Exponential Distribution - ML estimator of λ in τ parametrization












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I am stuck with the following problem: We know that $x_1 cdots x_n$ ~ Ex$(1/lambda)$ and we are given the sum $sum_{i=1}^n x_i$. We want to find the maximum likelihood estimate of $lambda$, as well as its variance. Also we want to find a $(1-alpha)$ - confidence interval.



I know that E $ [x_i] = 1/lambda $ and that the likelihood is $L(frac{1}{lambda} | x_1cdots x_n) = lambda^n e^{-sum_i x_ilambda}$. Therefore the ML estimate should be
begin{align}
hat{lambda}= frac{n}{sum_i x_i}.
end{align}
However, I don't know how to calculate the expectation value or the variance of $hat{lambda}$ as this involes $E[frac{1}{sum_i x_i}]$ and I know that normally
begin{align}
E [frac{1}{X}] > frac{1}{E[X]}.
end{align}
Therefore I guess that E$[hat{lambda}] > lambda$, but I don't know how I would calculate it. Thank you already.






statistics exponential-distribution







share|cite|improve this question






















  • $sum X_i sim Gamma(n,1/lambda)$, then find its variance.
    – John_Wick
    Nov 22 at 20:51












  • So var $[sum X_i ] = n/lambda^2$ ? I still don't see how this would help me as var $[1/X] neq 1/$var$[X]$, right?
    – F. Moe
    Nov 22 at 21:04










  • If $X$ is a random variable with support $Omega$ and pdf $f_X$, then $E(1/X)=int_Omega big(frac{1}{t}big)f_X(t)dt$, so like was mentioned use $Xsim Gamma(n,1/lambda)$
    – gd1035
    Nov 22 at 22:04


















0














I am stuck with the following problem: We know that $x_1 cdots x_n$ ~ Ex$(1/lambda)$ and we are given the sum $sum_{i=1}^n x_i$. We want to find the maximum likelihood estimate of $lambda$, as well as its variance. Also we want to find a $(1-alpha)$ - confidence interval.



I know that E $ [x_i] = 1/lambda $ and that the likelihood is $L(frac{1}{lambda} | x_1cdots x_n) = lambda^n e^{-sum_i x_ilambda}$. Therefore the ML estimate should be
begin{align}
hat{lambda}= frac{n}{sum_i x_i}.
end{align}
However, I don't know how to calculate the expectation value or the variance of $hat{lambda}$ as this involes $E[frac{1}{sum_i x_i}]$ and I know that normally
begin{align}
E [frac{1}{X}] > frac{1}{E[X]}.
end{align}
Therefore I guess that E$[hat{lambda}] > lambda$, but I don't know how I would calculate it. Thank you already.






statistics exponential-distribution







share|cite|improve this question






















  • $sum X_i sim Gamma(n,1/lambda)$, then find its variance.
    – John_Wick
    Nov 22 at 20:51












  • So var $[sum X_i ] = n/lambda^2$ ? I still don't see how this would help me as var $[1/X] neq 1/$var$[X]$, right?
    – F. Moe
    Nov 22 at 21:04










  • If $X$ is a random variable with support $Omega$ and pdf $f_X$, then $E(1/X)=int_Omega big(frac{1}{t}big)f_X(t)dt$, so like was mentioned use $Xsim Gamma(n,1/lambda)$
    – gd1035
    Nov 22 at 22:04
















0












0








0







I am stuck with the following problem: We know that $x_1 cdots x_n$ ~ Ex$(1/lambda)$ and we are given the sum $sum_{i=1}^n x_i$. We want to find the maximum likelihood estimate of $lambda$, as well as its variance. Also we want to find a $(1-alpha)$ - confidence interval.



