Bayesian Probability question — Pointwise Probability












1












$begingroup$


I am stuck in this question:



if $a = 1$ then $m sim U(0.2,1)$ else if $a=0$ then $m sim U(0,0.5)$. The question is if $m$ is $0.3$ what is the probability that $a$ equals to 1?



My thought is to compute $p(a=1mid m=0.3)$ and $p(a=0mid m=0.3)$ and whichever class gives the higher probability then it is the answer. However, when I am executing this thought I have a problem of computing $p(m=0.3mid a=1)$ which is supposed to be zero since it follows $U(0.2,1)$. I feel like I can use the density function to compute this probability but I am not sure why?






probability bayesian







share|cite|improve this question











$endgroup$












  • $begingroup$
    Big hint: Bayes Rule. Think of the probability you want to compute, and the probabilities you already know.
    $endgroup$
    – Cliff AB
    8 hours ago










  • $begingroup$
    As I mentioned, I am already implementing the Bayes rule. but I have to compute p(m=0.3|a=1) which follows U(0.2,1). In other words p(u=m=0.3) which is zero (because it is a continuous pdf). But I know it should be not zero
    $endgroup$
    – prony
    8 hours ago












  • $begingroup$
    oh sorry, I get your question now. I'll write up an answer.
    $endgroup$
    – Cliff AB
    7 hours ago
















1












$begingroup$


I am stuck in this question:



if $a = 1$ then $m sim U(0.2,1)$ else if $a=0$ then $m sim U(0,0.5)$. The question is if $m$ is $0.3$ what is the probability that $a$ equals to 1?



My thought is to compute $p(a=1mid m=0.3)$ and $p(a=0mid m=0.3)$ and whichever class gives the higher probability then it is the answer. However, when I am executing this thought I have a problem of computing $p(m=0.3mid a=1)$ which is supposed to be zero since it follows $U(0.2,1)$. I feel like I can use the density function to compute this probability but I am not sure why?






probability bayesian







share|cite|improve this question











$endgroup$












  • $begingroup$
    Big hint: Bayes Rule. Think of the probability you want to compute, and the probabilities you already know.
    $endgroup$
    – Cliff AB
    8 hours ago










  • $begingroup$
    As I mentioned, I am already implementing the Bayes rule. but I have to compute p(m=0.3|a=1) which follows U(0.2,1). In other words p(u=m=0.3) which is zero (because it is a continuous pdf). But I know it should be not zero
    $endgroup$
    – prony
    8 hours ago












  • $begingroup$
    oh sorry, I get your question now. I'll write up an answer.
    $endgroup$
    – Cliff AB
    7 hours ago














1












1








1





$begingroup$


I am stuck in this question:



if $a = 1$ then $m sim U(0.2,1)$ else if $a=0$ then $m sim U(0,0.5)$. The question is if $m$ is $0.3$ what is the probability that $a$ equals to 1?



My thought is to compute $p(a=1mid m=0.3)$ and $p(a=0mid m=0.3)$ and whichever class gives the higher probability then it is the answer. However, when I am executing this thought I have a problem of computing $p(m=0.3mid a=1)$ which is supposed to be zero since it follows $U(0.2,1)$. I feel like I can use the density function to compute this probability but I am not sure why?






probability bayesian







share|cite|improve this question











$endgroup$




I am stuck in this question:



if $a = 1$ then $m sim U(0.2,1)$ else if $a=0$ then $m sim U(0,0.5)$. The question is if $m$ is $0.3$ what is the probability that $a$ equals to 1?



