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Why is this estimator biased?



The Next CEO of Stack OverflowHow do I check for bias of an estimator?Bootstrap confidence interval for a biased estimatorConceptual question about bias of an estimatorEstimator, unbiased or biasedCalculating risk for a density estimatorLinear model with biased estimatorBias Correction for Estimator with known biasWhy is OLS estimator of AR(1) coefficient biased?Cramer-Rao lower bound for biased estimatorVariance of distribution for maximum likelihood estimator










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 $X sim operatornamePoisson(mu)$ is
$$
E[e^s X] = exp(mu(e^s - 1))
$$
for all $s in mathbbR$



Proof.
We just compute:
$$
beginaligned
E[e^s X]
&= sum_k=0^infty e^s k e^-mu fracmu^kk! \
&= e^-mu sum_k=0^infty fracleft(mu e^sright)^kk! \
&= e^-mu e^mu e^s \
&= exp(mu(e^s - 1)).
endaligned
$$
We will use this result with $s = -1/n$.



If $X_1, ldots, X_n sim operatornamePoisson(mu)$ are i.i.d. and
$$
widehattheta
= expleft(-frac1n sum_i=1^n X_iright)
= prod_i=1^n expleft(-fracX_inright),
$$
then, using independence,
$$
beginaligned
E[widehattheta]
&= prod_i=1^n Eleft[e^-X_i / nright] \
&= prod_i=1^n exp(mu(e^-1/n - 1)) \
&= exp(n mu(e^-1/n - 1)) \
&neq e^-mu,
endaligned
$$
so $widehattheta$ 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









$endgroup$













    Your Answer





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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 $X sim operatornamePoisson(mu)$ is
    $$
    E[e^s X] = exp(mu(e^s - 1))
    $$
    for all $s in mathbbR$



    Proof.
    We just compute:
    $$
    beginaligned
    E[e^s X]
    &= sum_k=0^infty e^s k e^-mu fracmu^kk! \
    &= e^-mu sum_k=0^infty fracleft(mu e^sright)^kk! \
    &= e^-mu e^mu e^s \
    &= exp(mu(e^s - 1)).
    endaligned
    $$
    We will use this result with $s = -1/n$.



    If $X_1, ldots, X_n sim operatornamePoisson(mu)$ are i.i.d. and
    $$
    widehattheta
    = expleft(-frac1n sum_i=1^n X_iright)
    = prod_i=1^n expleft(-fracX_inright),
    $$
    then, using independence,
    $$
    beginaligned
    E[widehattheta]
    &= prod_i=1^n Eleft[e^-X_i / nright] \
    &= prod_i=1^n exp(mu(e^-1/n - 1)) \
    &= exp(n mu(e^-1/n - 1)) \
    &neq e^-mu,
    endaligned
    $$
    so $widehattheta$ 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 $X sim operatornamePoisson(mu)$ is
    $$
    E[e^s X] = exp(mu(e^s - 1))
    $$
    for all $s in mathbbR$



    Proof.
    We just compute:
    $$
    beginaligned
    E[e^s X]
    &= sum_k=0^infty e^s k e^-mu fracmu^kk! \
    &= e^-mu sum_k=0^infty fracleft(mu e^sright)^kk! \
    &= e^-mu e^mu e^s \
    &= exp(mu(e^s - 1)).
    endaligned
    $$
    We will use this result with $s = -1/n$.



    If $X_1, ldots, X_n sim operatornamePoisson(mu)$ are i.i.d. and
    $$
    widehattheta
    = expleft(-frac1n sum_i=1^n X_iright)
    = prod_i=1^n expleft(-fracX_inright),
    $$
    then, using independence,
    $$
    beginaligned
    E[widehattheta]
    &= prod_i=1^n Eleft[e^-X_i / nright] \
    &= prod_i=1^n exp(mu(e^-1/n - 1)) \
    &= exp(n mu(e^-1/n - 1)) \
    &neq e^-mu,
    endaligned
    $$
    so $widehattheta$ 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 $X sim operatornamePoisson(mu)$ is
    $$
    E[e^s X] = exp(mu(e^s - 1))
    $$
    for all $s in mathbbR$



    Proof.
    We just compute:
    $$
    beginaligned
    E[e^s X]
    &= sum_k=0^infty e^s k e^-mu fracmu^kk! \
    &= e^-mu sum_k=0^infty fracleft(mu e^sright)^kk! \
    &= e^-mu e^mu e^s \
    &= exp(mu(e^s - 1)).
    endaligned
    $$
    We will use this result with $s = -1/n$.



    If $X_1, ldots, X_n sim operatornamePoisson(mu)$ are i.i.d. and
    $$
    widehattheta
    = expleft(-frac1n sum_i=1^n X_iright)
    = prod_i=1^n expleft(-fracX_inright),
    $$
    then, using independence,
    $$
    beginaligned
    E[widehattheta]
    &= prod_i=1^n Eleft[e^-X_i / nright] \
    &= prod_i=1^n exp(mu(e^-1/n - 1)) \
    &= exp(n mu(e^-1/n - 1)) \
    &neq e^-mu,
    endaligned
    $$
    so $widehattheta$ is not an unbiased estimator of $e^-mu$






    share|cite|improve this answer









    $endgroup$



    Recall that the moment generating function of $X sim operatornamePoisson(mu)$ is
    $$
    E[e^s X] = exp(mu(e^s - 1))
    $$
    for all $s in mathbbR$



    Proof.
    We just compute:
    $$
    beginaligned
    E[e^s X]
    &= sum_k=0^infty e^s k e^-mu fracmu^kk! \
    &= e^-mu sum_k=0^infty fracleft(mu e^sright)^kk! \
    &= e^-mu e^mu e^s \
    &= exp(mu(e^s - 1)).
    endaligned
    $$
    We will use this result with $s = -1/n$.



    If $X_1, ldots, X_n sim operatornamePoisson(mu)$ are i.i.d. and
    $$
    widehattheta
    = expleft(-frac1n sum_i=1^n X_iright)
    = prod_i=1^n expleft(-fracX_inright),
    $$
    then, using independence,
    $$
    beginaligned
    E[widehattheta]
    &= prod_i=1^n Eleft[e^-X_i / nright] \
    &= prod_i=1^n exp(mu(e^-1/n - 1)) \
    &= exp(n mu(e^-1/n - 1)) \
    &neq e^-mu,
    endaligned
    $$
    so $widehattheta$ 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

    1,201710




    1,201710











    • $begingroup$
      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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