Estimation of parameter by the method of moments












1












$begingroup$


Let $X_1,X_2,dots X_n$ be a random sample from the density $$f(x;theta)=e^{-(x-theta)} e^{-e^{-(x-theta)}}, quad -infty<x<infty ,quad -infty<theta<infty$$
Find the method of moments estimator of $theta$.



What I attempted:-



Here $E(X)=int_{-infty}^{infty} xe^{-(x-theta)} e^{-{e^{-(x-theta)}}}dx=int_{-infty}^{infty} (x-theta)e^{-(x-theta)} e^{-{e^{-(x-theta)}}}dx+theta$.

The integral of the final expression can be written as begin{equation}
begin{aligned}
int_{-infty}^{infty} (x-theta)e^{-(x-theta)} e^{-{e^{-(x-theta)}}}dx&=
int_{-infty}^{infty}t e^{-t} e^{-{e^{-t}}}dt quad mbox{where $t=x-theta$}\
&=int_{0}^{infty} log y hspace{1mm} e^{-y}dy quad mbox{where $y=e^{-t}$}\
&=-log y hspace{1mm} e^{-y}|_0^{infty}+int_{0}^{infty}frac{e^{-y}}{y}dy
end{aligned}
end{equation}



I was not able to proceed beyond that. I don't think that the integral exist.



However, it can be shown that $e^{-X}$ is an exponential variate with mean $e^{-theta}$. Is this information of any help?






integration statistics parameter-estimation







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    yeah $$int_0^infty y^{-1}e^{-y}dy=Gamma(0)$$ which is undefined
    $endgroup$
    – clathratus
    Jan 8 at 5:19










  • $begingroup$
    So ultimately $E(X)$ does not exist.
    $endgroup$
    – user440191
    Jan 8 at 5:21










  • $begingroup$
    Yeah I think so. The integrand $e^{-x}log(x)$ is undefined at $x=0$ Which I'm pretty sure implies divergence
    $endgroup$
    – clathratus
    Jan 8 at 5:23






  • 2




    $begingroup$
    This is a Gumbel distribution with scale parameter unity. So expected value of $X$ should turn out to be $theta+gamma$, $gamma$ being the Euler-Mascheroni constant, as elaborated in the answer below.
    $endgroup$
    – StubbornAtom
    Jan 8 at 6:37
















1












$begingroup$


Let $X_1,X_2,dots X_n$ be a random sample from the density $$f(x;theta)=e^{-(x-theta)} e^{-e^{-(x-theta)}}, quad -infty<x<infty ,quad -infty<theta<infty$$
Find the method of moments estimator of $theta$.



What I attempted:-



Here $E(X)=int_{-infty}^{infty} xe^{-(x-theta)} e^{-{e^{-(x-theta)}}}dx=int_{-infty}^{infty} (x-theta)e^{-(x-theta)} e^{-{e^{-(x-theta)}}}dx+theta$.

The integral of the final expression can be written as begin{equation}
begin{aligned}
int_{-infty}^{infty} (x-theta)e^{-(x-theta)} e^{-{e^{-(x-theta)}}}dx&=
int_{-infty}^{infty}t e^{-t} e^{-{e^{-t}}}dt quad mbox{where $t=x-theta$}\
&=int_{0}^{infty} log y hspace{1mm} e^{-y}dy quad mbox{where $y=e^{-t}$}\
&=-log y hspace{1mm} e^{-y}|_0^{infty}+int_{0}^{infty}frac{e^{-y}}{y}dy
end{aligned}
end{equation}



I was not able to proceed beyond that. I don't think that the integral exist.



However, it can be shown that $e^{-X}$ is an exponential variate with mean $e^{-theta}$. Is this information of any help?






integration statistics parameter-estimation







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    yeah $$int_0^infty y^{-1}e^{-y}dy=Gamma(0)$$ which is undefined
    $endgroup$
    – clathratus
    Jan 8 at 5:19










  • $begingroup$
    So ultimately $E(X)$ does not exist.
    $endgroup$
    – user440191
    Jan 8 at 5:21










  • $begingroup$
    Yeah I think so. The integrand $e^{-x}log(x)$ is undefined at $x=0$ Which I'm pretty sure implies divergence
    $endgroup$
    – clathratus
    Jan 8 at 5:23






  • 2




    $begingroup$
    This is a Gumbel distribution with scale parameter unity. So expected value of $X$ should turn out to be $theta+gamma$, $gamma$ being the Euler-Mascheroni constant, as elaborated in the answer below.
    $endgroup$
    – StubbornAtom
    Jan 8 at 6:37














1












1








1





$begingroup$


Let $X_1,X_2,dots X_n$ be a random sample from the density $$f(x;theta)=e^{-(x-theta)} e^{-e^{-(x-theta)}}, quad -infty<x<infty ,quad -infty<theta<infty$$
Find the method of moments estimator of $theta$.



