What does the value of a PDF mean?












3












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I understand that the integral of a PDF provides tangible value --i.e., the integral of a PDF allows one to see the probability of a value or less than that value, under a particular distribution, occurring. But, what does the value of just the output of the PDF provide? In other words, what does the PDF of the standard normal distribution at x=0.5 mean?










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  • 2




    $begingroup$
    The PDF is the rate of change of the CDF. No more, no less.
    $endgroup$
    – JimB
    1 hour ago










  • $begingroup$
    Possible duplicate of What does the value of a probability density function (PDF) at some x indicate?
    $endgroup$
    – Victoria M
    1 hour ago
















3












$begingroup$


I understand that the integral of a PDF provides tangible value --i.e., the integral of a PDF allows one to see the probability of a value or less than that value, under a particular distribution, occurring. But, what does the value of just the output of the PDF provide? In other words, what does the PDF of the standard normal distribution at x=0.5 mean?










share|cite|improve this question







New contributor




Lea England is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 2




    $begingroup$
    The PDF is the rate of change of the CDF. No more, no less.
    $endgroup$
    – JimB
    1 hour ago










  • $begingroup$
    Possible duplicate of What does the value of a probability density function (PDF) at some x indicate?
    $endgroup$
    – Victoria M
    1 hour ago














3












3








3





$begingroup$


I understand that the integral of a PDF provides tangible value --i.e., the integral of a PDF allows one to see the probability of a value or less than that value, under a particular distribution, occurring. But, what does the value of just the output of the PDF provide? In other words, what does the PDF of the standard normal distribution at x=0.5 mean?










share|cite|improve this question







New contributor




Lea England is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I understand that the integral of a PDF provides tangible value --i.e., the integral of a PDF allows one to see the probability of a value or less than that value, under a particular distribution, occurring. But, what does the value of just the output of the PDF provide? In other words, what does the PDF of the standard normal distribution at x=0.5 mean?







probability






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asked 1 hour ago









Lea EnglandLea England

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  • 2




    $begingroup$
    The PDF is the rate of change of the CDF. No more, no less.
    $endgroup$
    – JimB
    1 hour ago










  • $begingroup$
    Possible duplicate of What does the value of a probability density function (PDF) at some x indicate?
    $endgroup$
    – Victoria M
    1 hour ago














  • 2




    $begingroup$
    The PDF is the rate of change of the CDF. No more, no less.
    $endgroup$
    – JimB
    1 hour ago










  • $begingroup$
    Possible duplicate of What does the value of a probability density function (PDF) at some x indicate?
    $endgroup$
    – Victoria M
    1 hour ago








2




2




$begingroup$
The PDF is the rate of change of the CDF. No more, no less.
$endgroup$
– JimB
1 hour ago




$begingroup$
The PDF is the rate of change of the CDF. No more, no less.
$endgroup$
– JimB
1 hour ago












$begingroup$
Possible duplicate of What does the value of a probability density function (PDF) at some x indicate?
$endgroup$
– Victoria M
1 hour ago




$begingroup$
Possible duplicate of What does the value of a probability density function (PDF) at some x indicate?
$endgroup$
– Victoria M
1 hour ago










3 Answers
3






active

oldest

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3












$begingroup$

I would first say that you should try to be OK with it not having a super down-to-earth meaning. It can sometimes help to let yourself be abstract and use the mathematical tools you have to get somewhere you might not have gotten by purely reasoning through with statements you might use to describe other things in your daily life. On the other hand, it is often helpful to have somewhat concrete ways of thinking about things that are abstract, so it's good that you're investigating this.



But as the PDF is the derivative of the CDF, the value of PDF $p(x)$ at $x$ is the (slope of the) best linear approximation of the CDF near $x$, i.e., that if $epsilon$ is super small, then if you want to approximate the probability that an event between $x$ and $x+epsilon$ happens linearly in $epsilon$, your best bet is to say $epsilon cdot p(x)$.






share|cite|improve this answer










New contributor




csprun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$





















    1












    $begingroup$

    The value of the PDF is sort of like the "weighting" that each event gets for the CDF.






    share|cite|improve this answer








    New contributor




    Jacob Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$





















      1












      $begingroup$


      Definition (Gut, 2005): A distribution function F is




      • discrete iff for some countable set of numbers {$x_j$} and point masses {$p_j$}, $$F(x) = sum_{x_jleq x} p_j, quad text{for all x} in Bbb R.$$ The function p is called the probability function.

      • continuous iff it is continuous for all x.

      • absolutely continuous iff there exists a non-negative, Lebesgue integrable function f, such that $$F(b) - F(a) = int_a^b f(x)text{ dx} quad {for space all space x < b.}$$ The function f is called the density of F.

      • singular iff F $neq$ 0, F' exists and equals 0 a.e.




      By the definition, evaluating a probability distribution function F is simply an integration of the density of said distribution. As mentioned, in the case for a $<$ b, this gives us the probability for a given range of values.



