Show that the Normal distribution is a member of the exponential family












0












$begingroup$


I want to show that the Normal distribution is a member of the exponential family.



I have been working under the assumption that a distribution is a member of the exponential family if its pdf/pmf can be transformed into the form:



$f(x|theta) = h(x)c(theta)exp{sumlimits_{i=1}^{k} w_{i}(theta)t_{i}(x)}$



This is my approach:



$f(x|mu, sigma^2) = frac{1}{sqrt{2pi sigma^2}}exp{-frac{(x-mu)^2}{2 sigma^2}}$



Taking the logs:



$log f(x|mu, sigma^2) = -frac{1}{2}log(2pisigma^2) - frac{(x-mu)^2}{2 sigma^2}$



Taking the exponential:



$f(x|mu, sigma^2) = exp{-frac{1}{2}log(2pisigma^2)-frac{(x-mu)^2}{2sigma^2}}$



= $exp{-frac{1}{2}log(2pisigma^2)-frac{(x^2 -2mu + mu^2)}{2sigma^2}}$



= $exp{-frac{1}{2}log(2pisigma^2)-frac{x^2}{2sigma^2} + frac{2xmu}{2sigma^2} - frac{mu^2}{2sigma^2}}$



= $exp{-frac{1}{2}log(2pisigma^2)} exp{-frac{x^2}{2sigma^2} + frac{xmu}{sigma^2} - frac{mu^2}{2sigma^2}}$



= $frac{1}{sqrt{2pisigma^2}} exp{-frac{x^2}{2sigma^2} + frac{xmu}{sigma^2} - frac{mu^2}{2sigma^2}}$



= $frac{1}{sqrt{2pisigma^2}} exp{-frac{mu^2}{2sigma^2}} exp{-frac{x^2}{2sigma^2} + frac{xmu}{sigma^2}}$



Now I have:



$c(theta) = frac{1}{sqrt{2pisigma^2}} exp{-frac{mu^2}{2sigma^2}}$,



$w_{1}(theta) = -frac{1}{2sigma^2}$,



$t_{1}(x) = x^2$,



$w_{2}(theta) = frac{mu}{sigma^2}$,



$t_{1}(x) = x$



But I am missing $h(x)$.



Am I missing something?






statistics normal-distribution







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    $h(x)$ is allowed to be an identity function (value of 1 $forall x$)
    $endgroup$
    – Easymode44
    Dec 17 '18 at 21:47






  • 1




    $begingroup$
    ...is a member of an exponential family.
    $endgroup$
    – Did
    Dec 17 '18 at 22:26
















0












$begingroup$


I want to show that the Normal distribution is a member of the exponential family.



I have been working under the assumption that a distribution is a member of the exponential family if its pdf/pmf can be transformed into the form:



$f(x|theta) = h(x)c(theta)exp{sumlimits_{i=1}^{k} w_{i}(theta)t_{i}(x)}$



This is my approach:



$f(x|mu, sigma^2) = frac{1}{sqrt{2pi sigma^2}}exp{-frac{(x-mu)^2}{2 sigma^2}}$



Taking the logs:



$log f(x|mu, sigma^2) = -frac{1}{2}log(2pisigma^2) - frac{(x-mu)^2}{2 sigma^2}$



Taking the exponential:



$f(x|mu, sigma^2) = exp{-frac{1}{2}log(2pisigma^2)-frac{(x-mu)^2}{2sigma^2}}$



= $exp{-frac{1}{2}log(2pisigma^2)-frac{(x^2 -2mu + mu^2)}{2sigma^2}}$



= $exp{-frac{1}{2}log(2pisigma^2)-frac{x^2}{2sigma^2} + frac{2xmu}{2sigma^2} - frac{mu^2}{2sigma^2}}$



= $exp{-frac{1}{2}log(2pisigma^2)} exp{-frac{x^2}{2sigma^2} + frac{xmu}{sigma^2} - frac{mu^2}{2sigma^2}}$



= $frac{1}{sqrt{2pisigma^2}} exp{-frac{x^2}{2sigma^2} + frac{xmu}{sigma^2} - frac{mu^2}{2sigma^2}}$



= $frac{1}{sqrt{2pisigma^2}} exp{-frac{mu^2}{2sigma^2}} exp{-frac{x^2}{2sigma^2} + frac{xmu}{sigma^2}}$



Now I have:



$c(theta) = frac{1}{sqrt{2pisigma^2}} exp{-frac{mu^2}{2sigma^2}}$,



$w_{1}(theta) = -frac{1}{2sigma^2}$,



$t_{1}(x) = x^2$,



$w_{2}(theta) = frac{mu}{sigma^2}$,



$t_{1}(x) = x$



But I am missing $h(x)$.



Am I missing something?






statistics normal-distribution







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    $h(x)$ is allowed to be an identity function (value of 1 $forall x$)
    $endgroup$
    – Easymode44
    Dec 17 '18 at 21:47






  • 1




    $begingroup$
    ...is a member of an exponential family.
    $endgroup$
    – Did
    Dec 17 '18 at 22:26














0












0








0





$begingroup$


I want to show that the Normal distribution is a member of the exponential family.



