Mixing
Understanding the solution of a question about sufficient statistics
$begingroup$
So this is a question and solution to an old exam I was practicing for my statistics course.In the solution for 1a), they say that $L(mathbf X;a,k)=k^ne^{sum_{i=1}^nX_i}e^{nka} mathbf 1(X_{(1)} geq a)$. Why do they use the $mathbf 1$ there, and what does that $X_{(1)}$ mean?
I calculated $L(mathbf X;a,k)=k^ne^{sum_{i=1}^nX_i}e^{nka}(X_i geq a)$, but than you get in trouble for the sufficient statistic for $a$.
Thanks in advance!
statistics proof-verification
asked Jan 24 at 9:37
Noah_KNoah_K
1117
$endgroup$
add a comment |
$begingroup$
So this is a question and solution to an old exam I was practicing for my statistics course.In the solution for 1a), they say that $L(mathbf X;a,k)=k^ne^{sum_{i=1}^nX_i}e^{nka} mathbf 1(X_{(1)} geq a)$. Why do they use the $mathbf 1$ there, and what does that $X_{(1)}$ mean?
I calculated $L(mathbf X;a,k)=k^ne^{sum_{i=1}^nX_i}e^{nka}(X_i geq a)$, but than you get in trouble for the sufficient statistic for $a$.
Thanks in advance!
statistics proof-verification
asked Jan 24 at 9:37
Noah_KNoah_K
1117
$endgroup$
1
$begingroup$
I suspect the "1" is actually the indicator function.
$endgroup$
– Paul
Jan 24 at 9:49
add a comment |
$begingroup$
So this is a question and solution to an old exam I was practicing for my statistics course.In the solution for 1a), they say that $L(mathbf X;a,k)=k^ne^{sum_{i=1}^nX_i}e^{nka} mathbf 1(X_{(1)} geq a)$. Why do they use the $mathbf 1$ there, and what does that $X_{(1)}$ mean?
I calculated $L(mathbf X;a,k)=k^ne^{sum_{i=1}^nX_i}e^{nka}(X_i geq a)$, but than you get in trouble for the sufficient statistic for $a$.
Thanks in advance!
statistics proof-verification
asked Jan 24 at 9:37
Noah_KNoah_K
1117
$endgroup$
So this is a question and solution to an old exam I was practicing for my statistics course.In the solution for 1a), they say that $L(mathbf X;a,k)=k^ne^{sum_{i=1}^nX_i}e^{nka} mathbf 1(X_{(1)} geq a)$. Why do they use the $mathbf 1$ there, and what does that $X_{(1)}$ mean?
I calculated $L(mathbf X;a,k)=k^ne^{sum_{i=1}^nX_i}e^{nka}(X_i geq a)$, but than you get in trouble for the sufficient statistic for $a$.
Thanks in advance!
statistics proof-verification
statistics proof-verification
asked Jan 24 at 9:37
Noah_KNoah_K
1117
asked Jan 24 at 9:37
Noah_KNoah_K
1117
asked Jan 24 at 9:37
Noah_KNoah_K
1117
asked Jan 24 at 9:37
Noah_KNoah_K
1117
asked Jan 24 at 9:37
Noah_KNoah_K
1117
1117
1
$begingroup$
I suspect the "1" is actually the indicator function.
$endgroup$
– Paul
Jan 24 at 9:49
add a comment |
1
$begingroup$
I suspect the "1" is actually the indicator function.
$endgroup$
– Paul
Jan 24 at 9:49
1
1
$begingroup$
I suspect the "1" is actually the indicator function.
$endgroup$
– Paul
Jan 24 at 9:49
$begingroup$
I suspect the "1" is actually the indicator function.
$endgroup$
– Paul
Jan 24 at 9:49
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
$X_{(1)}$ means the first order statistic of the $X_i$, in other words the minimum of the observations; similarly $X_{(n)}$ would the $n$th order statistic i.e. the $n$ lowest value
$mathbf 1(X_{(1)} geq a)$ is an indicator function taking the value $1$ when the event in the bracket occurs and the value $0$ when it does not, so here is $1$ when the minimum (and so all) $X_i$ are greater than equal to $a$ but $0$ when any of the $X_i$ are less than $a$
answered Jan 24 at 11:20
HenryHenry
101k481168
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add a comment |
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$begingroup$
$X_{(1)}$ means the first order statistic of the $X_i$, in other words the minimum of the observations; similarly $X_{(n)}$ would the $n$th order statistic i.e. the $n$ lowest value
$mathbf 1(X_{(1)} geq a)$ is an indicator function taking the value $1$ when the event in the bracket occurs and the value $0$ when it does not, so here is $1$ when the minimum (and so all) $X_i$ are greater than equal to $a$ but $0$ when any of the $X_i$ are less than $a$
answered Jan 24 at 11:20
HenryHenry
101k481168
$endgroup$
add a comment |
$begingroup$
$X_{(1)}$ means the first order statistic of the $X_i$, in other words the minimum of the observations; similarly $X_{(n)}$ would the $n$th order statistic i.e. the $n$ lowest value
$mathbf 1(X_{(1)} geq a)$ is an indicator function taking the value $1$ when the event in the bracket occurs and the value $0$ when it does not, so here is $1$ when the minimum (and so all) $X_i$ are greater than equal to $a$ but $0$ when any of the $X_i$ are less than $a$
answered Jan 24 at 11:20
HenryHenry
101k481168
$endgroup$
add a comment |
$begingroup$
$X_{(1)}$ means the first order statistic of the $X_i$, in other words the minimum of the observations; similarly $X_{(n)}$ would the $n$th order statistic i.e. the $n$ lowest value
$mathbf 1(X_{(1)} geq a)$ is an indicator function taking the value $1$ when the event in the bracket occurs and the value $0$ when it does not, so here is $1$ when the minimum (and so all) $X_i$ are greater than equal to $a$ but $0$ when any of the $X_i$ are less than $a$
answered Jan 24 at 11:20
HenryHenry
101k481168
$endgroup$
$X_{(1)}$ means the first order statistic of the $X_i$, in other words the minimum of the observations; similarly $X_{(n)}$ would the $n$th order statistic i.e. the $n$ lowest value
$mathbf 1(X_{(1)} geq a)$ is an indicator function taking the value $1$ when the event in the bracket occurs and the value $0$ when it does not, so here is $1$ when the minimum (and so all) $X_i$ are greater than equal to $a$ but $0$ when any of the $X_i$ are less than $a$
answered Jan 24 at 11:20
HenryHenry
101k481168
answered Jan 24 at 11:20
HenryHenry
101k481168
answered Jan 24 at 11:20
HenryHenry
101k481168
answered Jan 24 at 11:20
HenryHenry
101k481168
101k481168
add a comment |
add a comment |
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$begingroup$
I suspect the "1" is actually the indicator function.
$endgroup$
– Paul
Jan 24 at 9:49