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Find the conditional probability density function of $Y$ given $X=x$

Probability Density Function - GammaFinding quantiles based on probability density functionsProbability density function of random variable $X-Y$Find the conditional probability of a probability density functionConditional Distribution from Conditional densityFind the marginal and conditional densities without explicitly having the joint density?Find the conditional probability density function of $X$ given $Y=y$Computing probability function of $Y$ in terms of $f_X$How can I get the probability density function?Compute the probability density function of Y

0

$begingroup$

Suppose $Y$ is a continuous random variable with probability density function $f(y)=192over y^4$, for $ygeq 4$ ($0$ otherwise). If the conditional distribution of X given $Y = y$ is a uniform distribution on $[0, y]$. Find the conditional probability density function of $Y$ given $X = x$.

My clue is that:
$$f_Y(x|y)=f_X,Y(x,y)overf_Y(y)=1over y$$
then we get $$f_X,Y(x,y)=192over y^5$$
$$f_Y(y|x)=f_X,Y(x,y)overf_X(x)$$
But I don't know how to find the domain of $Y$ in terms of $X$ to get $f_X(x)$.

Am I on the right track?

probability

share|cite|improve this question

edited Mar 14 at 6:15

Rócherz

2,9863821

asked Oct 1 '18 at 18:06

Yibei HeYibei He

3139

$endgroup$

add a comment | 

0

$begingroup$

Suppose $Y$ is a continuous random variable with probability density function $f(y)=192over y^4$, for $ygeq 4$ ($0$ otherwise). If the conditional distribution of X given $Y = y$ is a uniform distribution on $[0, y]$. Find the conditional probability density function of $Y$ given $X = x$.

My clue is that:
$$f_Y(x|y)=f_X,Y(x,y)overf_Y(y)=1over y$$
then we get $$f_X,Y(x,y)=192over y^5$$
$$f_Y(y|x)=f_X,Y(x,y)overf_X(x)$$
But I don't know how to find the domain of $Y$ in terms of $X$ to get $f_X(x)$.

Am I on the right track?

probability

share|cite|improve this question

edited Mar 14 at 6:15

Rócherz

2,9863821

asked Oct 1 '18 at 18:06

Yibei HeYibei He

3139

$endgroup$

add a comment | 

$begingroup$

Suppose $Y$ is a continuous random variable with probability density function $f(y)=192over y^4$, for $ygeq 4$ ($0$ otherwise). If the conditional distribution of X given $Y = y$ is a uniform distribution on $[0, y]$. Find the conditional probability density function of $Y$ given $X = x$.

My clue is that:
$$f_Y(x|y)=f_X,Y(x,y)overf_Y(y)=1over y$$
then we get $$f_X,Y(x,y)=192over y^5$$
$$f_Y(y|x)=f_X,Y(x,y)overf_X(x)$$
But I don't know how to find the domain of $Y$ in terms of $X$ to get $f_X(x)$.

Am I on the right track?

probability

share|cite|improve this question

edited Mar 14 at 6:15

Rócherz

2,9863821

asked Oct 1 '18 at 18:06

Yibei HeYibei He

3139

$endgroup$

share|cite|improve this question

edited Mar 14 at 6:15

Rócherz

2,9863821

asked Oct 1 '18 at 18:06

Yibei HeYibei He

3139

share|cite|improve this question

edited Mar 14 at 6:15

Rócherz

2,9863821

asked Oct 1 '18 at 18:06

Yibei HeYibei He

3139

edited Mar 14 at 6:15

Rócherz

2,9863821

edited Mar 14 at 6:15

Rócherz

2,9863821

asked Oct 1 '18 at 18:06

Yibei HeYibei He

3139

asked Oct 1 '18 at 18:06

Yibei HeYibei He

3139

1 Answer
1

active

oldest

votes

1

$begingroup$

Note that the expression for the joint density $f_X,Y(x,y)$ that you computed only holds for $x$ and $y$ satisfying $x ge 0$ and $y ge maxx, 4$; it is zero otherwise. Thus for $x ge 0$ we have
$$f_X(x) = int_-infty^infty f_X,Y(x,z) , dz = int_maxx, 4^infty frac192z^5 , dz.$$

---

Continuing the computation leads to
$$f_X(x) = 48 maxx, 4^-4 = begincases48 / 4^4 & x in [0, 4] \ 48 / x^4 & x > 4endcases$$
So,
$$int_0^infty f_X(x) , dx = 4 cdot frac484^4 + int_4^infty frac48x^4 , dx
= frac34 + frac164^3 = 1.$$

share|cite|improve this answer

edited Oct 2 '18 at 2:51

answered Oct 1 '18 at 18:10

angryavianangryavian

42.2k23481

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So when you integral the joint pdf from x to ∞, you assumed that x is greater than 4, right? Otherwise it will be from 4 to ∞. But does it contradict with 0≤x?
  $endgroup$
  – Yibei He
  Oct 1 '18 at 20:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YibeiHe Good catch; I've edited my answer.
  $endgroup$
  – angryavian
  Oct 1 '18 at 22:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm still quite confused about the domain. Do you mean $f_X(x)$ will have two case, one is x=[0,4] and the other is x=[4,∞]? If I get $f_X(x)$ in this way, when I integral $f_X(x)$ in both case, I can't get the summation to 1. How does that happen?
  $endgroup$
  – Yibei He
  Oct 2 '18 at 2:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YibeiHe See my edit.
  $endgroup$
  – angryavian
  Oct 2 '18 at 2:51
  