I have been working under the assumption that a distribution is a member of the exponential family if its pdf/pmf can be transformed into the form:



$f(x|theta) = h(x)c(theta)exp{sumlimits_{i=1}^{k} w_{i}(theta)t_{i}(x)}$



This is my approach:



$f(x|mu, sigma^2) = frac{1}{sqrt{2pi sigma^2}}exp{-frac{(x-mu)^2}{2 sigma^2}}$



Taking the logs:



$log f(x|mu, sigma^2) = -frac{1}{2}log(2pisigma^2) - frac{(x-mu)^2}{2 sigma^2}$



Taking the exponential:



$f(x|mu, sigma^2) = exp{-frac{1}{2}log(2pisigma^2)-frac{(x-mu)^2}{2sigma^2}}$



= $exp{-frac{1}{2}log(2pisigma^2)-frac{(x^2 -2mu + mu^2)}{2sigma^2}}$



= $exp{-frac{1}{2}log(2pisigma^2)-frac{x^2}{2sigma^2} + frac{2xmu}{2sigma^2} - frac{mu^2}{2sigma^2}}$



= $exp{-frac{1}{2}log(2pisigma^2)} exp{-frac{x^2}{2sigma^2} + frac{xmu}{sigma^2} - frac{mu^2}{2sigma^2}}$



= $frac{1}{sqrt{2pisigma^2}} exp{-frac{x^2}{2sigma^2} + frac{xmu}{sigma^2} - frac{mu^2}{2sigma^2}}$



= $frac{1}{sqrt{2pisigma^2}} exp{-frac{mu^2}{2sigma^2}} exp{-frac{x^2}{2sigma^2} + frac{xmu}{sigma^2}}$



Now I have:



$c(theta) = frac{1}{sqrt{2pisigma^2}} exp{-frac{mu^2}{2sigma^2}}$,



$w_{1}(theta) = -frac{1}{2sigma^2}$,



$t_{1}(x) = x^2$,



$w_{2}(theta) = frac{mu}{sigma^2}$,



$t_{1}(x) = x$



But I am missing $h(x)$.



Am I missing something?






statistics normal-distribution







share|cite|improve this question









$endgroup$




I want to show that the Normal distribution is a member of the exponential family.



I have been working under the assumption that a distribution is a member of the exponential family if its pdf/pmf can be transformed into the form:



$f(x|theta) = h(x)c(theta)exp{sumlimits_{i=1}^{k} w_{i}(theta)t_{i}(x)}$



This is my approach:



$f(x|mu, sigma^2) = frac{1}{sqrt{2pi sigma^2}}exp{-frac{(x-mu)^2}{2 sigma^2}}$



Taking the logs:



$log f(x|mu, sigma^2) = -frac{1}{2}log(2pisigma^2) - frac{(x-mu)^2}{2 sigma^2}$



Taking the exponential:



$f(x|mu, sigma^2) = exp{-frac{1}{2}log(2pisigma^2)-frac{(x-mu)^2}{2sigma^2}}$



= $exp{-frac{1}{2}log(2pisigma^2)-frac{(x^2 -2mu + mu^2)}{2sigma^2}}$



= $exp{-frac{1}{2}log(2pisigma^2)-frac{x^2}{2sigma^2} + frac{2xmu}{2sigma^2} - frac{mu^2}{2sigma^2}}$



= $exp{-frac{1}{2}log(2pisigma^2)} exp{-frac{x^2}{2sigma^2} + frac{xmu}{sigma^2} - frac{mu^2}{2sigma^2}}$



= $frac{1}{sqrt{2pisigma^2}} exp{-frac{x^2}{2sigma^2} + frac{xmu}{sigma^2} - frac{mu^2}{2sigma^2}}$



= $frac{1}{sqrt{2pisigma^2}} exp{-frac{mu^2}{2sigma^2}} exp{-frac{x^2}{2sigma^2} + frac{xmu}{sigma^2}}$



Now I have:



$c(theta) = frac{1}{sqrt{2pisigma^2}} exp{-frac{mu^2}{2sigma^2}}$,



$w_{1}(theta) = -frac{1}{2sigma^2}$,



$t_{1}(x) = x^2$,



$w_{2}(theta) = frac{mu}{sigma^2}$,



$t_{1}(x) = x$



But I am missing $h(x)$.



Am I missing something?






statistics normal-distribution




statistics normal-distribution






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 17 '18 at 21:17









cmplx96cmplx96

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1224








  • 2




    $begingroup$
    $h(x)$ is allowed to be an identity function (value of 1 $forall x$)
    $endgroup$
    – Easymode44
    Dec 17 '18 at 21:47






  • 1




    $begingroup$
    ...is a member of an exponential family.
    $endgroup$
    – Did
    Dec 17 '18 at 22:26














  • 2




    $begingroup$
    $h(x)$ is allowed to be an identity function (value of 1 $forall x$)
    $endgroup$
    – Easymode44
    Dec 17 '18 at 21:47






  • 1




    $begingroup$
    ...is a member of an exponential family.
    $endgroup$
    – Did
    Dec 17 '18 at 22:26








2




2




$begingroup$
$h(x)$ is allowed to be an identity function (value of 1 $forall x$)
$endgroup$
– Easymode44
Dec 17 '18 at 21:47




$begingroup$
$h(x)$ is allowed to be an identity function (value of 1 $forall x$)
$endgroup$
– Easymode44
Dec 17 '18 at 21:47




1




1




$begingroup$
...is a member of an exponential family.
$endgroup$
– Did
Dec 17 '18 at 22:26




$begingroup$
...is a member of an exponential family.
$endgroup$
– Did
Dec 17 '18 at 22:26










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