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Why is this estimator biased?

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1

1

$begingroup$

$X_1,X_2,..,X_n$ 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

asked 1 hour ago

QualityQuality

29019

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  E[f(X)] != f(E[X]) in general.
  $endgroup$
  – The Laconic
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  58 mins ago
  

  
  
  
  

add a comment | 

1

1

$begingroup$

$X_1,X_2,..,X_n$ 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

asked 1 hour ago

QualityQuality

29019

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  E[f(X)] != f(E[X]) in general.
  $endgroup$
  – The Laconic
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  58 mins ago
  

  
  
  
  

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

share|cite|improve this question

asked 1 hour ago

QualityQuality

29019

$endgroup$

share|cite|improve this question

asked 1 hour ago

QualityQuality

29019

share|cite|improve this question

asked 1 hour ago

QualityQuality

29019

asked 1 hour ago

QualityQuality

29019

asked 1 hour ago

QualityQuality

29019

2 Answers
2

active

oldest

votes

3

$begingroup$

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

answered 47 mins ago

Artem MavrinArtem Mavrin

1,066710

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, makes perfect sense
  $endgroup$
  – Quality
  43 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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

answered 42 mins ago

StubbornAtomStubbornAtom

2,7671532

$endgroup$

add a comment | 

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3

$begingroup$

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

answered 47 mins ago

Artem MavrinArtem Mavrin

1,066710

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, makes perfect sense
  $endgroup$
  – Quality
  43 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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

answered 47 mins ago

Artem MavrinArtem Mavrin

1,066710

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, makes perfect sense
  $endgroup$
  – Quality
  43 mins ago
  

  
  
  
  

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 $e^-mu$

share|cite|improve this answer

answered 47 mins ago

Artem MavrinArtem Mavrin

1,066710

$endgroup$

share|cite|improve this answer

answered 47 mins ago

Artem MavrinArtem Mavrin

1,066710

answered 47 mins ago

Artem MavrinArtem Mavrin

1,066710

answered 47 mins ago

Artem MavrinArtem Mavrin

1,066710

2

$begingroup$

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

answered 42 mins ago

StubbornAtomStubbornAtom

2,7671532

$endgroup$

add a comment | 

2

$begingroup$

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

answered 42 mins ago

StubbornAtomStubbornAtom

2,7671532

$endgroup$

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.

share|cite|improve this answer

answered 42 mins ago

StubbornAtomStubbornAtom

2,7671532

$endgroup$

share|cite|improve this answer

answered 42 mins ago

StubbornAtomStubbornAtom

2,7671532

answered 42 mins ago

StubbornAtomStubbornAtom

2,7671532

answered 42 mins ago

StubbornAtomStubbornAtom

2,7671532