How to get a PDF from other two PDFs

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

At a bus stop, suppose the interval between two buses (bus headway) is $x$, PDF of $x$ is $f(x)$. The in-stop waiting time for a passenger is $y$. For a bus interval $x$, suppose the arrival of passengers satisfies uniform distribution with PDF:

$$
g(y|x) =
begincases
1/x, & textif $0 le yle x$\
0, & textotherwise
endcases.
$$

To derive the PDF of $y$ without the intermedia $x$, there is
$$p(y) = int_0^+inftyg(y|x)f(x)dx
= int_0^xfrac1xf(x)dx $$

Here is the problem. $p(y)$ should be a function of $y$, but why the rightmost of the equation above is only a function of $x$, not $y$?

probability probability-theory probability-distributions

share|cite|improve this question

edited Jul 30 at 2:59

asked Jul 29 at 8:10

Macer

174

• 
  
  
  

  
  

  
  
  
  
  
  

  
  en.wikipedia.org/wiki/Bayes%27_theorem#Extended_form_2
  – Did
  Jul 29 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you help me with the updated question? thanks.
  – Macer
  Jul 30 at 3:01
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

At a bus stop, suppose the interval between two buses (bus headway) is $x$, PDF of $x$ is $f(x)$. The in-stop waiting time for a passenger is $y$. For a bus interval $x$, suppose the arrival of passengers satisfies uniform distribution with PDF:

$$
g(y|x) =
begincases
1/x, & textif $0 le yle x$\
0, & textotherwise
endcases.
$$

To derive the PDF of $y$ without the intermedia $x$, there is
$$p(y) = int_0^+inftyg(y|x)f(x)dx
= int_0^xfrac1xf(x)dx $$

Here is the problem. $p(y)$ should be a function of $y$, but why the rightmost of the equation above is only a function of $x$, not $y$?

probability probability-theory probability-distributions

share|cite|improve this question

edited Jul 30 at 2:59

asked Jul 29 at 8:10

Macer

174

• 
  
  
  

  
  

  
  
  
  
  
  

  
  en.wikipedia.org/wiki/Bayes%27_theorem#Extended_form_2
  – Did
  Jul 29 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you help me with the updated question? thanks.
  – Macer
  Jul 30 at 3:01
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

At a bus stop, suppose the interval between two buses (bus headway) is $x$, PDF of $x$ is $f(x)$. The in-stop waiting time for a passenger is $y$. For a bus interval $x$, suppose the arrival of passengers satisfies uniform distribution with PDF:

$$
g(y|x) =
begincases
1/x, & textif $0 le yle x$\
0, & textotherwise
endcases.
$$

To derive the PDF of $y$ without the intermedia $x$, there is
$$p(y) = int_0^+inftyg(y|x)f(x)dx
= int_0^xfrac1xf(x)dx $$

Here is the problem. $p(y)$ should be a function of $y$, but why the rightmost of the equation above is only a function of $x$, not $y$?

probability probability-theory probability-distributions

share|cite|improve this question

edited Jul 30 at 2:59

asked Jul 29 at 8:10

Macer

174

At a bus stop, suppose the interval between two buses (bus headway) is $x$, PDF of $x$ is $f(x)$. The in-stop waiting time for a passenger is $y$. For a bus interval $x$, suppose the arrival of passengers satisfies uniform distribution with PDF:

$$
g(y|x) =
begincases
1/x, & textif $0 le yle x$\
0, & textotherwise
endcases.
$$

To derive the PDF of $y$ without the intermedia $x$, there is
$$p(y) = int_0^+inftyg(y|x)f(x)dx
= int_0^xfrac1xf(x)dx $$

Here is the problem. $p(y)$ should be a function of $y$, but why the rightmost of the equation above is only a function of $x$, not $y$?

probability probability-theory probability-distributions

share|cite|improve this question

edited Jul 30 at 2:59

asked Jul 29 at 8:10

Macer

174

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 2:59

edited Jul 30 at 2:59

edited Jul 30 at 2:59

asked Jul 29 at 8:10

Macer

174

asked Jul 29 at 8:10

Macer

174

asked Jul 29 at 8:10

Macer

174

174

• 
  
  
  

  
  

  
  
  
  
  
  

  
  en.wikipedia.org/wiki/Bayes%27_theorem#Extended_form_2
  – Did
  Jul 29 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you help me with the updated question? thanks.
  – Macer
  Jul 30 at 3:01
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  en.wikipedia.org/wiki/Bayes%27_theorem#Extended_form_2
  – Did
  Jul 29 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you help me with the updated question? thanks.
  – Macer
  Jul 30 at 3:01
  

  
  
  
  

en.wikipedia.org/wiki/Bayes%27_theorem#Extended_form_2
– Did
Jul 29 at 8:23

en.wikipedia.org/wiki/Bayes%27_theorem#Extended_form_2
– Did
Jul 29 at 8:23

Can you help me with the updated question? thanks.
– Macer
Jul 30 at 3:01

Can you help me with the updated question? thanks.
– Macer
Jul 30 at 3:01

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

Yes, it should be a function of $y$, so clearly you have the bounds muddled up.   $x$ should not appear there.

