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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
$begingroup$
$X_1,X_2,..,X_n$ 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
asked Mar 22 at 20:47
QualityQuality
30019
$endgroup$
add a comment |
$begingroup$
$X_1,X_2,..,X_n$ 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
asked Mar 22 at 20:47
QualityQuality
30019
$endgroup$
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
add a comment |
$begingroup$
$X_1,X_2,..,X_n$ 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
asked Mar 22 at 20:47
QualityQuality
30019
$endgroup$
$X_1,X_2,..,X_n$ 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
asked Mar 22 at 20:47
QualityQuality
30019
asked Mar 22 at 20:47
QualityQuality
30019
asked Mar 22 at 20:47
QualityQuality
30019
asked Mar 22 at 20:47
QualityQuality
30019
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $e^-mu$
answered Mar 22 at 21:12
Artem MavrinArtem Mavrin
1,201710
$endgroup$
$begingroup$
Thank you, makes perfect sense
$endgroup$
– Quality
Mar 22 at 21:16
add a comment |
$begingroup$
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.
answered Mar 22 at 21:16
StubbornAtomStubbornAtom
2,8621532
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $e^-mu$
answered Mar 22 at 21:12
Artem MavrinArtem Mavrin
1,201710
$endgroup$
$begingroup$
Thank you, makes perfect sense
$endgroup$
– Quality
Mar 22 at 21:16
add a comment |
$begingroup$
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 $e^-mu$
answered Mar 22 at 21:12
Artem MavrinArtem Mavrin
1,201710
$endgroup$
$begingroup$
Thank you, makes perfect sense
$endgroup$
– Quality
Mar 22 at 21:16
add a comment |
$begingroup$
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 $e^-mu$
answered Mar 22 at 21:12
Artem MavrinArtem Mavrin
1,201710
$endgroup$
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 $e^-mu$
answered Mar 22 at 21:12
Artem MavrinArtem Mavrin
1,201710
answered Mar 22 at 21:12
Artem MavrinArtem Mavrin
1,201710
answered Mar 22 at 21:12
Artem MavrinArtem Mavrin
1,201710
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
answered Mar 22 at 21:16
StubbornAtomStubbornAtom
2,8621532
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 22 at 21:16
StubbornAtomStubbornAtom
2,8621532
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 22 at 21:16
StubbornAtomStubbornAtom
2,8621532
$endgroup$
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.
answered Mar 22 at 21:16
StubbornAtomStubbornAtom
2,8621532
answered Mar 22 at 21:16
StubbornAtomStubbornAtom
2,8621532
answered Mar 22 at 21:16
StubbornAtomStubbornAtom
2,8621532
answered Mar 22 at 21:16
StubbornAtomStubbornAtom
2,8621532
2,8621532
add a comment |
add a comment |
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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