My thought is to compute $p(a=1mid m=0.3)$ and $p(a=0mid m=0.3)$ and whichever class gives the higher probability then it is the answer. However, when I am executing this thought I have a problem of computing $p(m=0.3mid a=1)$ which is supposed to be zero since it follows $U(0.2,1)$. I feel like I can use the density function to compute this probability but I am not sure why?






probability bayesian




probability bayesian






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









Michael Hardy

3,7901430




3,7901430










asked 8 hours ago









pronyprony

225




225












  • $begingroup$
    Big hint: Bayes Rule. Think of the probability you want to compute, and the probabilities you already know.
    $endgroup$
    – Cliff AB
    8 hours ago










  • $begingroup$
    As I mentioned, I am already implementing the Bayes rule. but I have to compute p(m=0.3|a=1) which follows U(0.2,1). In other words p(u=m=0.3) which is zero (because it is a continuous pdf). But I know it should be not zero
    $endgroup$
    – prony
    8 hours ago












  • $begingroup$
    oh sorry, I get your question now. I'll write up an answer.
    $endgroup$
    – Cliff AB
    7 hours ago


















  • $begingroup$
    Big hint: Bayes Rule. Think of the probability you want to compute, and the probabilities you already know.
    $endgroup$
    – Cliff AB
    8 hours ago










  • $begingroup$
    As I mentioned, I am already implementing the Bayes rule. but I have to compute p(m=0.3|a=1) which follows U(0.2,1). In other words p(u=m=0.3) which is zero (because it is a continuous pdf). But I know it should be not zero
    $endgroup$
    – prony
    8 hours ago












  • $begingroup$
    oh sorry, I get your question now. I'll write up an answer.
    $endgroup$
    – Cliff AB
    7 hours ago
















$begingroup$
Big hint: Bayes Rule. Think of the probability you want to compute, and the probabilities you already know.
$endgroup$
– Cliff AB
8 hours ago




$begingroup$
Big hint: Bayes Rule. Think of the probability you want to compute, and the probabilities you already know.
$endgroup$
– Cliff AB
8 hours ago












$begingroup$
As I mentioned, I am already implementing the Bayes rule. but I have to compute p(m=0.3|a=1) which follows U(0.2,1). In other words p(u=m=0.3) which is zero (because it is a continuous pdf). But I know it should be not zero
$endgroup$
– prony
8 hours ago






$begingroup$
As I mentioned, I am already implementing the Bayes rule. but I have to compute p(m=0.3|a=1) which follows U(0.2,1). In other words p(u=m=0.3) which is zero (because it is a continuous pdf). But I know it should be not zero
$endgroup$
– prony
8 hours ago














$begingroup$
oh sorry, I get your question now. I'll write up an answer.
$endgroup$
– Cliff AB
7 hours ago




$begingroup$
oh sorry, I get your question now. I'll write up an answer.
$endgroup$
– Cliff AB
7 hours ago










2 Answers
2






active

oldest

votes


















1












$begingroup$

If $a=1$ then the density function of $m$ is $displaystyle f_m(x) = begin{cases} 1/(1-0.2) = 1.25 & text{if } 0.2le xle 1, \ 0 & text{if } x<0.2 text{ or } x>1. end{cases}$



If $a=0,$ it is $displaystyle f_m(x) = begin{cases} 1/(0.5-0) = 2 & text{if } 0le xle0.5, \ 0 & text{if } x<0 text{ or } x>0.5. end{cases}$



Thus the likelihood function is
$$
begin{cases} L( 1 mid m=0.3) = 1.25, \ L(0 mid m=0.3) = 2. end{cases}
$$



Hence the posterior probability distribution is
$$
begin{cases} Pr(a=1mid m=0.3) = ctimes 1.25timesPr(a=1), \[5pt] Pr(a=0mid m = 0.3) = ctimes 2 times Pr(a=0). end{cases} tag 1
$$
The normalizing constant is
$$
c = frac 1 {1.25Pr(a=1) + 2Pr(a=0)}
$$
(that is what $c$ must be to make the sum of the two probabilities in line $(1)$ above equal to $1.$)



So for example, if $Pr(a=1)=Pr(a=0) = dfrac 1 2$ then $Pr(a=1mid m=0.3) = dfrac 5 {13}.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot for your answer. Can you elaborate on how you constructed the likelihood from $f_m(x)$. I mean how $P(x=0.3|a=1) = f_m(x=0.3|a=1)$ becomes equal to $L(1|m=0.3)$
    $endgroup$
    – prony
    5 hours ago








  • 1




    $begingroup$
    @prony : You have a density function $xmapsto f_n(x mid a=alpha),$ where $alphain{0,1},$ which is a function of $x,$ with $alpha$ fixed, and the likelihood function $alphamapsto L(alphamid m = x) = f_m(xmid a=alpha),$ which is a function of $alphain{0,1}$ with $x$ fixed at the observed value, which is $0.3. qquad$
    $endgroup$
    – Michael Hardy
    5 hours ago





















2












$begingroup$

Instead of Bayes rule, we should look closely at the definition of conditional probability. as Bayes rule is simply a small transformation of the definition of conditional probability.