What I attempted:-



Here $E(X)=int_{-infty}^{infty} xe^{-(x-theta)} e^{-{e^{-(x-theta)}}}dx=int_{-infty}^{infty} (x-theta)e^{-(x-theta)} e^{-{e^{-(x-theta)}}}dx+theta$.

The integral of the final expression can be written as begin{equation}
begin{aligned}
int_{-infty}^{infty} (x-theta)e^{-(x-theta)} e^{-{e^{-(x-theta)}}}dx&=
int_{-infty}^{infty}t e^{-t} e^{-{e^{-t}}}dt quad mbox{where $t=x-theta$}\
&=int_{0}^{infty} log y hspace{1mm} e^{-y}dy quad mbox{where $y=e^{-t}$}\
&=-log y hspace{1mm} e^{-y}|_0^{infty}+int_{0}^{infty}frac{e^{-y}}{y}dy
end{aligned}
end{equation}



I was not able to proceed beyond that. I don't think that the integral exist.



However, it can be shown that $e^{-X}$ is an exponential variate with mean $e^{-theta}$. Is this information of any help?






integration statistics parameter-estimation







share|cite|improve this question









$endgroup$




Let $X_1,X_2,dots X_n$ be a random sample from the density $$f(x;theta)=e^{-(x-theta)} e^{-e^{-(x-theta)}}, quad -infty<x<infty ,quad -infty<theta<infty$$
Find the method of moments estimator of $theta$.



What I attempted:-



Here $E(X)=int_{-infty}^{infty} xe^{-(x-theta)} e^{-{e^{-(x-theta)}}}dx=int_{-infty}^{infty} (x-theta)e^{-(x-theta)} e^{-{e^{-(x-theta)}}}dx+theta$.

The integral of the final expression can be written as begin{equation}
begin{aligned}
int_{-infty}^{infty} (x-theta)e^{-(x-theta)} e^{-{e^{-(x-theta)}}}dx&=
int_{-infty}^{infty}t e^{-t} e^{-{e^{-t}}}dt quad mbox{where $t=x-theta$}\
&=int_{0}^{infty} log y hspace{1mm} e^{-y}dy quad mbox{where $y=e^{-t}$}\
&=-log y hspace{1mm} e^{-y}|_0^{infty}+int_{0}^{infty}frac{e^{-y}}{y}dy
end{aligned}
end{equation}



I was not able to proceed beyond that. I don't think that the integral exist.



However, it can be shown that $e^{-X}$ is an exponential variate with mean $e^{-theta}$. Is this information of any help?






integration statistics parameter-estimation




integration statistics parameter-estimation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 8 at 5:12







user440191















  • 1




    $begingroup$
    yeah $$int_0^infty y^{-1}e^{-y}dy=Gamma(0)$$ which is undefined
    $endgroup$
    – clathratus
    Jan 8 at 5:19










  • $begingroup$
    So ultimately $E(X)$ does not exist.
    $endgroup$
    – user440191
    Jan 8 at 5:21










  • $begingroup$
    Yeah I think so. The integrand $e^{-x}log(x)$ is undefined at $x=0$ Which I'm pretty sure implies divergence
    $endgroup$
    – clathratus
    Jan 8 at 5:23






  • 2




    $begingroup$
    This is a Gumbel distribution with scale parameter unity. So expected value of $X$ should turn out to be $theta+gamma$, $gamma$ being the Euler-Mascheroni constant, as elaborated in the answer below.
    $endgroup$
    – StubbornAtom
    Jan 8 at 6:37














  • 1




    $begingroup$
    yeah $$int_0^infty y^{-1}e^{-y}dy=Gamma(0)$$ which is undefined
    $endgroup$
    – clathratus
    Jan 8 at 5:19










  • $begingroup$
    So ultimately $E(X)$ does not exist.
    $endgroup$
    – user440191
    Jan 8 at 5:21










  • $begingroup$
    Yeah I think so. The integrand $e^{-x}log(x)$ is undefined at $x=0$ Which I'm pretty sure implies divergence
    $endgroup$
    – clathratus
    Jan 8 at 5:23






  • 2




    $begingroup$
    This is a Gumbel distribution with scale parameter unity. So expected value of $X$ should turn out to be $theta+gamma$, $gamma$ being the Euler-Mascheroni constant, as elaborated in the answer below.
    $endgroup$
    – StubbornAtom
    Jan 8 at 6:37