I know that E $ [x_i] = 1/lambda $ and that the likelihood is $L(frac{1}{lambda} | x_1cdots x_n) = lambda^n e^{-sum_i x_ilambda}$. Therefore the ML estimate should be
begin{align}
hat{lambda}= frac{n}{sum_i x_i}.
end{align}
However, I don't know how to calculate the expectation value or the variance of $hat{lambda}$ as this involes $E[frac{1}{sum_i x_i}]$ and I know that normally
begin{align}
E [frac{1}{X}] > frac{1}{E[X]}.
end{align}
Therefore I guess that E$[hat{lambda}] > lambda$, but I don't know how I would calculate it. Thank you already.






statistics exponential-distribution







share|cite|improve this question













I am stuck with the following problem: We know that $x_1 cdots x_n$ ~ Ex$(1/lambda)$ and we are given the sum $sum_{i=1}^n x_i$. We want to find the maximum likelihood estimate of $lambda$, as well as its variance. Also we want to find a $(1-alpha)$ - confidence interval.



I know that E $ [x_i] = 1/lambda $ and that the likelihood is $L(frac{1}{lambda} | x_1cdots x_n) = lambda^n e^{-sum_i x_ilambda}$. Therefore the ML estimate should be
begin{align}
hat{lambda}= frac{n}{sum_i x_i}.
end{align}
However, I don't know how to calculate the expectation value or the variance of $hat{lambda}$ as this involes $E[frac{1}{sum_i x_i}]$ and I know that normally
begin{align}
E [frac{1}{X}] > frac{1}{E[X]}.
end{align}
Therefore I guess that E$[hat{lambda}] > lambda$, but I don't know how I would calculate it. Thank you already.






statistics exponential-distribution




statistics exponential-distribution






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 at 20:47









F. Moe

205




205












  • $sum X_i sim Gamma(n,1/lambda)$, then find its variance.
    – John_Wick
    Nov 22 at 20:51












  • So var $[sum X_i ] = n/lambda^2$ ? I still don't see how this would help me as var $[1/X] neq 1/$var$[X]$, right?
    – F. Moe
    Nov 22 at 21:04










  • If $X$ is a random variable with support $Omega$ and pdf $f_X$, then $E(1/X)=int_Omega big(frac{1}{t}big)f_X(t)dt$, so like was mentioned use $Xsim Gamma(n,1/lambda)$
    – gd1035
    Nov 22 at 22:04




















  • $sum X_i sim Gamma(n,1/lambda)$, then find its variance.
    – John_Wick
    Nov 22 at 20:51












  • So var $[sum X_i ] = n/lambda^2$ ? I still don't see how this would help me as var $[1/X] neq 1/$var$[X]$, right?
    – F. Moe
    Nov 22 at 21:04










  • If $X$ is a random variable with support $Omega$ and pdf $f_X$, then $E(1/X)=int_Omega big(frac{1}{t}big)f_X(t)dt$, so like was mentioned use $Xsim Gamma(n,1/lambda)$
    – gd1035
    Nov 22 at 22:04


















$sum X_i sim Gamma(n,1/lambda)$, then find its variance.
– John_Wick
Nov 22 at 20:51






$sum X_i sim Gamma(n,1/lambda)$, then find its variance.
– John_Wick
Nov 22 at 20:51














So var $[sum X_i ] = n/lambda^2$ ? I still don't see how this would help me as var $[1/X] neq 1/$var$[X]$, right?
– F. Moe
Nov 22 at 21:04




So var $[sum X_i ] = n/lambda^2$ ? I still don't see how this would help me as var $[1/X] neq 1/$var$[X]$, right?
– F. Moe
Nov 22 at 21:04












If $X$ is a random variable with support $Omega$ and pdf $f_X$, then $E(1/X)=int_Omega big(frac{1}{t}big)f_X(t)dt$, so like was mentioned use $Xsim Gamma(n,1/lambda)$
– gd1035
Nov 22 at 22:04






If $X$ is a random variable with support $Omega$ and pdf $f_X$, then $E(1/X)=int_Omega big(frac{1}{t}big)f_X(t)dt$, so like was mentioned use $Xsim Gamma(n,1/lambda)$
– gd1035
Nov 22 at 22:04