  
  
  
  

add a comment | 

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1

$begingroup$

Note that the expression for the joint density $f_X,Y(x,y)$ that you computed only holds for $x$ and $y$ satisfying $x ge 0$ and $y ge maxx, 4$; it is zero otherwise. Thus for $x ge 0$ we have
$$f_X(x) = int_-infty^infty f_X,Y(x,z) , dz = int_maxx, 4^infty frac192z^5 , dz.$$

---

Continuing the computation leads to
$$f_X(x) = 48 maxx, 4^-4 = begincases48 / 4^4 & x in [0, 4] \ 48 / x^4 & x > 4endcases$$
So,
$$int_0^infty f_X(x) , dx = 4 cdot frac484^4 + int_4^infty frac48x^4 , dx
= frac34 + frac164^3 = 1.$$

share|cite|improve this answer

edited Oct 2 '18 at 2:51

answered Oct 1 '18 at 18:10

angryavianangryavian

42.2k23481

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So when you integral the joint pdf from x to ∞, you assumed that x is greater than 4, right? Otherwise it will be from 4 to ∞. But does it contradict with 0≤x?
  $endgroup$
  – Yibei He
  Oct 1 '18 at 20:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YibeiHe Good catch; I've edited my answer.
  $endgroup$
  – angryavian
  Oct 1 '18 at 22:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm still quite confused about the domain. Do you mean $f_X(x)$ will have two case, one is x=[0,4] and the other is x=[4,∞]? If I get $f_X(x)$ in this way, when I integral $f_X(x)$ in both case, I can't get the summation to 1. How does that happen?
  $endgroup$
  – Yibei He
  Oct 2 '18 at 2:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YibeiHe See my edit.
  $endgroup$
  – angryavian
  Oct 2 '18 at 2:51
  

  
  
  
  

add a comment | 

1

$begingroup$

Note that the expression for the joint density $f_X,Y(x,y)$ that you computed only holds for $x$ and $y$ satisfying $x ge 0$ and $y ge maxx, 4$; it is zero otherwise. Thus for $x ge 0$ we have
$$f_X(x) = int_-infty^infty f_X,Y(x,z) , dz = int_maxx, 4^infty frac192z^5 , dz.$$

---

Continuing the computation leads to
$$f_X(x) = 48 maxx, 4^-4 = begincases48 / 4^4 & x in [0, 4] \ 48 / x^4 & x > 4endcases$$
So,
$$int_0^infty f_X(x) , dx = 4 cdot frac484^4 + int_4^infty frac48x^4 , dx
= frac34 + frac164^3 = 1.$$

share|cite|improve this answer

edited Oct 2 '18 at 2:51

answered Oct 1 '18 at 18:10

angryavianangryavian

42.2k23481

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So when you integral the joint pdf from x to ∞, you assumed that x is greater than 4, right? Otherwise it will be from 4 to ∞. But does it contradict with 0≤x?
  $endgroup$
  – Yibei He
  Oct 1 '18 at 20:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YibeiHe Good catch; I've edited my answer.
  $endgroup$
  – angryavian
  Oct 1 '18 at 22:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm still quite confused about the domain. Do you mean $f_X(x)$ will have two case, one is x=[0,4] and the other is x=[4,∞]? If I get $f_X(x)$ in this way, when I integral $f_X(x)$ in both case, I can't get the summation to 1. How does that happen?
  $endgroup$
  – Yibei He
  Oct 2 '18 at 2:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YibeiHe See my edit.
  $endgroup$
  – angryavian
  Oct 2 '18 at 2:51
  

  
  
  
  

add a comment | 

$begingroup$

Note that the expression for the joint density $f_X,Y(x,y)$ that you computed only holds for $x$ and $y$ satisfying $x ge 0$ and $y ge maxx, 4$; it is zero otherwise. Thus for $x ge 0$ we have
$$f_X(x) = int_-infty^infty f_X,Y(x,z) , dz = int_maxx, 4^infty frac192z^5 , dz.$$

---

Continuing the computation leads to
$$f_X(x) = 48 maxx, 4^-4 = begincases48 / 4^4 & x in [0, 4] \ 48 / x^4 & x > 4endcases$$
So,
$$int_0^infty f_X(x) , dx = 4 cdot frac484^4 + int_4^infty frac48x^4 , dx
= frac34 + frac164^3 = 1.$$

share|cite|improve this answer

edited Oct 2 '18 at 2:51

answered Oct 1 '18 at 18:10

angryavianangryavian

42.2k23481

$endgroup$

share|cite|improve this answer

edited Oct 2 '18 at 2:51

answered Oct 1 '18 at 18:10

angryavianangryavian

42.2k23481

answered Oct 1 '18 at 18:10

angryavianangryavian

42.2k23481

answered Oct 1 '18 at 18:10

angryavianangryavian

42.2k23481