With discrete probabilities, it is simple to define



$$P(A = a mid B = b) = frac{P(A = acap B = b)}{P(B = b)}$$



As you pointed out, this would be ill-defined if both $A$ and $B$ were continuous, as it would result $0/0.$



Instead, let's think about $P(A in a pm varepsilon mid B in b pm varepsilon)$ for a continuous distribution. Then the value



$$ frac{P(A in a pm varepsilon cap B in b pm varepsilon)}{P(B in b pm varepsilon) }$$



is properly defined for all $epsilon$, as long as $int_{b - varepsilon}^{b + varepsilon} f_b(x) , dx > 0 $. Finally, we just define the conditional distribution of $A|B$ as



$$lim_{varepsilon rightarrow 0} frac{P(A in a pm varepsilon cap B in b pm varepsilon) / varepsilon}{P(B in b pm varepsilon) / varepsilon } $$



By definition, this is



$$frac{ f_{A, B}(a, b) }{ f_B(b)}$$






share|cite|improve this answer











$endgroup$













    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: "65"
    };
    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',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f391519%2fbayesian-probability-question-pointwise-probability%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    If $a=1$ then the density function of $m$ is $displaystyle f_m(x) = begin{cases} 1/(1-0.2) = 1.25 & text{if } 0.2le xle 1, \ 0 & text{if } x<0.2 text{ or } x>1. end{cases}$



    If $a=0,$ it is $displaystyle f_m(x) = begin{cases} 1/(0.5-0) = 2 & text{if } 0le xle0.5, \ 0 & text{if } x<0 text{ or } x>0.5. end{cases}$



    Thus the likelihood function is
    $$
    begin{cases} L( 1 mid m=0.3) = 1.25, \ L(0 mid m=0.3) = 2. end{cases}
    $$



    Hence the posterior probability distribution is
    $$
    begin{cases} Pr(a=1mid m=0.3) = ctimes 1.25timesPr(a=1), \[5pt] Pr(a=0mid m = 0.3) = ctimes 2 times Pr(a=0). end{cases} tag 1
    $$
    The normalizing constant is
    $$
    c = frac 1 {1.25Pr(a=1) + 2Pr(a=0)}
    $$
    (that is what $c$ must be to make the sum of the two probabilities in line $(1)$ above equal to $1.$)



    So for example, if $Pr(a=1)=Pr(a=0) = dfrac 1 2$ then $Pr(a=1mid m=0.3) = dfrac 5 {13}.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks a lot for your answer. Can you elaborate on how you constructed the likelihood from $f_m(x)$. I mean how $P(x=0.3|a=1) = f_m(x=0.3|a=1)$ becomes equal to $L(1|m=0.3)$
      $endgroup$
      – prony
      5 hours ago








    • 1




      $begingroup$
      @prony : You have a density function $xmapsto f_n(x mid a=alpha),$ where $alphain{0,1},$ which is a function of $x,$ with $alpha$ fixed, and the likelihood function $alphamapsto L(alphamid m = x) = f_m(xmid a=alpha),$ which is a function of $alphain{0,1}$ with $x$ fixed at the observed value, which is $0.3. qquad$
      $endgroup$
      – Michael Hardy
      5 hours ago


















    1












    $begingroup$

    If $a=1$ then the density function of $m$ is $displaystyle f_m(x) = begin{cases} 1/(1-0.2) = 1.25 & text{if } 0.2le xle 1, \ 0 & text{if } x<0.2 text{ or } x>1. end{cases}$