1 Answer
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$S=sum X_i sim Gamma(n, 1/lambda).$ If $Ssim Gamma(n,a)$ then $E(S^t)=int_0^infty x^t frac{1}{Gamma(n)a^n}e^{-x/a}x^{n-1}dx=frac{Gamma(n+t)a^{n+t}}{Gamma(n)a^n}int_0^infty frac{1}{Gamma(n+t)a^{n+t}}e^{-x/a}x^{n+t-1}dx=frac{Gamma(n+t)a^{n+t}}{Gamma(n)a^n}$ for all $t>-n$.
$Var(n/S)=E(n^2/S^2)-E^2(n/S)=n^2left(frac{Gamma(n-2)a^{n-2}}{Gamma(n)a^n}-left(frac{Gamma(n-1)a^{n-1}}{Gamma(n)a^n}right)^2right)=n^2left(frac{1}{(n-1)(n-2)a^2}-left(frac{1}{(n-1)a}right)^2right)=frac{n^2}{a^2(n-1)^2(n-2)}.$






share|cite|improve this answer























  • Thank you for clarifying! In your calculation of $E(S^t)$ you forgot the $a$ in the exponent, right? So it would be $e^{-x/a}$. Am I right that $E(hat{lambda})$ then is $E(hat{lambda}) = n frac{Gamma(n-1)}{Gamma(n)} lambda$?
    – F. Moe
    Nov 22 at 22:53












  • Yes, that is correct.
    – John_Wick
    Nov 22 at 22:55










  • Great! Just one more question: will the confidence interval for $lambda$ be just taken from the regular Gamma quantiles, i.e. from Gamma$(1,frac{1}{lambda})$ as Ex$(frac{1}{lambda}) = $Gamma$(1,frac{1}{lambda})$ ?
    – F. Moe
    Nov 22 at 23:08












  • Yes, you can find the bounds for CI using quantiles for Gamma.
    – John_Wick
    Nov 22 at 23:10










  • Thanks again. You really helped me clarify all this in my mind!
    – F. Moe
    Nov 22 at 23:12











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$S=sum X_i sim Gamma(n, 1/lambda).$ If $Ssim Gamma(n,a)$ then $E(S^t)=int_0^infty x^t frac{1}{Gamma(n)a^n}e^{-x/a}x^{n-1}dx=frac{Gamma(n+t)a^{n+t}}{Gamma(n)a^n}int_0^infty frac{1}{Gamma(n+t)a^{n+t}}e^{-x/a}x^{n+t-1}dx=frac{Gamma(n+t)a^{n+t}}{Gamma(n)a^n}$ for all $t>-n$.
$Var(n/S)=E(n^2/S^2)-E^2(n/S)=n^2left(frac{Gamma(n-2)a^{n-2}}{Gamma(n)a^n}-left(frac{Gamma(n-1)a^{n-1}}{Gamma(n)a^n}right)^2right)=n^2left(frac{1}{(n-1)(n-2)a^2}-left(frac{1}{(n-1)a}right)^2right)=frac{n^2}{a^2(n-1)^2(n-2)}.$






share|cite|improve this answer























  • Thank you for clarifying! In your calculation of $E(S^t)$ you forgot the $a$ in the exponent, right? So it would be $e^{-x/a}$. Am I right that $E(hat{lambda})$ then is $E(hat{lambda}) = n frac{Gamma(n-1)}{Gamma(n)} lambda$?
    – F. Moe
    Nov 22 at 22:53












  • Yes, that is correct.
    – John_Wick
    Nov 22 at 22:55










  • Great! Just one more question: will the confidence interval for $lambda$ be just taken from the regular Gamma quantiles, i.e. from Gamma$(1,frac{1}{lambda})$ as Ex$(frac{1}{lambda}) = $Gamma$(1,frac{1}{lambda})$ ?
    – F. Moe
    Nov 22 at 23:08












  • Yes, you can find the bounds for CI using quantiles for Gamma.
    – John_Wick
    Nov 22 at 23:10










  • Thanks again. You really helped me clarify all this in my mind!
    – F. Moe
    Nov 22 at 23:12
















