The mean of positive half Gaussian distribution? [on hold]

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There is a Gaussian distribution $g(x)$, its mean is $mu=0$ and the variance is $sigma=1$.
So what is the mean value of $f(x)=max(0, g(x))$?



Sorry for the wrong discription. I meant the Half-normal distribution. Sorry for the troubles. Thank you for your help.




probability




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put on hold as off-topic by amWhy, Shaun, José Carlos Santos, StubbornAtom, Delta-u 20 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shaun, José Carlos Santos, StubbornAtom, Delta-u
If this question can be reworded to fit the rules in the help center, please edit the question.












  • $max(0,g(x))$ is just $g(x)$ (since from what you've written it appears that $g$, as a function, is the density of a positive measure).
    – Saucy O'Path
    21 hours ago











  • What are your thoughts about that? What have you tried so far?
    – Taroccoesbrocco
    21 hours ago










  • @Yikouniao, As the question is currently worded it is trivial (possibly nonsensical) because g(x) is always positive. However, if you intended what i have written about in my answer, please edit your question to make it clear to other readers and people who may contribute answers.
    – Martin Roberts
    20 hours ago















up vote
-1
down vote

favorite












There is a Gaussian distribution $g(x)$, its mean is $mu=0$ and the variance is $sigma=1$.
So what is the mean value of $f(x)=max(0, g(x))$?



Sorry for the wrong discription. I meant the Half-normal distribution. Sorry for the troubles. Thank you for your help.




probability




share|cite|improve this question













put on hold as off-topic by amWhy, Shaun, José Carlos Santos, StubbornAtom, Delta-u 20 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shaun, José Carlos Santos, StubbornAtom, Delta-u
If this question can be reworded to fit the rules in the help center, please edit the question.












  • $max(0,g(x))$ is just $g(x)$ (since from what you've written it appears that $g$, as a function, is the density of a positive measure).
    – Saucy O'Path
    21 hours ago











  • What are your thoughts about that? What have you tried so far?
    – Taroccoesbrocco
    21 hours ago










  • @Yikouniao, As the question is currently worded it is trivial (possibly nonsensical) because g(x) is always positive. However, if you intended what i have written about in my answer, please edit your question to make it clear to other readers and people who may contribute answers.
    – Martin Roberts
    20 hours ago













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











There is a Gaussian distribution $g(x)$, its mean is $mu=0$ and the variance is $sigma=1$.
So what is the mean value of $f(x)=max(0, g(x))$?



Sorry for the wrong discription. I meant the Half-normal distribution. Sorry for the troubles. Thank you for your help.




probability




share|cite|improve this question













There is a Gaussian distribution $g(x)$, its mean is $mu=0$ and the variance is $sigma=1$.
So what is the mean value of $f(x)=max(0, g(x))$?



Sorry for the wrong discription. I meant the Half-normal distribution. Sorry for the troubles. Thank you for your help.





probability





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 19 hours ago
























asked 21 hours ago









yikouniao

1073




1073




put on hold as off-topic by amWhy, Shaun, José Carlos Santos, StubbornAtom, Delta-u 20 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shaun, José Carlos Santos, StubbornAtom, Delta-u
If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by amWhy, Shaun, José Carlos Santos, StubbornAtom, Delta-u 20 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shaun, José Carlos Santos, StubbornAtom, Delta-u
If this question can be reworded to fit the rules in the help center, please edit the question.











  • $max(0,g(x))$ is just $g(x)$ (since from what you've written it appears that $g$, as a function, is the density of a positive measure).
    – Saucy O'Path
    21 hours ago











  • What are your thoughts about that? What have you tried so far?
    – Taroccoesbrocco
    21 hours ago










  • @Yikouniao, As the question is currently worded it is trivial (possibly nonsensical) because g(x) is always positive. However, if you intended what i have written about in my answer, please edit your question to make it clear to other readers and people who may contribute answers.
    – Martin Roberts
    20 hours ago

















  • $max(0,g(x))$ is just $g(x)$ (since from what you've written it appears that $g$, as a function, is the density of a positive measure).
    – Saucy O'Path
    21 hours ago