    If $a=0,$ it is $displaystyle f_m(x) = begin{cases} 1/(0.5-0) = 2 & text{if } 0le xle0.5, \ 0 & text{if } x<0 text{ or } x>0.5. end{cases}$



    Thus the likelihood function is
    $$
    begin{cases} L( 1 mid m=0.3) = 1.25, \ L(0 mid m=0.3) = 2. end{cases}
    $$



    Hence the posterior probability distribution is
    $$
    begin{cases} Pr(a=1mid m=0.3) = ctimes 1.25timesPr(a=1), \[5pt] Pr(a=0mid m = 0.3) = ctimes 2 times Pr(a=0). end{cases} tag 1
    $$
    The normalizing constant is
    $$
    c = frac 1 {1.25Pr(a=1) + 2Pr(a=0)}
    $$
    (that is what $c$ must be to make the sum of the two probabilities in line $(1)$ above equal to $1.$)



    So for example, if $Pr(a=1)=Pr(a=0) = dfrac 1 2$ then $Pr(a=1mid m=0.3) = dfrac 5 {13}.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks a lot for your answer. Can you elaborate on how you constructed the likelihood from $f_m(x)$. I mean how $P(x=0.3|a=1) = f_m(x=0.3|a=1)$ becomes equal to $L(1|m=0.3)$
      $endgroup$
      – prony
      5 hours ago








    • 1




      $begingroup$
      @prony : You have a density function $xmapsto f_n(x mid a=alpha),$ where $alphain{0,1},$ which is a function of $x,$ with $alpha$ fixed, and the likelihood function $alphamapsto L(alphamid m = x) = f_m(xmid a=alpha),$ which is a function of $alphain{0,1}$ with $x$ fixed at the observed value, which is $0.3. qquad$
      $endgroup$
      – Michael Hardy
      5 hours ago
















    1












    1








    1





    $begingroup$

    If $a=1$ then the density function of $m$ is $displaystyle f_m(x) = begin{cases} 1/(1-0.2) = 1.25 & text{if } 0.2le xle 1, \ 0 & text{if } x<0.2 text{ or } x>1. end{cases}$



    If $a=0,$ it is $displaystyle f_m(x) = begin{cases} 1/(0.5-0) = 2 & text{if } 0le xle0.5, \ 0 & text{if } x<0 text{ or } x>0.5. end{cases}$



    Thus the likelihood function is
    $$
    begin{cases} L( 1 mid m=0.3) = 1.25, \ L(0 mid m=0.3) = 2. end{cases}
    $$



    Hence the posterior probability distribution is
    $$
    begin{cases} Pr(a=1mid m=0.3) = ctimes 1.25timesPr(a=1), \[5pt] Pr(a=0mid m = 0.3) = ctimes 2 times Pr(a=0). end{cases} tag 1
    $$
    The normalizing constant is
    $$
    c = frac 1 {1.25Pr(a=1) + 2Pr(a=0)}
    $$
    (that is what $c$ must be to make the sum of the two probabilities in line $(1)$ above equal to $1.$)



    So for example, if $Pr(a=1)=Pr(a=0) = dfrac 1 2$ then $Pr(a=1mid m=0.3) = dfrac 5 {13}.$






    share|cite|improve this answer









    $endgroup$



    If $a=1$ then the density function of $m$ is $displaystyle f_m(x) = begin{cases} 1/(1-0.2) = 1.25 & text{if } 0.2le xle 1, \ 0 & text{if } x<0.2 text{ or } x>1. end{cases}$



    If $a=0,$ it is $displaystyle f_m(x) = begin{cases} 1/(0.5-0) = 2 & text{if } 0le xle0.5, \ 0 & text{if } x<0 text{ or } x>0.5. end{cases}$



    Thus the likelihood function is
    $$
    begin{cases} L( 1 mid m=0.3) = 1.25, \ L(0 mid m=0.3) = 2. end{cases}
    $$