Look to the support for $g(ymid x)$, it is $0leq yleq x$, so when "integrating out" the $x$, that leaves: $0leq y$ as the support, and the integral is over $x$ from $y$ to infinity.

$$p(y)~=int_0^infty left(frac 1xmathbf 1_0leq yleq xright) f(x)mathsf d x\= mathbf 1_0leq y int_y^infty fracf(x)x,mathsf d x$$

Of course, also the support for $f(x)$ itself needs to be considered.

share|cite|improve this answer

answered Jul 30 at 3:10

Graham Kemp

80k43275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you, but I don't really understand how to manipulate form the first line of your equation to the second line.
  – Macer
  Jul 30 at 3:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are integrating $x$ over the support $0leq yleq xlt infty$ for a particular value of $y$, which is non-negative. The integral's variable, $x$, clearly 'lives' between that $y$ and infinity. $$int_Bbb R h(x,y)~mathbf 1_0leq yleq xlt infty~mathrm d x ~ =~ int_Bbb Rmathbf 1_yleq xltinfty~mathbf 1_0leq ylt infty~h(x,y)~mathsf d x \ =~mathbf 1_0leq ylt infty~int_yleq xltinfty~h(x,y)~mathsf d x$$
  – Graham Kemp
  Jul 30 at 4:27
  
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865883%2fhow-to-get-a-pdf-from-other-two-pdfs%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Yes, it should be a function of $y$, so clearly you have the bounds muddled up.   $x$ should not appear there.

Look to the support for $g(ymid x)$, it is $0leq yleq x$, so when "integrating out" the $x$, that leaves: $0leq y$ as the support, and the integral is over $x$ from $y$ to infinity.

$$p(y)~=int_0^infty left(frac 1xmathbf 1_0leq yleq xright) f(x)mathsf d x\= mathbf 1_0leq y int_y^infty fracf(x)x,mathsf d x$$

Of course, also the support for $f(x)$ itself needs to be considered.

share|cite|improve this answer

answered Jul 30 at 3:10

Graham Kemp

80k43275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you, but I don't really understand how to manipulate form the first line of your equation to the second line.
  – Macer
  Jul 30 at 3:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are integrating $x$ over the support $0leq yleq xlt infty$ for a particular value of $y$, which is non-negative. The integral's variable, $x$, clearly 'lives' between that $y$ and infinity. $$int_Bbb R h(x,y)~mathbf 1_0leq yleq xlt infty~mathrm d x ~ =~ int_Bbb Rmathbf 1_yleq xltinfty~mathbf 1_0leq ylt infty~h(x,y)~mathsf d x \ =~mathbf 1_0leq ylt infty~int_yleq xltinfty~h(x,y)~mathsf d x$$
  – Graham Kemp
  Jul 30 at 4:27
  
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

Yes, it should be a function of $y$, so clearly you have the bounds muddled up.   $x$ should not appear there.

Look to the support for $g(ymid x)$, it is $0leq yleq x$, so when "integrating out" the $x$, that leaves: $0leq y$ as the support, and the integral is over $x$ from $y$ to infinity.

$$p(y)~=int_0^infty left(frac 1xmathbf 1_0leq yleq xright) f(x)mathsf d x\= mathbf 1_0leq y int_y^infty fracf(x)x,mathsf d x$$

Of course, also the support for $f(x)$ itself needs to be considered.

share|cite|improve this answer

answered Jul 30 at 3:10

Graham Kemp

80k43275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you, but I don't really understand how to manipulate form the first line of your equation to the second line.
  – Macer
  Jul 30 at 3:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are integrating $x$ over the support $0leq yleq xlt infty$ for a particular value of $y$, which is non-negative. The integral's variable, $x$, clearly 'lives' between that $y$ and infinity. $$int_Bbb R h(x,y)~mathbf 1_0leq yleq xlt infty~mathrm d x ~ =~ int_Bbb Rmathbf 1_yleq xltinfty~mathbf 1_0leq ylt infty~h(x,y)~mathsf d x \ =~mathbf 1_0leq ylt infty~int_yleq xltinfty~h(x,y)~mathsf d x$$
  – Graham Kemp
  Jul 30 at 4:27
  
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Yes, it should be a function of $y$, so clearly you have the bounds muddled up.   $x$ should not appear there.

