Why is this estimator biased?












2












$begingroup$


$X_{1},X_{2},..,X_{n}$ are iid $sim Poisson(mu)$



than the MLE for $theta=e^{-mu}$ is $hat theta =e^{-bar x}$



Why is this considered to be biased for $theta$?



Is $E[hat theta]$ not $theta$ ?



as



$E[bar x]= mu$






estimation poisson-distribution bias







share|cite|improve this question









$endgroup$








  • 6




    $begingroup$
    E[f(X)] != f(E[X]) in general.
    $endgroup$
    – The Laconic
    Mar 22 at 20:54












  • $begingroup$
    I know that general result but I am still confused how to do it here since the sum of poisson is also poisson with mean $n mu$
    $endgroup$
    – Quality
    Mar 22 at 21:01
















2












$begingroup$


$X_{1},X_{2},..,X_{n}$ are iid $sim Poisson(mu)$



than the MLE for $theta=e^{-mu}$ is $hat theta =e^{-bar x}$



Why is this considered to be biased for $theta$?



Is $E[hat theta]$ not $theta$ ?



as



$E[bar x]= mu$






estimation poisson-distribution bias







share|cite|improve this question









$endgroup$








  • 6




    $begingroup$
    E[f(X)] != f(E[X]) in general.
    $endgroup$
    – The Laconic
    Mar 22 at 20:54












  • $begingroup$
    I know that general result but I am still confused how to do it here since the sum of poisson is also poisson with mean $n mu$
    $endgroup$
    – Quality
    Mar 22 at 21:01














2












2








2


2



$begingroup$


$X_{1},X_{2},..,X_{n}$ are iid $sim Poisson(mu)$



than the MLE for $theta=e^{-mu}$ is $hat theta =e^{-bar x}$



Why is this considered to be biased for $theta$?



Is $E[hat theta]$ not $theta$ ?



as



$E[bar x]= mu$






estimation poisson-distribution bias







share|cite|improve this question









$endgroup$




$X_{1},X_{2},..,X_{n}$ are iid $sim Poisson(mu)$



than the MLE for $theta=e^{-mu}$ is $hat theta =e^{-bar x}$



Why is this considered to be biased for $theta$?



Is $E[hat theta]$ not $theta$ ?



as



$E[bar x]= mu$






estimation poisson-distribution bias




estimation poisson-distribution bias






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 22 at 20:47









QualityQuality

30019




30019








  • 6




    $begingroup$
    E[f(X)] != f(E[X]) in general.
    $endgroup$
    – The Laconic
    Mar 22 at 20:54












  • $begingroup$
    I know that general result but I am still confused how to do it here since the sum of poisson is also poisson with mean $n mu$
    $endgroup$
    – Quality
    Mar 22 at 21:01














  • 6




    $begingroup$
    E[f(X)] != f(E[X]) in general.
    $endgroup$
    – The Laconic
    Mar 22 at 20:54












  • $begingroup$
    I know that general result but I am still confused how to do it here since the sum of poisson is also poisson with mean $n mu$
    $endgroup$
    – Quality
    Mar 22 at 21:01








6




6




$begingroup$
E[f(X)] != f(E[X]) in general.
$endgroup$
– The Laconic
Mar 22 at 20:54






$begingroup$
E[f(X)] != f(E[X]) in general.
$endgroup$
– The Laconic
Mar 22 at 20:54














$begingroup$
I know that general result but I am still confused how to do it here since the sum of poisson is also poisson with mean $n mu$
$endgroup$
– Quality
Mar 22 at 21:01




$begingroup$
I know that general result but I am still confused how to do it here since the sum of poisson is also poisson with mean $n mu$
$endgroup$
– Quality
Mar 22 at 21:01










2 Answers
2






active

oldest

votes


















5












$begingroup$

Recall that the moment generating function of $X sim operatorname{Poisson}(mu)$ is
$$
E[e^{s X}] = exp(mu(e^s - 1))
$$
for all $s in mathbb{R}$



Proof.
We just compute:
$$
begin{aligned}
E[e^{s X}]
&= sum_{k=0}^infty e^{s k} e^{-mu} frac{mu^k}{k!} \
&= e^{-mu} sum_{k=0}^infty frac{left(mu e^sright)^k}{k!} \
&= e^{-mu} e^{mu e^s} \
&= exp(mu(e^s - 1)).
end{aligned}
$$
We will use this result with $s = -1/n$.



