Is there a name for this average?

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Let



$lambda = fracsumlimits_n=1^N lambda_n B_n e^-lambda_n t_0sumlimits_n=1^N B_n e^-lambda_n t_0,,$



with $lambda_n$, $B_n$, $t_0$ real numbers.



I interpret $lambda$ as some sort of average. Is there a name for it?




terminology average




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  • A "weighted average", perhaps
    – Omnomnomnom
    Aug 2 at 10:07














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Let



$lambda = fracsumlimits_n=1^N lambda_n B_n e^-lambda_n t_0sumlimits_n=1^N B_n e^-lambda_n t_0,,$



with $lambda_n$, $B_n$, $t_0$ real numbers.



I interpret $lambda$ as some sort of average. Is there a name for it?




terminology average




share|cite|improve this question



















  • A "weighted average", perhaps
    – Omnomnomnom
    Aug 2 at 10:07












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let



$lambda = fracsumlimits_n=1^N lambda_n B_n e^-lambda_n t_0sumlimits_n=1^N B_n e^-lambda_n t_0,,$



with $lambda_n$, $B_n$, $t_0$ real numbers.



I interpret $lambda$ as some sort of average. Is there a name for it?




terminology average




share|cite|improve this question











Let



$lambda = fracsumlimits_n=1^N lambda_n B_n e^-lambda_n t_0sumlimits_n=1^N B_n e^-lambda_n t_0,,$



with $lambda_n$, $B_n$, $t_0$ real numbers.



I interpret $lambda$ as some sort of average. Is there a name for it?





terminology average





share|cite|improve this question










share|cite|improve this question




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asked Aug 2 at 10:05









toliveira

597312




597312











  • A "weighted average", perhaps
    – Omnomnomnom
    Aug 2 at 10:07
















  • A "weighted average", perhaps
    – Omnomnomnom
    Aug 2 at 10:07















A "weighted average", perhaps
– Omnomnomnom
Aug 2 at 10:07




A "weighted average", perhaps
– Omnomnomnom
Aug 2 at 10:07










1 Answer
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2
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Generally,



$$
lambda=fracsum_nw_nlambda_nsum_nw_n
$$



is called a weighted average of the $lambda_n$ with weights $w_n$. In your case, the weights $w_n=B_nmathrm e^-lambda_nt_0$ depend on the quantities $lambda_n$ being averaged, which isn't usually the case, but one might still call it a weighted average in a wider sense.



Specifically,



$$
E=fracsum_nB_nE_nmathrm e^-beta E_nsum_nB_nmathrm e^-beta E_n
$$



is the mean energy in a system described by the Boltzmann distribution of statistical mechanics, where $E_n$ is the energy of state $n$, $B_n$ is its multiplicity, and $beta=frac1kT$ is the inverse temperature. An average taken with respect to the Boltzmann distribution is sometimes called a Boltzmann average.






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    1 Answer
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    active

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    1 Answer
    1






    active

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    active

    oldest

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    active

    oldest

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    up vote
    2
    down vote













    Generally,



    $$
    lambda=fracsum_nw_nlambda_nsum_nw_n
    $$



    is called a weighted average of the $lambda_n$ with weights $w_n$. In your case, the weights $w_n=B_nmathrm e^-lambda_nt_0$ depend on the quantities $lambda_n$ being averaged, which isn't usually the case, but one might still call it a weighted average in a wider sense.



    Specifically,



    $$
    E=fracsum_nB_nE_nmathrm e^-beta E_nsum_nB_nmathrm e^-beta E_n
    $$



    is the mean energy in a system described by the Boltzmann distribution of statistical mechanics, where $E_n$ is the energy of state $n$, $B_n$ is its multiplicity, and $beta=frac1kT$ is the inverse temperature. An average taken with respect to the Boltzmann distribution is sometimes called a Boltzmann average.






    share|cite|improve this answer

























      up vote
      2
      down vote













      Generally,



      $$
      lambda=fracsum_nw_nlambda_nsum_nw_n
      $$



      is called a weighted average of the $lambda_n$ with weights $w_n$. In your case, the weights $w_n=B_nmathrm e^-lambda_nt_0$ depend on the quantities $lambda_n$ being averaged, which isn't usually the case, but one might still call it a weighted average in a wider sense.



      Specifically,



      $$
      E=fracsum_nB_nE_nmathrm e^-beta E_nsum_nB_nmathrm e^-beta E_n
      $$



      is the mean energy in a system described by the Boltzmann distribution of statistical mechanics, where $E_n$ is the energy of state $n$, $B_n$ is its multiplicity, and $beta=frac1kT$ is the inverse temperature. An average taken with respect to the Boltzmann distribution is sometimes called a Boltzmann average.






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        Generally,



        $$
        lambda=fracsum_nw_nlambda_nsum_nw_n
        $$



        is called a weighted average of the $lambda_n$ with weights $w_n$. In your case, the weights $w_n=B_nmathrm e^-lambda_nt_0$ depend on the quantities $lambda_n$ being averaged, which isn't usually the case, but one might still call it a weighted average in a wider sense.



        Specifically,



        $$
        E=fracsum_nB_nE_nmathrm e^-beta E_nsum_nB_nmathrm e^-beta E_n
        $$



        is the mean energy in a system described by the Boltzmann distribution of statistical mechanics, where $E_n$ is the energy of state $n$, $B_n$ is its multiplicity, and $beta=frac1kT$ is the inverse temperature. An average taken with respect to the Boltzmann distribution is sometimes called a Boltzmann average.






        share|cite|improve this answer













        Generally,



        $$
        lambda=fracsum_nw_nlambda_nsum_nw_n
        $$



        is called a weighted average of the $lambda_n$ with weights $w_n$. In your case, the weights $w_n=B_nmathrm e^-lambda_nt_0$ depend on the quantities $lambda_n$ being averaged, which isn't usually the case, but one might still call it a weighted average in a wider sense.



        Specifically,



        $$
        E=fracsum_nB_nE_nmathrm e^-beta E_nsum_nB_nmathrm e^-beta E_n
        $$



        is the mean energy in a system described by the Boltzmann distribution of statistical mechanics, where $E_n$ is the energy of state $n$, $B_n$ is its multiplicity, and $beta=frac1kT$ is the inverse temperature. An average taken with respect to the Boltzmann distribution is sometimes called a Boltzmann average.







        share|cite|improve this answer













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        share|cite|improve this answer











        answered Aug 2 at 10:50









        joriki

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