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Why is this estimator biased?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

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

asked Mar 22 at 20:47

QualityQuality

309110

$endgroup$

• 
  
  
  

  
  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 | 

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

asked Mar 22 at 20:47

QualityQuality

309110

$endgroup$

• 
  
  
  

  
  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

share|cite|improve this question

asked Mar 22 at 20:47

QualityQuality

309110

$endgroup$

share|cite|improve this question

asked Mar 22 at 20:47

QualityQuality

309110

share|cite|improve this question

asked Mar 22 at 20:47

QualityQuality

309110

asked Mar 22 at 20:47

QualityQuality

309110

asked Mar 22 at 20:47

QualityQuality

309110

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

answered Mar 22 at 21:12

Artem MavrinArtem Mavrin

1,256712

$endgroup$

add a comment | 

7

$begingroup$

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 Mar 22 at 21:16

StubbornAtomStubbornAtom

3,0581535

$endgroup$

add a comment | 

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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

answered Mar 22 at 21:12

Artem MavrinArtem Mavrin

1,256712

$endgroup$

add a comment | 

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

answered Mar 22 at 21:12

Artem MavrinArtem Mavrin

1,256712

$endgroup$

add a comment | 

$begingroup$

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

answered Mar 22 at 21:12

Artem MavrinArtem Mavrin

1,256712

$endgroup$

share|cite|improve this answer

answered Mar 22 at 21:12

Artem MavrinArtem Mavrin

1,256712

answered Mar 22 at 21:12

Artem MavrinArtem Mavrin

1,256712

answered Mar 22 at 21:12

Artem MavrinArtem Mavrin

1,256712

7

$begingroup$

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 Mar 22 at 21:16

StubbornAtomStubbornAtom

3,0581535

$endgroup$

add a comment | 

7

$begingroup$

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 Mar 22 at 21:16

StubbornAtomStubbornAtom

3,0581535

$endgroup$

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 Mar 22 at 21:16

StubbornAtomStubbornAtom

3,0581535

$endgroup$

share|cite|improve this answer

answered Mar 22 at 21:16

StubbornAtomStubbornAtom

3,0581535

answered Mar 22 at 21:16

StubbornAtomStubbornAtom

3,0581535

answered Mar 22 at 21:16

StubbornAtomStubbornAtom

3,0581535