If $X_1, ldots, X_n sim operatorname{Poisson}(mu)$ are i.i.d. and
$$
widehat{theta}
= expleft(-frac{1}{n} sum_{i=1}^n X_iright)
= prod_{i=1}^n expleft(-frac{X_i}{n}right),
$$
then, using independence,
$$
begin{aligned}
E[widehat{theta}]
&= prod_{i=1}^n Eleft[e^{-X_i / n}right] \
&= prod_{i=1}^n exp(mu(e^{-1/n} - 1)) \
&= exp(n mu(e^{-1/n} - 1)) \
&neq e^{-mu},
end{aligned}
$$
so $widehat{theta}$ is not an unbiased estimator of $e^{-mu}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, makes perfect sense
    $endgroup$
    – Quality
    Mar 22 at 21:16



















6












$begingroup$

As an aside from finding the exact expectation, you can use Jensen's inequality, which says that for a random variable $X$ and a convex function $g$,



$$Eleft[g(X)right]ge gleft(Eleft[Xright]right)$$



, provided the expectations exist.



Verify that $g(x)=e^{-x}$ is a convex function from the fact that $g''>0$.



The equality does not hold because $g$ is not an affine function or a constant function.






share|cite|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    Recall that the moment generating function of $X sim operatorname{Poisson}(mu)$ is
    $$
    E[e^{s X}] = exp(mu(e^s - 1))
    $$
    for all $s in mathbb{R}$



    Proof.
    We just compute:
    $$
    begin{aligned}
    E[e^{s X}]
    &= sum_{k=0}^infty e^{s k} e^{-mu} frac{mu^k}{k!} \
    &= e^{-mu} sum_{k=0}^infty frac{left(mu e^sright)^k}{k!} \
    &= e^{-mu} e^{mu e^s} \
    &= exp(mu(e^s - 1)).
    end{aligned}
    $$
    We will use this result with $s = -1/n$.



    If $X_1, ldots, X_n sim operatorname{Poisson}(mu)$ are i.i.d. and
    $$
    widehat{theta}
    = expleft(-frac{1}{n} sum_{i=1}^n X_iright)
    = prod_{i=1}^n expleft(-frac{X_i}{n}right),
    $$
    then, using independence,
    $$
    begin{aligned}
    E[widehat{theta}]
    &= prod_{i=1}^n Eleft[e^{-X_i / n}right] \
    &= prod_{i=1}^n exp(mu(e^{-1/n} - 1)) \
    &= exp(n mu(e^{-1/n} - 1)) \
    &neq e^{-mu},
    end{aligned}
    $$
    so $widehat{theta}$ is not an unbiased estimator of $e^{-mu}$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, makes perfect sense
      $endgroup$
      – Quality
      Mar 22 at 21:16
















    5












    $begingroup$

    Recall that the moment generating function of $X sim operatorname{Poisson}(mu)$ is
    $$
    E[e^{s X}] = exp(mu(e^s - 1))
    $$
    for all $s in mathbb{R}$



    Proof.
    We just compute:
    $$
    begin{aligned}
    E[e^{s X}]
    &= sum_{k=0}^infty e^{s k} e^{-mu} frac{mu^k}{k!} \
    &= e^{-mu} sum_{k=0}^infty frac{left(mu e^sright)^k}{k!} \
    &= e^{-mu} e^{mu e^s} \
    &= exp(mu(e^s - 1)).
    end{aligned}
    $$
    We will use this result with $s = -1/n$.



    If $X_1, ldots, X_n sim operatorname{Poisson}(mu)$ are i.i.d. and
    $$
    widehat{theta}
    = expleft(-frac{1}{n} sum_{i=1}^n X_iright)
    = prod_{i=1}^n expleft(-frac{X_i}{n}right),
    $$
    then, using independence,
    $$
    begin{aligned}
    E[widehat{theta}]
    &= prod_{i=1}^n Eleft[e^{-X_i / n}right] \
    &= prod_{i=1}^n exp(mu(e^{-1/n} - 1)) \
    &= exp(n mu(e^{-1/n} - 1)) \
    &neq e^{-mu},
    end{aligned}
    $$
    so $widehat{theta}$ is not an unbiased estimator of $e^{-mu}$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, makes perfect sense
      $endgroup$
      – Quality
      Mar 22 at 21:16














    5












    5








    5





    $begingroup$

    Recall that the moment generating function of $X sim operatorname{Poisson}(mu)$ is
    $$
    E[e^{s X}] = exp(mu(e^s - 1))
    $$
    for all $s in mathbb{R}$



    Proof.
    We just compute:
    $$
    begin{aligned}
    E[e^{s X}]
    &= sum_{k=0}^infty e^{s k} e^{-mu} frac{mu^k}{k!} \
    &= e^{-mu} sum_{k=0}^infty frac{left(mu e^sright)^k}{k!} \
    &= e^{-mu} e^{mu e^s} \
    &= exp(mu(e^s - 1)).
    end{aligned}
    $$
    We will use this result with $s = -1/n$.



