Showing the sample mean is a sufficient statistics from an exponential distribution
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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:
$$L(theta mid x_1,ldots,x_{20})=prod_{i=1}^{20}theta e^{-theta overline x} = theta^{20}e^{-20 theta overline x}$$
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
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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:
$$L(theta mid x_1,ldots,x_{20})=prod_{i=1}^{20}theta e^{-theta overline x} = theta^{20}e^{-20 theta overline x}$$
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
edited 4 hours ago
Community♦
1
asked 4 hours ago
hkj447hkj447
102
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hkj447 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
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:
$$L(theta mid x_1,ldots,x_{20})=prod_{i=1}^{20}theta e^{-theta overline x} = theta^{20}e^{-20 theta overline x}$$
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
edited 4 hours ago
Community♦
1
asked 4 hours ago
hkj447hkj447
102
New contributor
hkj447 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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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:
$$L(theta mid x_1,ldots,x_{20})=prod_{i=1}^{20}theta e^{-theta overline x} = theta^{20}e^{-20 theta overline x}$$
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
likelihood sufficient-statistics factorisation-theorem
edited 4 hours ago
Community♦
1
asked 4 hours ago
hkj447hkj447
102
New contributor
hkj447 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 4 hours ago
Community♦
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asked 4 hours ago
hkj447hkj447
102
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edited 4 hours ago
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edited 4 hours ago
Community♦
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edited 4 hours ago
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asked 4 hours ago
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asked 4 hours ago
hkj447hkj447
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asked 4 hours ago
hkj447hkj447
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102
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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.)
answered 4 hours ago
Michael HardyMichael Hardy
4,2051430
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$begingroup$
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.)
answered 4 hours ago
Michael HardyMichael Hardy
4,2051430
$endgroup$
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.)
answered 4 hours ago
Michael HardyMichael Hardy
4,2051430
$endgroup$
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.)
answered 4 hours ago
Michael HardyMichael Hardy
4,2051430
$endgroup$
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.)
answered 4 hours ago
Michael HardyMichael Hardy
4,2051430
answered 4 hours ago
Michael HardyMichael Hardy
4,2051430
answered 4 hours ago
Michael HardyMichael Hardy
4,2051430
answered 4 hours ago
Michael HardyMichael Hardy
4,2051430
4,2051430
add a comment |
add a comment |
hkj447 is a new contributor. Be nice, and check out our Code of Conduct.
hkj447 is a new contributor. Be nice, and check out our Code of Conduct.
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