Jul 19 at 22:46

spaceisdarkgreen

27.5k21547

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also do the same for (b). @Basbas
  – Graham Kemp
  Jul 19 at 22:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so, x to 4 for b?
  – Bas bas
  Jul 19 at 22:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Basbas No. What is the support of $f(ymid X=3)?$
  – spaceisdarkgreen
  Jul 19 at 22:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh. So, 3 to 4.
  – Bas bas
  Jul 19 at 23:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Basbas yes, the support of $f(ymid x)$ is $(x,5).$
  – spaceisdarkgreen
  Jul 19 at 23:09
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

The pdf only has support for $y>x,$ so actually $$ f_X(x) = int_x^5cxy;dy.$$

share|cite|improve this answer

answered Jul 19 at 22:46

spaceisdarkgreen

27.5k21547

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also do the same for (b). @Basbas
  – Graham Kemp
  Jul 19 at 22:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so, x to 4 for b?
  – Bas bas
  Jul 19 at 22:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Basbas No. What is the support of $f(ymid X=3)?$
  – spaceisdarkgreen
  Jul 19 at 22:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh. So, 3 to 4.
  – Bas bas
  Jul 19 at 23:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Basbas yes, the support of $f(ymid x)$ is $(x,5).$
  – spaceisdarkgreen
  Jul 19 at 23:09
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

The pdf only has support for $y>x,$ so actually $$ f_X(x) = int_x^5cxy;dy.$$

share|cite|improve this answer

answered Jul 19 at 22:46

spaceisdarkgreen

27.5k21547

The pdf only has support for $y>x,$ so actually $$ f_X(x) = int_x^5cxy;dy.$$

share|cite|improve this answer

answered Jul 19 at 22:46

spaceisdarkgreen

27.5k21547

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 22:46

spaceisdarkgreen

27.5k21547

answered Jul 19 at 22:46

spaceisdarkgreen

27.5k21547

answered Jul 19 at 22:46

spaceisdarkgreen

27.5k21547

27.5k21547

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also do the same for (b). @Basbas
  – Graham Kemp
  Jul 19 at 22:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so, x to 4 for b?
  – Bas bas
  Jul 19 at 22:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Basbas No. What is the support of $f(ymid X=3)?$
  – spaceisdarkgreen
  Jul 19 at 22:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh. So, 3 to 4.
  – Bas bas
  Jul 19 at 23:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Basbas yes, the support of $f(ymid x)$ is $(x,5).$
  – spaceisdarkgreen
  Jul 19 at 23:09
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also do the same for (b). @Basbas
  – Graham Kemp
  Jul 19 at 22:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so, x to 4 for b?
  – Bas bas
  Jul 19 at 22:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Basbas No. What is the support of $f(ymid X=3)?$
  – spaceisdarkgreen
  Jul 19 at 22:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh. So, 3 to 4.
  – Bas bas
  Jul 19 at 23:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Basbas yes, the support of $f(ymid x)$ is $(x,5).$
  – spaceisdarkgreen
  Jul 19 at 23:09
  
  

  
  
  
  

Also do the same for (b). @Basbas
– Graham Kemp
Jul 19 at 22:51

Also do the same for (b). @Basbas
– Graham Kemp
Jul 19 at 22:51

so, x to 4 for b?
– Bas bas
Jul 19 at 22:55

so, x to 4 for b?
– Bas bas
Jul 19 at 22:55

@Basbas No. What is the support of $f(ymid X=3)?$
– spaceisdarkgreen
Jul 19 at 22:59

@Basbas No. What is the support of $f(ymid X=3)?$
– spaceisdarkgreen
Jul 19 at 22:59

Oh. So, 3 to 4.
– Bas bas
Jul 19 at 23:02

Oh. So, 3 to 4.
– Bas bas
Jul 19 at 23:02

@Basbas yes, the support of $f(ymid x)$ is $(x,5).$
– spaceisdarkgreen
Jul 19 at 23:09

@Basbas yes, the support of $f(ymid x)$ is $(x,5).$
– spaceisdarkgreen
Jul 19 at 23:09

add a comment | 

 

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