1














$S=sum X_i sim Gamma(n, 1/lambda).$ If $Ssim Gamma(n,a)$ then $E(S^t)=int_0^infty x^t frac{1}{Gamma(n)a^n}e^{-x/a}x^{n-1}dx=frac{Gamma(n+t)a^{n+t}}{Gamma(n)a^n}int_0^infty frac{1}{Gamma(n+t)a^{n+t}}e^{-x/a}x^{n+t-1}dx=frac{Gamma(n+t)a^{n+t}}{Gamma(n)a^n}$ for all $t>-n$.
$Var(n/S)=E(n^2/S^2)-E^2(n/S)=n^2left(frac{Gamma(n-2)a^{n-2}}{Gamma(n)a^n}-left(frac{Gamma(n-1)a^{n-1}}{Gamma(n)a^n}right)^2right)=n^2left(frac{1}{(n-1)(n-2)a^2}-left(frac{1}{(n-1)a}right)^2right)=frac{n^2}{a^2(n-1)^2(n-2)}.$






share|cite|improve this answer























  • Thank you for clarifying! In your calculation of $E(S^t)$ you forgot the $a$ in the exponent, right? So it would be $e^{-x/a}$. Am I right that $E(hat{lambda})$ then is $E(hat{lambda}) = n frac{Gamma(n-1)}{Gamma(n)} lambda$?
    – F. Moe
    Nov 22 at 22:53












  • Yes, that is correct.
    – John_Wick
    Nov 22 at 22:55










  • Great! Just one more question: will the confidence interval for $lambda$ be just taken from the regular Gamma quantiles, i.e. from Gamma$(1,frac{1}{lambda})$ as Ex$(frac{1}{lambda}) = $Gamma$(1,frac{1}{lambda})$ ?
    – F. Moe
    Nov 22 at 23:08












  • Yes, you can find the bounds for CI using quantiles for Gamma.
    – John_Wick
    Nov 22 at 23:10










  • Thanks again. You really helped me clarify all this in my mind!
    – F. Moe
    Nov 22 at 23:12














1












1








1






$S=sum X_i sim Gamma(n, 1/lambda).$ If $Ssim Gamma(n,a)$ then $E(S^t)=int_0^infty x^t frac{1}{Gamma(n)a^n}e^{-x/a}x^{n-1}dx=frac{Gamma(n+t)a^{n+t}}{Gamma(n)a^n}int_0^infty frac{1}{Gamma(n+t)a^{n+t}}e^{-x/a}x^{n+t-1}dx=frac{Gamma(n+t)a^{n+t}}{Gamma(n)a^n}$ for all $t>-n$.
$Var(n/S)=E(n^2/S^2)-E^2(n/S)=n^2left(frac{Gamma(n-2)a^{n-2}}{Gamma(n)a^n}-left(frac{Gamma(n-1)a^{n-1}}{Gamma(n)a^n}right)^2right)=n^2left(frac{1}{(n-1)(n-2)a^2}-left(frac{1}{(n-1)a}right)^2right)=frac{n^2}{a^2(n-1)^2(n-2)}.$






share|cite|improve this answer














$S=sum X_i sim Gamma(n, 1/lambda).$ If $Ssim Gamma(n,a)$ then $E(S^t)=int_0^infty x^t frac{1}{Gamma(n)a^n}e^{-x/a}x^{n-1}dx=frac{Gamma(n+t)a^{n+t}}{Gamma(n)a^n}int_0^infty frac{1}{Gamma(n+t)a^{n+t}}e^{-x/a}x^{n+t-1}dx=frac{Gamma(n+t)a^{n+t}}{Gamma(n)a^n}$ for all $t>-n$.
$Var(n/S)=E(n^2/S^2)-E^2(n/S)=n^2left(frac{Gamma(n-2)a^{n-2}}{Gamma(n)a^n}-left(frac{Gamma(n-1)a^{n-1}}{Gamma(n)a^n}right)^2right)=n^2left(frac{1}{(n-1)(n-2)a^2}-left(frac{1}{(n-1)a}right)^2right)=frac{n^2}{a^2(n-1)^2(n-2)}.$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 22 at 22:56

























answered Nov 22 at 21:19









John_Wick

1,346111




1,346111












  • Thank you for clarifying! In your calculation of $E(S^t)$ you forgot the $a$ in the exponent, right? So it would be $e^{-x/a}$. Am I right that $E(hat{lambda})$ then is $E(hat{lambda}) = n frac{Gamma(n-1)}{Gamma(n)} lambda$?
    – F. Moe
    Nov 22 at 22:53