    If $X_1, ldots, X_n sim operatorname{Poisson}(mu)$ are i.i.d. and
    $$
    widehat{theta}
    = expleft(-frac{1}{n} sum_{i=1}^n X_iright)
    = prod_{i=1}^n expleft(-frac{X_i}{n}right),
    $$
    then, using independence,
    $$
    begin{aligned}
    E[widehat{theta}]
    &= prod_{i=1}^n Eleft[e^{-X_i / n}right] \
    &= prod_{i=1}^n exp(mu(e^{-1/n} - 1)) \
    &= exp(n mu(e^{-1/n} - 1)) \
    &neq e^{-mu},
    end{aligned}
    $$
    so $widehat{theta}$ is not an unbiased estimator of $e^{-mu}$






    share|cite|improve this answer









    $endgroup$



    Recall that the moment generating function of $X sim operatorname{Poisson}(mu)$ is
    $$
    E[e^{s X}] = exp(mu(e^s - 1))
    $$
    for all $s in mathbb{R}$



    Proof.
    We just compute:
    $$
    begin{aligned}
    E[e^{s X}]
    &= sum_{k=0}^infty e^{s k} e^{-mu} frac{mu^k}{k!} \
    &= e^{-mu} sum_{k=0}^infty frac{left(mu e^sright)^k}{k!} \
    &= e^{-mu} e^{mu e^s} \
    &= exp(mu(e^s - 1)).
    end{aligned}
    $$
    We will use this result with $s = -1/n$.



    If $X_1, ldots, X_n sim operatorname{Poisson}(mu)$ are i.i.d. and
    $$
    widehat{theta}
    = expleft(-frac{1}{n} sum_{i=1}^n X_iright)
    = prod_{i=1}^n expleft(-frac{X_i}{n}right),
    $$
    then, using independence,
    $$
    begin{aligned}
    E[widehat{theta}]
    &= prod_{i=1}^n Eleft[e^{-X_i / n}right] \
    &= prod_{i=1}^n exp(mu(e^{-1/n} - 1)) \
    &= exp(n mu(e^{-1/n} - 1)) \
    &neq e^{-mu},
    end{aligned}
    $$
    so $widehat{theta}$ is not an unbiased estimator of $e^{-mu}$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 22 at 21:12









    Artem MavrinArtem Mavrin

    1,211710




    1,211710












    • $begingroup$
      Thank you, makes perfect sense
      $endgroup$
      – Quality
      Mar 22 at 21:16


















    • $begingroup$
      Thank you, makes perfect sense
      $endgroup$
      – Quality
      Mar 22 at 21:16
















    $begingroup$
    Thank you, makes perfect sense
    $endgroup$
    – Quality
    Mar 22 at 21:16




    $begingroup$
    Thank you, makes perfect sense
    $endgroup$
    – Quality
    Mar 22 at 21:16













    6












    $begingroup$

    As an aside from finding the exact expectation, you can use Jensen's inequality, which says that for a random variable $X$ and a convex function $g$,



    $$Eleft[g(X)right]ge gleft(Eleft[Xright]right)$$



    , provided the expectations exist.



    Verify that $g(x)=e^{-x}$ is a convex function from the fact that $g''>0$.



    The equality does not hold because $g$ is not an affine function or a constant function.






    share|cite|improve this answer









    $endgroup$


















      6












      $begingroup$

      As an aside from finding the exact expectation, you can use Jensen's inequality, which says that for a random variable $X$ and a convex function $g$,



      $$Eleft[g(X)right]ge gleft(Eleft[Xright]right)$$



      , provided the expectations exist.



      Verify that $g(x)=e^{-x}$ is a convex function from the fact that $g''>0$.



      The equality does not hold because $g$ is not an affine function or a constant function.






      share|cite|improve this answer









      $endgroup$
















        6












        6








        6





        $begingroup$

        As an aside from finding the exact expectation, you can use Jensen's inequality, which says that for a random variable $X$ and a convex function $g$,



        $$Eleft[g(X)right]ge gleft(Eleft[Xright]right)$$



        , provided the expectations exist.



        Verify that $g(x)=e^{-x}$ is a convex function from the fact that $g''>0$.



        The equality does not hold because $g$ is not an affine function or a constant function.






        share|cite|improve this answer









        $endgroup$



        As an aside from finding the exact expectation, you can use Jensen's inequality, which says that for a random variable $X$ and a convex function $g$,



        $$Eleft[g(X)right]ge gleft(Eleft[Xright]right)$$



        , provided the expectations exist.



        Verify that $g(x)=e^{-x}$ is a convex function from the fact that $g''>0$.



        The equality does not hold because $g$ is not an affine function or a constant function.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 22 at 21:16









        StubbornAtomStubbornAtom

        2,8621532




        2,8621532






























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