      However, if evaluating the distribution function F at a single value, the integral clearly vanishes. To find the probability of a single value, F must be discrete thereby allowing us to compute $$F(a) = sum_{a in X} p_j(a).$$ By taking the limit on $int_a^bf(x)text{dx}$ to F(a + $epsilon$) - F(a - $epsilon$) = $int_{a-epsilon}^{a+epsilon}f(x)text{dx}$, we then thus find the relative probability of F(a) occuring.






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        3 Answers
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        3 Answers
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        active

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        $begingroup$

        I would first say that you should try to be OK with it not having a super down-to-earth meaning. It can sometimes help to let yourself be abstract and use the mathematical tools you have to get somewhere you might not have gotten by purely reasoning through with statements you might use to describe other things in your daily life. On the other hand, it is often helpful to have somewhat concrete ways of thinking about things that are abstract, so it's good that you're investigating this.



        But as the PDF is the derivative of the CDF, the value of PDF $p(x)$ at $x$ is the (slope of the) best linear approximation of the CDF near $x$, i.e., that if $epsilon$ is super small, then if you want to approximate the probability that an event between $x$ and $x+epsilon$ happens linearly in $epsilon$, your best bet is to say $epsilon cdot p(x)$.






        share|cite|improve this answer










        New contributor




        csprun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$


















          3












          $begingroup$

          I would first say that you should try to be OK with it not having a super down-to-earth meaning. It can sometimes help to let yourself be abstract and use the mathematical tools you have to get somewhere you might not have gotten by purely reasoning through with statements you might use to describe other things in your daily life. On the other hand, it is often helpful to have somewhat concrete ways of thinking about things that are abstract, so it's good that you're investigating this.



          But as the PDF is the derivative of the CDF, the value of PDF $p(x)$ at $x$ is the (slope of the) best linear approximation of the CDF near $x$, i.e., that if $epsilon$ is super small, then if you want to approximate the probability that an event between $x$ and $x+epsilon$ happens linearly in $epsilon$, your best bet is to say $epsilon cdot p(x)$.






          share|cite|improve this answer










          New contributor




          csprun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$
















            3












            3








            3





            $begingroup$

            I would first say that you should try to be OK with it not having a super down-to-earth meaning. It can sometimes help to let yourself be abstract and use the mathematical tools you have to get somewhere you might not have gotten by purely reasoning through with statements you might use to describe other things in your daily life. On the other hand, it is often helpful to have somewhat concrete ways of thinking about things that are abstract, so it's good that you're investigating this.



            But as the PDF is the derivative of the CDF, the value of PDF $p(x)$ at $x$ is the (slope of the) best linear approximation of the CDF near $x$, i.e., that if $epsilon$ is super small, then if you want to approximate the probability that an event between $x$ and $x+epsilon$ happens linearly in $epsilon$, your best bet is to say $epsilon cdot p(x)$.






            share|cite|improve this answer










            New contributor




            csprun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$



            I would first say that you should try to be OK with it not having a super down-to-earth meaning. It can sometimes help to let yourself be abstract and use the mathematical tools you have to get somewhere you might not have gotten by purely reasoning through with statements you might use to describe other things in your daily life. On the other hand, it is often helpful to have somewhat concrete ways of thinking about things that are abstract, so it's good that you're investigating this.



            But as the PDF is the derivative of the CDF, the value of PDF $p(x)$ at $x$ is the (slope of the) best linear approximation of the CDF near $x$, i.e., that if $epsilon$ is super small, then if you want to approximate the probability that an event between $x$ and $x+epsilon$ happens linearly in $epsilon$, your best bet is to say $epsilon cdot p(x)$.







            share|cite|improve this answer










            New contributor




            csprun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            share|cite|improve this answer



            share|cite|improve this answer








            edited 20 mins ago





















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            answered 1 hour ago









            cspruncsprun

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                1












                $begingroup$

                The value of the PDF is sort of like the "weighting" that each event gets for the CDF.






                share|cite|improve this answer








                New contributor




                Jacob Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                $endgroup$


















                  1












                  $begingroup$

                  The value of the PDF is sort of like the "weighting" that each event gets for the CDF.






                  share|cite|improve this answer








                  New contributor




                  Jacob Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    The value of the PDF is sort of like the "weighting" that each event gets for the CDF.






                    share|cite|improve this answer








                    New contributor




                    Jacob Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$



                    The value of the PDF is sort of like the "weighting" that each event gets for the CDF.







                    share|cite|improve this answer








                    New contributor




                    Jacob Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|cite|improve this answer



                    share|cite|improve this answer






                    New contributor




                    Jacob Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                    answered 1 hour ago









                    Jacob SmithJacob Smith

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                    111




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                        1












                        $begingroup$


                        Definition (Gut, 2005): A distribution function F is




                        • discrete iff for some countable set of numbers {$x_j$} and point masses {$p_j$}, $$F(x) = sum_{x_jleq x} p_j, quad text{for all x} in Bbb R.$$ The function p is called the probability function.