  • Yes, that is correct.
    – John_Wick
    Nov 22 at 22:55










  • Great! Just one more question: will the confidence interval for $lambda$ be just taken from the regular Gamma quantiles, i.e. from Gamma$(1,frac{1}{lambda})$ as Ex$(frac{1}{lambda}) = $Gamma$(1,frac{1}{lambda})$ ?
    – F. Moe
    Nov 22 at 23:08












  • Yes, you can find the bounds for CI using quantiles for Gamma.
    – John_Wick
    Nov 22 at 23:10










  • Thanks again. You really helped me clarify all this in my mind!
    – F. Moe
    Nov 22 at 23:12


















  • Thank you for clarifying! In your calculation of $E(S^t)$ you forgot the $a$ in the exponent, right? So it would be $e^{-x/a}$. Am I right that $E(hat{lambda})$ then is $E(hat{lambda}) = n frac{Gamma(n-1)}{Gamma(n)} lambda$?
    – F. Moe
    Nov 22 at 22:53












  • Yes, that is correct.
    – John_Wick
    Nov 22 at 22:55










  • Great! Just one more question: will the confidence interval for $lambda$ be just taken from the regular Gamma quantiles, i.e. from Gamma$(1,frac{1}{lambda})$ as Ex$(frac{1}{lambda}) = $Gamma$(1,frac{1}{lambda})$ ?
    – F. Moe
    Nov 22 at 23:08












  • Yes, you can find the bounds for CI using quantiles for Gamma.
    – John_Wick
    Nov 22 at 23:10










  • Thanks again. You really helped me clarify all this in my mind!
    – F. Moe
    Nov 22 at 23:12
















Thank you for clarifying! In your calculation of $E(S^t)$ you forgot the $a$ in the exponent, right? So it would be $e^{-x/a}$. Am I right that $E(hat{lambda})$ then is $E(hat{lambda}) = n frac{Gamma(n-1)}{Gamma(n)} lambda$?
– F. Moe
Nov 22 at 22:53






Thank you for clarifying! In your calculation of $E(S^t)$ you forgot the $a$ in the exponent, right? So it would be $e^{-x/a}$. Am I right that $E(hat{lambda})$ then is $E(hat{lambda}) = n frac{Gamma(n-1)}{Gamma(n)} lambda$?
– F. Moe
Nov 22 at 22:53














Yes, that is correct.
– John_Wick
Nov 22 at 22:55




Yes, that is correct.
– John_Wick
Nov 22 at 22:55












Great! Just one more question: will the confidence interval for $lambda$ be just taken from the regular Gamma quantiles, i.e. from Gamma$(1,frac{1}{lambda})$ as Ex$(frac{1}{lambda}) = $Gamma$(1,frac{1}{lambda})$ ?
– F. Moe
Nov 22 at 23:08






Great! Just one more question: will the confidence interval for $lambda$ be just taken from the regular Gamma quantiles, i.e. from Gamma$(1,frac{1}{lambda})$ as Ex$(frac{1}{lambda}) = $Gamma$(1,frac{1}{lambda})$ ?
– F. Moe
Nov 22 at 23:08














Yes, you can find the bounds for CI using quantiles for Gamma.
– John_Wick
Nov 22 at 23:10




Yes, you can find the bounds for CI using quantiles for Gamma.
– John_Wick
Nov 22 at 23:10












Thanks again. You really helped me clarify all this in my mind!
– F. Moe
Nov 22 at 23:12




Thanks again. You really helped me clarify all this in my mind!
– F. Moe
Nov 22 at 23:12


















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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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