    Hence the posterior probability distribution is
    $$
    begin{cases} Pr(a=1mid m=0.3) = ctimes 1.25timesPr(a=1), \[5pt] Pr(a=0mid m = 0.3) = ctimes 2 times Pr(a=0). end{cases} tag 1
    $$
    The normalizing constant is
    $$
    c = frac 1 {1.25Pr(a=1) + 2Pr(a=0)}
    $$
    (that is what $c$ must be to make the sum of the two probabilities in line $(1)$ above equal to $1.$)



    So for example, if $Pr(a=1)=Pr(a=0) = dfrac 1 2$ then $Pr(a=1mid m=0.3) = dfrac 5 {13}.$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 6 hours ago









    Michael HardyMichael Hardy

    3,7901430




    3,7901430












    • $begingroup$
      Thanks a lot for your answer. Can you elaborate on how you constructed the likelihood from $f_m(x)$. I mean how $P(x=0.3|a=1) = f_m(x=0.3|a=1)$ becomes equal to $L(1|m=0.3)$
      $endgroup$
      – prony
      5 hours ago








    • 1




      $begingroup$
      @prony : You have a density function $xmapsto f_n(x mid a=alpha),$ where $alphain{0,1},$ which is a function of $x,$ with $alpha$ fixed, and the likelihood function $alphamapsto L(alphamid m = x) = f_m(xmid a=alpha),$ which is a function of $alphain{0,1}$ with $x$ fixed at the observed value, which is $0.3. qquad$
      $endgroup$
      – Michael Hardy
      5 hours ago




















    • $begingroup$
      Thanks a lot for your answer. Can you elaborate on how you constructed the likelihood from $f_m(x)$. I mean how $P(x=0.3|a=1) = f_m(x=0.3|a=1)$ becomes equal to $L(1|m=0.3)$
      $endgroup$
      – prony
      5 hours ago








    • 1




      $begingroup$
      @prony : You have a density function $xmapsto f_n(x mid a=alpha),$ where $alphain{0,1},$ which is a function of $x,$ with $alpha$ fixed, and the likelihood function $alphamapsto L(alphamid m = x) = f_m(xmid a=alpha),$ which is a function of $alphain{0,1}$ with $x$ fixed at the observed value, which is $0.3. qquad$
      $endgroup$
      – Michael Hardy
      5 hours ago


















    $begingroup$
    Thanks a lot for your answer. Can you elaborate on how you constructed the likelihood from $f_m(x)$. I mean how $P(x=0.3|a=1) = f_m(x=0.3|a=1)$ becomes equal to $L(1|m=0.3)$
    $endgroup$
    – prony
    5 hours ago






    $begingroup$
    Thanks a lot for your answer. Can you elaborate on how you constructed the likelihood from $f_m(x)$. I mean how $P(x=0.3|a=1) = f_m(x=0.3|a=1)$ becomes equal to $L(1|m=0.3)$
    $endgroup$
    – prony
    5 hours ago






    1




    1




    $begingroup$
    @prony : You have a density function $xmapsto f_n(x mid a=alpha),$ where $alphain{0,1},$ which is a function of $x,$ with $alpha$ fixed, and the likelihood function $alphamapsto L(alphamid m = x) = f_m(xmid a=alpha),$ which is a function of $alphain{0,1}$ with $x$ fixed at the observed value, which is $0.3. qquad$
    $endgroup$
    – Michael Hardy
    5 hours ago






    $begingroup$
    @prony : You have a density function $xmapsto f_n(x mid a=alpha),$ where $alphain{0,1},$ which is a function of $x,$ with $alpha$ fixed, and the likelihood function $alphamapsto L(alphamid m = x) = f_m(xmid a=alpha),$ which is a function of $alphain{0,1}$ with $x$ fixed at the observed value, which is $0.3. qquad$
    $endgroup$
    – Michael Hardy
    5 hours ago















    2












    $begingroup$

    Instead of Bayes rule, we should look closely at the definition of conditional probability. as Bayes rule is simply a small transformation of the definition of conditional probability.