1




1




$begingroup$
yeah $$int_0^infty y^{-1}e^{-y}dy=Gamma(0)$$ which is undefined
$endgroup$
– clathratus
Jan 8 at 5:19




$begingroup$
yeah $$int_0^infty y^{-1}e^{-y}dy=Gamma(0)$$ which is undefined
$endgroup$
– clathratus
Jan 8 at 5:19












$begingroup$
So ultimately $E(X)$ does not exist.
$endgroup$
– user440191
Jan 8 at 5:21




$begingroup$
So ultimately $E(X)$ does not exist.
$endgroup$
– user440191
Jan 8 at 5:21












$begingroup$
Yeah I think so. The integrand $e^{-x}log(x)$ is undefined at $x=0$ Which I'm pretty sure implies divergence
$endgroup$
– clathratus
Jan 8 at 5:23




$begingroup$
Yeah I think so. The integrand $e^{-x}log(x)$ is undefined at $x=0$ Which I'm pretty sure implies divergence
$endgroup$
– clathratus
Jan 8 at 5:23




2




2




$begingroup$
This is a Gumbel distribution with scale parameter unity. So expected value of $X$ should turn out to be $theta+gamma$, $gamma$ being the Euler-Mascheroni constant, as elaborated in the answer below.
$endgroup$
– StubbornAtom
Jan 8 at 6:37




$begingroup$
This is a Gumbel distribution with scale parameter unity. So expected value of $X$ should turn out to be $theta+gamma$, $gamma$ being the Euler-Mascheroni constant, as elaborated in the answer below.
$endgroup$
– StubbornAtom
Jan 8 at 6:37










1 Answer
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oldest

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$begingroup$

Don't use integration by parts on $int_0^infty log (y)e^{-y}dy$. Look closely and you'll find that integration by parts leads to the indeterminate form $-infty +infty$. In fact the integral
$$int_0^infty log( y) e^{-y}dy$$
converges; its value is well known to be the opposite of the Euler-Mascheroni constant.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Note also that you've lost a minus sign in your calculation, occurring during the substitution $y=e^{-t}$.
    $endgroup$
    – grand_chat
    Jan 8 at 6:39












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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

Don't use integration by parts on $int_0^infty log (y)e^{-y}dy$. Look closely and you'll find that integration by parts leads to the indeterminate form $-infty +infty$. In fact the integral
$$int_0^infty log( y) e^{-y}dy$$
converges; its value is well known to be the opposite of the Euler-Mascheroni constant.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Note also that you've lost a minus sign in your calculation, occurring during the substitution $y=e^{-t}$.
    $endgroup$
    – grand_chat
    Jan 8 at 6:39
















2












$begingroup$

Don't use integration by parts on $int_0^infty log (y)e^{-y}dy$. Look closely and you'll find that integration by parts leads to the indeterminate form $-infty +infty$. In fact the integral
$$int_0^infty log( y) e^{-y}dy$$
converges; its value is well known to be the opposite of the Euler-Mascheroni constant.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Note also that you've lost a minus sign in your calculation, occurring during the substitution $y=e^{-t}$.
    $endgroup$
    – grand_chat
    Jan 8 at 6:39














2












2








2





$begingroup$

Don't use integration by parts on $int_0^infty log (y)e^{-y}dy$. Look closely and you'll find that integration by parts leads to the indeterminate form $-infty +infty$. In fact the integral
$$int_0^infty log( y) e^{-y}dy$$
converges; its value is well known to be the opposite of the Euler-Mascheroni constant.






share|cite|improve this answer











$endgroup$



Don't use integration by parts on $int_0^infty log (y)e^{-y}dy$. Look closely and you'll find that integration by parts leads to the indeterminate form $-infty +infty$. In fact the integral
$$int_0^infty log( y) e^{-y}dy$$
converges; its value is well known to be the opposite of the Euler-Mascheroni constant.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 8 at 6:35

























answered Jan 8 at 6:30









grand_chatgrand_chat

20.5k11327




20.5k11327








  • 1




    $begingroup$
    Note also that you've lost a minus sign in your calculation, occurring during the substitution $y=e^{-t}$.
    $endgroup$
    – grand_chat
    Jan 8 at 6:39














  • 1




    $begingroup$
    Note also that you've lost a minus sign in your calculation, occurring during the substitution $y=e^{-t}$.
    $endgroup$
    – grand_chat
    Jan 8 at 6:39








1




1




$begingroup$
Note also that you've lost a minus sign in your calculation, occurring during the substitution $y=e^{-t}$.
$endgroup$
– grand_chat
Jan 8 at 6:39




$begingroup$
Note also that you've lost a minus sign in your calculation, occurring during the substitution $y=e^{-t}$.
$endgroup$
– grand_chat
Jan 8 at 6:39


















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