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Showing the sample mean is a sufficient statistics from an exponential distribution

Proof that n-order statistics are sufficient for a sample of size nSufficient Statistic for non-exponential family distributionLikelihood Ratio Test for Exponential Distribution with a Limited Parameter SpaceDefinition of sufficient statistic when the support of the statistic depends on the unknown parameter?For the family of distributions, $f_theta(x) = theta x^theta-1$, what is the sufficient statistic corresponding to the monotone likelihood ratio?self-study on sufficient and minimal statistic and linear estimator for Pareto distributionComplete and Sufficient Statistics from Two SamplesSufficient statistic function for $f(x) = theta x^-2, ; ; 0 < theta leq x < infty$Formulating the likelihood in an LRT involving geometric and exp distributionBinomial distributed random sample: find the least variance from the set of all unbiased estimators of $theta$

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2

$begingroup$

Suppose that the lifelengths ( in thousands of hours) of light bulbs are distributed Exponential($theta$), where $theta>0$ is unknown. If we observe $overline x = 5.2$ for a sample of $20$ light bulbs, record a representative likelihood function. Why is it that we only need to observe the sample average to obtain a representative likelihood?

The likelihood is pretty straightforward to find:
$$L(theta mid x_1,ldots,x_20)=prod_i=1^20theta e^-theta overline x = theta^20e^-20 theta overline x$$

I am having an issue showing this is a sufficient statistic however, as I cannot seem to factor it in the form of $h(overline x)g_theta (T(overline x))$

likelihood sufficient-statistics factorisation-theorem

share|cite|improve this question

edited Apr 30 at 5:04

Community♦

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asked Apr 30 at 4:21

hkj447hkj447

153

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2

$begingroup$

Suppose that the lifelengths ( in thousands of hours) of light bulbs are distributed Exponential($theta$), where $theta>0$ is unknown. If we observe $overline x = 5.2$ for a sample of $20$ light bulbs, record a representative likelihood function. Why is it that we only need to observe the sample average to obtain a representative likelihood?

The likelihood is pretty straightforward to find:
$$L(theta mid x_1,ldots,x_20)=prod_i=1^20theta e^-theta overline x = theta^20e^-20 theta overline x$$

I am having an issue showing this is a sufficient statistic however, as I cannot seem to factor it in the form of $h(overline x)g_theta (T(overline x))$

likelihood sufficient-statistics factorisation-theorem

share|cite|improve this question

edited Apr 30 at 5:04

Community♦

1

asked Apr 30 at 4:21

hkj447hkj447

153

$endgroup$

add a comment | 

$begingroup$

Suppose that the lifelengths ( in thousands of hours) of light bulbs are distributed Exponential($theta$), where $theta>0$ is unknown. If we observe $overline x = 5.2$ for a sample of $20$ light bulbs, record a representative likelihood function. Why is it that we only need to observe the sample average to obtain a representative likelihood?

The likelihood is pretty straightforward to find:
$$L(theta mid x_1,ldots,x_20)=prod_i=1^20theta e^-theta overline x = theta^20e^-20 theta overline x$$

I am having an issue showing this is a sufficient statistic however, as I cannot seem to factor it in the form of $h(overline x)g_theta (T(overline x))$

likelihood sufficient-statistics factorisation-theorem

share|cite|improve this question

edited Apr 30 at 5:04

Community♦

1

asked Apr 30 at 4:21

hkj447hkj447

153

$endgroup$

share|cite|improve this question

edited Apr 30 at 5:04

Community♦

1

asked Apr 30 at 4:21

hkj447hkj447

153

share|cite|improve this question

edited Apr 30 at 5:04

Community♦

1

asked Apr 30 at 4:21

hkj447hkj447

153

edited Apr 30 at 5:04

Community♦

1

edited Apr 30 at 5:04

Community♦

1

asked Apr 30 at 4:21

hkj447hkj447

153

asked Apr 30 at 4:21

hkj447hkj447

153

1 Answer
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4

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You've already done what you think you haven't done.

You should write it as $h(x_1,ldots,x_20) g_theta( (x_1 + cdots + x_20)/20),$ or in other words as $h(x_1,ldots,x_20) g_theta(T(x_1,ldots, x_20)).$

What's going on is camouflaged by the fact that $h(x_1,ldots,x_20) = 1$ regardless of the value of $(x_1,ldots,x_20).$

That often happens. (One situation where it does not happen is with an i.i.d. sample from the family of Poisson distributions.)

share|cite|improve this answer

answered Apr 30 at 5:07

Michael HardyMichael Hardy

4,2351430

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4

$begingroup$

You've already done what you think you haven't done.

You should write it as $h(x_1,ldots,x_20) g_theta( (x_1 + cdots + x_20)/20),$ or in other words as $h(x_1,ldots,x_20) g_theta(T(x_1,ldots, x_20)).$

What's going on is camouflaged by the fact that $h(x_1,ldots,x_20) = 1$ regardless of the value of $(x_1,ldots,x_20).$

That often happens. (One situation where it does not happen is with an i.i.d. sample from the family of Poisson distributions.)

share|cite|improve this answer

answered Apr 30 at 5:07

Michael HardyMichael Hardy

4,2351430

$endgroup$

add a comment | 

4

$begingroup$

You've already done what you think you haven't done.

You should write it as $h(x_1,ldots,x_20) g_theta( (x_1 + cdots + x_20)/20),$ or in other words as $h(x_1,ldots,x_20) g_theta(T(x_1,ldots, x_20)).$

What's going on is camouflaged by the fact that $h(x_1,ldots,x_20) = 1$ regardless of the value of $(x_1,ldots,x_20).$

That often happens. (One situation where it does not happen is with an i.i.d. sample from the family of Poisson distributions.)

share|cite|improve this answer

answered Apr 30 at 5:07

Michael HardyMichael Hardy

4,2351430

$endgroup$

add a comment | 

$begingroup$

You've already done what you think you haven't done.

You should write it as $h(x_1,ldots,x_20) g_theta( (x_1 + cdots + x_20)/20),$ or in other words as $h(x_1,ldots,x_20) g_theta(T(x_1,ldots, x_20)).$

What's going on is camouflaged by the fact that $h(x_1,ldots,x_20) = 1$ regardless of the value of $(x_1,ldots,x_20).$

That often happens. (One situation where it does not happen is with an i.i.d. sample from the family of Poisson distributions.)

share|cite|improve this answer

answered Apr 30 at 5:07

Michael HardyMichael Hardy

4,2351430

$endgroup$

share|cite|improve this answer

answered Apr 30 at 5:07

Michael HardyMichael Hardy

4,2351430

answered Apr 30 at 5:07

Michael HardyMichael Hardy

4,2351430

answered Apr 30 at 5:07

Michael HardyMichael Hardy

4,2351430