    With discrete probabilities, it is simple to define



    $$P(A = a mid B = b) = frac{P(A = acap B = b)}{P(B = b)}$$



    As you pointed out, this would be ill-defined if both $A$ and $B$ were continuous, as it would result $0/0.$



    Instead, let's think about $P(A in a pm varepsilon mid B in b pm varepsilon)$ for a continuous distribution. Then the value



    $$ frac{P(A in a pm varepsilon cap B in b pm varepsilon)}{P(B in b pm varepsilon) }$$



    is properly defined for all $epsilon$, as long as $int_{b - varepsilon}^{b + varepsilon} f_b(x) , dx > 0 $. Finally, we just define the conditional distribution of $A|B$ as



    $$lim_{varepsilon rightarrow 0} frac{P(A in a pm varepsilon cap B in b pm varepsilon) / varepsilon}{P(B in b pm varepsilon) / varepsilon } $$



    By definition, this is



    $$frac{ f_{A, B}(a, b) }{ f_B(b)}$$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Instead of Bayes rule, we should look closely at the definition of conditional probability. as Bayes rule is simply a small transformation of the definition of conditional probability.



      With discrete probabilities, it is simple to define



      $$P(A = a mid B = b) = frac{P(A = acap B = b)}{P(B = b)}$$



      As you pointed out, this would be ill-defined if both $A$ and $B$ were continuous, as it would result $0/0.$



      Instead, let's think about $P(A in a pm varepsilon mid B in b pm varepsilon)$ for a continuous distribution. Then the value



      $$ frac{P(A in a pm varepsilon cap B in b pm varepsilon)}{P(B in b pm varepsilon) }$$



      is properly defined for all $epsilon$, as long as $int_{b - varepsilon}^{b + varepsilon} f_b(x) , dx > 0 $. Finally, we just define the conditional distribution of $A|B$ as



      $$lim_{varepsilon rightarrow 0} frac{P(A in a pm varepsilon cap B in b pm varepsilon) / varepsilon}{P(B in b pm varepsilon) / varepsilon } $$



      By definition, this is



      $$frac{ f_{A, B}(a, b) }{ f_B(b)}$$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Instead of Bayes rule, we should look closely at the definition of conditional probability. as Bayes rule is simply a small transformation of the definition of conditional probability.



        With discrete probabilities, it is simple to define



        $$P(A = a mid B = b) = frac{P(A = acap B = b)}{P(B = b)}$$



        As you pointed out, this would be ill-defined if both $A$ and $B$ were continuous, as it would result $0/0.$



        Instead, let's think about $P(A in a pm varepsilon mid B in b pm varepsilon)$ for a continuous distribution. Then the value



        $$ frac{P(A in a pm varepsilon cap B in b pm varepsilon)}{P(B in b pm varepsilon) }$$



        is properly defined for all $epsilon$, as long as $int_{b - varepsilon}^{b + varepsilon} f_b(x) , dx > 0 $. Finally, we just define the conditional distribution of $A|B$ as



        $$lim_{varepsilon rightarrow 0} frac{P(A in a pm varepsilon cap B in b pm varepsilon) / varepsilon}{P(B in b pm varepsilon) / varepsilon } $$



        By definition, this is



        $$frac{ f_{A, B}(a, b) }{ f_B(b)}$$






        share|cite|improve this answer











        $endgroup$



        Instead of Bayes rule, we should look closely at the definition of conditional probability. as Bayes rule is simply a small transformation of the definition of conditional probability.



        With discrete probabilities, it is simple to define



        $$P(A = a mid B = b) = frac{P(A = acap B = b)}{P(B = b)}$$



        As you pointed out, this would be ill-defined if both $A$ and $B$ were continuous, as it would result $0/0.$



        Instead, let's think about $P(A in a pm varepsilon mid B in b pm varepsilon)$ for a continuous distribution. Then the value



        $$ frac{P(A in a pm varepsilon cap B in b pm varepsilon)}{P(B in b pm varepsilon) }$$



        is properly defined for all $epsilon$, as long as $int_{b - varepsilon}^{b + varepsilon} f_b(x) , dx > 0 $. Finally, we just define the conditional distribution of $A|B$ as



        $$lim_{varepsilon rightarrow 0} frac{P(A in a pm varepsilon cap B in b pm varepsilon) / varepsilon}{P(B in b pm varepsilon) / varepsilon } $$