                        • continuous iff it is continuous for all x.

                        • absolutely continuous iff there exists a non-negative, Lebesgue integrable function f, such that $$F(b) - F(a) = int_a^b f(x)text{ dx} quad {for space all space x < b.}$$ The function f is called the density of F.

                        • singular iff F $neq$ 0, F' exists and equals 0 a.e.




                        By the definition, evaluating a probability distribution function F is simply an integration of the density of said distribution. As mentioned, in the case for a $<$ b, this gives us the probability for a given range of values.



                        However, if evaluating the distribution function F at a single value, the integral clearly vanishes. To find the probability of a single value, F must be discrete thereby allowing us to compute $$F(a) = sum_{a in X} p_j(a).$$ By taking the limit on $int_a^bf(x)text{dx}$ to F(a + $epsilon$) - F(a - $epsilon$) = $int_{a-epsilon}^{a+epsilon}f(x)text{dx}$, we then thus find the relative probability of F(a) occuring.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$


                          Definition (Gut, 2005): A distribution function F is




                          • discrete iff for some countable set of numbers {$x_j$} and point masses {$p_j$}, $$F(x) = sum_{x_jleq x} p_j, quad text{for all x} in Bbb R.$$ The function p is called the probability function.

                          • continuous iff it is continuous for all x.

                          • absolutely continuous iff there exists a non-negative, Lebesgue integrable function f, such that $$F(b) - F(a) = int_a^b f(x)text{ dx} quad {for space all space x < b.}$$ The function f is called the density of F.

                          • singular iff F $neq$ 0, F' exists and equals 0 a.e.




                          By the definition, evaluating a probability distribution function F is simply an integration of the density of said distribution. As mentioned, in the case for a $<$ b, this gives us the probability for a given range of values.



                          However, if evaluating the distribution function F at a single value, the integral clearly vanishes. To find the probability of a single value, F must be discrete thereby allowing us to compute $$F(a) = sum_{a in X} p_j(a).$$ By taking the limit on $int_a^bf(x)text{dx}$ to F(a + $epsilon$) - F(a - $epsilon$) = $int_{a-epsilon}^{a+epsilon}f(x)text{dx}$, we then thus find the relative probability of F(a) occuring.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$


                            Definition (Gut, 2005): A distribution function F is




                            • discrete iff for some countable set of numbers {$x_j$} and point masses {$p_j$}, $$F(x) = sum_{x_jleq x} p_j, quad text{for all x} in Bbb R.$$ The function p is called the probability function.

                            • continuous iff it is continuous for all x.

                            • absolutely continuous iff there exists a non-negative, Lebesgue integrable function f, such that $$F(b) - F(a) = int_a^b f(x)text{ dx} quad {for space all space x < b.}$$ The function f is called the density of F.

                            • singular iff F $neq$ 0, F' exists and equals 0 a.e.




                            By the definition, evaluating a probability distribution function F is simply an integration of the density of said distribution. As mentioned, in the case for a $<$ b, this gives us the probability for a given range of values.



                            However, if evaluating the distribution function F at a single value, the integral clearly vanishes. To find the probability of a single value, F must be discrete thereby allowing us to compute $$F(a) = sum_{a in X} p_j(a).$$ By taking the limit on $int_a^bf(x)text{dx}$ to F(a + $epsilon$) - F(a - $epsilon$) = $int_{a-epsilon}^{a+epsilon}f(x)text{dx}$, we then thus find the relative probability of F(a) occuring.






                            share|cite|improve this answer









                            $endgroup$




                            Definition (Gut, 2005): A distribution function F is




                            • discrete iff for some countable set of numbers {$x_j$} and point masses {$p_j$}, $$F(x) = sum_{x_jleq x} p_j, quad text{for all x} in Bbb R.$$ The function p is called the probability function.

                            • continuous iff it is continuous for all x.

                            • absolutely continuous iff there exists a non-negative, Lebesgue integrable function f, such that $$F(b) - F(a) = int_a^b f(x)text{ dx} quad {for space all space x < b.}$$ The function f is called the density of F.

                            • singular iff F $neq$ 0, F' exists and equals 0 a.e.




                            By the definition, evaluating a probability distribution function F is simply an integration of the density of said distribution. As mentioned, in the case for a $<$ b, this gives us the probability for a given range of values.



                            However, if evaluating the distribution function F at a single value, the integral clearly vanishes. To find the probability of a single value, F must be discrete thereby allowing us to compute $$F(a) = sum_{a in X} p_j(a).$$ By taking the limit on $int_a^bf(x)text{dx}$ to F(a + $epsilon$) - F(a - $epsilon$) = $int_{a-epsilon}^{a+epsilon}f(x)text{dx}$, we then thus find the relative probability of F(a) occuring.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 27 mins ago









                            Victoria MVictoria M

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