  • What are your thoughts about that? What have you tried so far?
    – Taroccoesbrocco
    21 hours ago










  • @Yikouniao, As the question is currently worded it is trivial (possibly nonsensical) because g(x) is always positive. However, if you intended what i have written about in my answer, please edit your question to make it clear to other readers and people who may contribute answers.
    – Martin Roberts
    20 hours ago
















$max(0,g(x))$ is just $g(x)$ (since from what you've written it appears that $g$, as a function, is the density of a positive measure).
– Saucy O'Path
21 hours ago





$max(0,g(x))$ is just $g(x)$ (since from what you've written it appears that $g$, as a function, is the density of a positive measure).
– Saucy O'Path
21 hours ago













What are your thoughts about that? What have you tried so far?
– Taroccoesbrocco
21 hours ago




What are your thoughts about that? What have you tried so far?
– Taroccoesbrocco
21 hours ago












@Yikouniao, As the question is currently worded it is trivial (possibly nonsensical) because g(x) is always positive. However, if you intended what i have written about in my answer, please edit your question to make it clear to other readers and people who may contribute answers.
– Martin Roberts
20 hours ago





@Yikouniao, As the question is currently worded it is trivial (possibly nonsensical) because g(x) is always positive. However, if you intended what i have written about in my answer, please edit your question to make it clear to other readers and people who may contribute answers.
– Martin Roberts
20 hours ago











1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










I assume that when you say 'the positive half', you are intending to refer to the positive values of independent variable $x$ of the Gaussian distribution, rather than the positive values of $g(x)$.



You can learn allow about the half-normal distribution at Wikipedia here. (Although the note that the conventional normalization constant for the half-normal distribution is double that of the standard normal distribution curve, to ensure that the integral of the half-normal distribution is still exactly unity).



Thus, using your notation, the mean of the positive half of the standard gaussian distribution is exactly $sqrtfrac12pi $.






share|cite|improve this answer






























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    I assume that when you say 'the positive half', you are intending to refer to the positive values of independent variable $x$ of the Gaussian distribution, rather than the positive values of $g(x)$.



    You can learn allow about the half-normal distribution at Wikipedia here. (Although the note that the conventional normalization constant for the half-normal distribution is double that of the standard normal distribution curve, to ensure that the integral of the half-normal distribution is still exactly unity).



    Thus, using your notation, the mean of the positive half of the standard gaussian distribution is exactly $sqrtfrac12pi $.






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      I assume that when you say 'the positive half', you are intending to refer to the positive values of independent variable $x$ of the Gaussian distribution, rather than the positive values of $g(x)$.



      You can learn allow about the half-normal distribution at Wikipedia here. (Although the note that the conventional normalization constant for the half-normal distribution is double that of the standard normal distribution curve, to ensure that the integral of the half-normal distribution is still exactly unity).



      Thus, using your notation, the mean of the positive half of the standard gaussian distribution is exactly $sqrtfrac12pi $.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        I assume that when you say 'the positive half', you are intending to refer to the positive values of independent variable $x$ of the Gaussian distribution, rather than the positive values of $g(x)$.



        You can learn allow about the half-normal distribution at Wikipedia here. (Although the note that the conventional normalization constant for the half-normal distribution is double that of the standard normal distribution curve, to ensure that the integral of the half-normal distribution is still exactly unity).



        Thus, using your notation, the mean of the positive half of the standard gaussian distribution is exactly $sqrtfrac12pi $.






        share|cite|improve this answer















        I assume that when you say 'the positive half', you are intending to refer to the positive values of independent variable $x$ of the Gaussian distribution, rather than the positive values of $g(x)$.



        You can learn allow about the half-normal distribution at Wikipedia here. (Although the note that the conventional normalization constant for the half-normal distribution is double that of the standard normal distribution curve, to ensure that the integral of the half-normal distribution is still exactly unity).



        Thus, using your notation, the mean of the positive half of the standard gaussian distribution is exactly $sqrtfrac12pi $.







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited 20 hours ago


























        answered 21 hours ago









        Martin Roberts

        1,194217




        1,194217












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