        By definition, this is



        $$frac{ f_{A, B}(a, b) }{ f_B(b)}$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 6 hours ago









        Michael Hardy

        3,7901430




        3,7901430










        answered 7 hours ago









        Cliff ABCliff AB

        12.7k12363




        12.7k12363






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Cross Validated!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f391519%2fbayesian-probability-question-pointwise-probability%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Ponta tanko

            Image
            Usona ponta tanko M60A1 Ponta tanko estas, simile kiel kirasita savveturilo, inter specialaj kirasitaj transportiloj. Kiel pontaj tankoj eblas konsideri ĉiujn memmovigaj rimedoj ebligantaj superi obstaklojn ĝis larĝeco 24 metroj rekte en spaco de bataloperacoj. Ties ĉefa tasko estas ebligi al la propraj taĉmentoj daŭrigi en rapida antaŭenirado sen senbezona reteniĝo. El tio pli-malpli rezultas taktikaj-teknikaj postuloj, kiujn ili devas plenumi. Precipe temas pri kapablo operacii komune kun unue-sinsekvaj taĉmentoj kaj transpontigi obstaklojn dum kelke da minutoj. This page is only for reference, If you need detailed information, please check here
            Read more

            Tantalo (mitologio)

            Image
            Ĉi tiu artikolo temas pri mitologio. Por aliaj signifoj vidu la paĝon Tantalo (apartigilo) . Tantalo , laŭ Gioacchino Assereto. Laŭ la helena mitologio, Tantalo (el la greka Τάνταλος , ŝancelanta aŭ tre malfeliĉa ) estis filo de Zeŭso kaj la oceanidino Pluto, laŭ la plej divastigata versio. Li estis reĝo de Frigio [1] aŭ de la monto Sipilo, en Lidio (Malgrandazio) [2] kaj post sia morto li suferis eternan torturadon ĉe la Tartaro, la plej profunda parto de Hadeso, rezervita por la plej perversaj homoj. Tamen, kelkaj rakontoj asertas ke li estis reĝo de Argo aŭ de Korinto [3] , aŭ ke li migris norden el la monto Sipilo por akiri la tronon de Paflagonio, el kie Ilo forpelis lin pro tio, ke Tantalo forkaptis kaj seksperfortis lian fraton Ganimedon [4] , kaj tiel ofendis la diojn. Verŝajne li ankaŭ regis urbon en okcidenta Anatolio, kiun oni nomis Tantalis [5] pro li, aŭ Sipilo , pro la monto sur kies flanko ankoraŭ staras ĝiaj ruinaĵoj [6] . Laŭ la geografiisto Strabo,
            Read more

            Erzsébet Schaár

            Image
            Erzsébet Schaár Memportreto de Erzsébet Schaár en Pécs Personaj informoj Naskiĝo 27-an de julio 1908 ( 1908-07-27 ) en Budafok Morto 29-an de aŭgusto 1975 ( 1975-08-29 ) (67-jara) en Budapeŝto Ŝtataneco Hungario Profesio skulptisto Familianoj Edz(in)o Tibor Vilt v   •   d   •   r Erzsébet Schaár [erĵEbet ŝAr], laŭ hungarlingve kutima sinsekvo Schaár Erzsébet estis hungara skulptisto. Ŝia edzo estis Tibor Vilt. Déryné Széppataki Róza en Miskolc fare de Erzsébet Schaár Erzsébet Schaár [1] naskiĝis la 27-an de julio 1908 en Budafok, ŝi mortis la 29-an de aŭgusto 1975 en Budapeŝto. Enhavo 1 Biografio 2 Elektitaj solaj ekspozicioj 3 Elektitaj kolektivaj ekspozicioj 4 Elektitaj skulptaĵoj 5 Elektitaj disĉiploj 6 Fontoj 7 Referencoj Biografio | Erzsébet Schaár studis en Industriarta Lernejo, poste en Reĝa Belarta Altlernejo (1924-1926). Ŝia ĉefinstruisto estis Zsig
            Read more