Look to the support for $g(ymid x)$, it is $0leq yleq x$, so when "integrating out" the $x$, that leaves: $0leq y$ as the support, and the integral is over $x$ from $y$ to infinity.

$$p(y)~=int_0^infty left(frac 1xmathbf 1_0leq yleq xright) f(x)mathsf d x\= mathbf 1_0leq y int_y^infty fracf(x)x,mathsf d x$$

Of course, also the support for $f(x)$ itself needs to be considered.

share|cite|improve this answer

answered Jul 30 at 3:10

Graham Kemp

80k43275

Yes, it should be a function of $y$, so clearly you have the bounds muddled up.   $x$ should not appear there.

Look to the support for $g(ymid x)$, it is $0leq yleq x$, so when "integrating out" the $x$, that leaves: $0leq y$ as the support, and the integral is over $x$ from $y$ to infinity.

$$p(y)~=int_0^infty left(frac 1xmathbf 1_0leq yleq xright) f(x)mathsf d x\= mathbf 1_0leq y int_y^infty fracf(x)x,mathsf d x$$

Of course, also the support for $f(x)$ itself needs to be considered.

share|cite|improve this answer

answered Jul 30 at 3:10

Graham Kemp

80k43275

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 3:10

Graham Kemp

80k43275

answered Jul 30 at 3:10

Graham Kemp

80k43275

answered Jul 30 at 3:10

Graham Kemp

80k43275

80k43275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you, but I don't really understand how to manipulate form the first line of your equation to the second line.
  – Macer
  Jul 30 at 3:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are integrating $x$ over the support $0leq yleq xlt infty$ for a particular value of $y$, which is non-negative. The integral's variable, $x$, clearly 'lives' between that $y$ and infinity. $$int_Bbb R h(x,y)~mathbf 1_0leq yleq xlt infty~mathrm d x ~ =~ int_Bbb Rmathbf 1_yleq xltinfty~mathbf 1_0leq ylt infty~h(x,y)~mathsf d x \ =~mathbf 1_0leq ylt infty~int_yleq xltinfty~h(x,y)~mathsf d x$$
  – Graham Kemp
  Jul 30 at 4:27
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you, but I don't really understand how to manipulate form the first line of your equation to the second line.
  – Macer
  Jul 30 at 3:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are integrating $x$ over the support $0leq yleq xlt infty$ for a particular value of $y$, which is non-negative. The integral's variable, $x$, clearly 'lives' between that $y$ and infinity. $$int_Bbb R h(x,y)~mathbf 1_0leq yleq xlt infty~mathrm d x ~ =~ int_Bbb Rmathbf 1_yleq xltinfty~mathbf 1_0leq ylt infty~h(x,y)~mathsf d x \ =~mathbf 1_0leq ylt infty~int_yleq xltinfty~h(x,y)~mathsf d x$$
  – Graham Kemp
  Jul 30 at 4:27
  
  

  
  
  
  

Thank you, but I don't really understand how to manipulate form the first line of your equation to the second line.
– Macer
Jul 30 at 3:59

Thank you, but I don't really understand how to manipulate form the first line of your equation to the second line.
– Macer
Jul 30 at 3:59

You are integrating $x$ over the support $0leq yleq xlt infty$ for a particular value of $y$, which is non-negative. The integral's variable, $x$, clearly 'lives' between that $y$ and infinity. $$int_Bbb R h(x,y)~mathbf 1_0leq yleq xlt infty~mathrm d x ~ =~ int_Bbb Rmathbf 1_yleq xltinfty~mathbf 1_0leq ylt infty~h(x,y)~mathsf d x \ =~mathbf 1_0leq ylt infty~int_yleq xltinfty~h(x,y)~mathsf d x$$
– Graham Kemp
Jul 30 at 4:27

You are integrating $x$ over the support $0leq yleq xlt infty$ for a particular value of $y$, which is non-negative. The integral's variable, $x$, clearly 'lives' between that $y$ and infinity. $$int_Bbb R h(x,y)~mathbf 1_0leq yleq xlt infty~mathrm d x ~ =~ int_Bbb Rmathbf 1_yleq xltinfty~mathbf 1_0leq ylt infty~h(x,y)~mathsf d x \ =~mathbf 1_0leq ylt infty~int_yleq xltinfty~h(x,y)~mathsf d x$$
– Graham Kemp
Jul 30 at 4:27

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865883%2fhow-to-get-a-pdf-from-other-two-pdfs%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email