How can I show $Y=frac{ln X_1}{ln X_2}$ is an ancillary statistic? [closed]
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Suppose $X_1,X_2$ be random sample with probability density function $f(x)=alpha x^{alpha-1}e^{{-x}^alpha}$, $x>alpha$, $alpha>0$. How can I show $Y=frac{ln X_1}{ln X_2}$ is an ancillary statistic.
statistics statistical-inference
edited Mar 12 '15 at 3:59
Michael Hardy
1
asked Mar 12 '15 at 3:29
hadihadi
22
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closed as off-topic by Did, Andrew, mrtaurho, NCh, Cesareo Dec 29 '18 at 0:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, mrtaurho, NCh
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$begingroup$
Suppose $X_1,X_2$ be random sample with probability density function $f(x)=alpha x^{alpha-1}e^{{-x}^alpha}$, $x>alpha$, $alpha>0$. How can I show $Y=frac{ln X_1}{ln X_2}$ is an ancillary statistic.
statistics statistical-inference
edited Mar 12 '15 at 3:59
Michael Hardy
1
asked Mar 12 '15 at 3:29
hadihadi
22
$endgroup$
closed as off-topic by Did, Andrew, mrtaurho, NCh, Cesareo Dec 29 '18 at 0:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, mrtaurho, NCh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Suppose $X_1,X_2$ be random sample with probability density function $f(x)=alpha x^{alpha-1}e^{{-x}^alpha}$, $x>alpha$, $alpha>0$. How can I show $Y=frac{ln X_1}{ln X_2}$ is an ancillary statistic.
statistics statistical-inference
edited Mar 12 '15 at 3:59
Michael Hardy
1
asked Mar 12 '15 at 3:29
hadihadi
22
$endgroup$
Suppose $X_1,X_2$ be random sample with probability density function $f(x)=alpha x^{alpha-1}e^{{-x}^alpha}$, $x>alpha$, $alpha>0$. How can I show $Y=frac{ln X_1}{ln X_2}$ is an ancillary statistic.
statistics statistical-inference
statistics statistical-inference
edited Mar 12 '15 at 3:59
Michael Hardy
1
asked Mar 12 '15 at 3:29
hadihadi
22
edited Mar 12 '15 at 3:59
Michael Hardy
1
asked Mar 12 '15 at 3:29
hadihadi
22
edited Mar 12 '15 at 3:59
Michael Hardy
1
edited Mar 12 '15 at 3:59
Michael Hardy
1
edited Mar 12 '15 at 3:59
Michael Hardy
1
1
asked Mar 12 '15 at 3:29
hadihadi
22
asked Mar 12 '15 at 3:29
hadihadi
22
asked Mar 12 '15 at 3:29
hadihadi
22
22
closed as off-topic by Did, Andrew, mrtaurho, NCh, Cesareo Dec 29 '18 at 0:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, mrtaurho, NCh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Andrew, mrtaurho, NCh, Cesareo Dec 29 '18 at 0:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, mrtaurho, NCh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
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votes
$begingroup$
You have $f(x),dx = e^{-x^alpha}Big( alpha x^{alpha-1},dxBig) = e^{-u},du$. This suggests we consider $U=X^alpha$, so that
begin{align}
f_U(u) & = frac d {du} Pr(Ule u) \[10pt]
& = left(frac d{dx} Pr(Ule u)right) cdot frac{dx}{du} \[10pt]
& = left(frac d{dx} Pr(X le x)right) cdot frac 1 {alpha x^{alpha-1}} \[10pt]
& = alpha x^{alpha-1} e^{-x^alpha}cdotfrac 1 {alpha x^{alpha-1}} \[10pt]
& = e^{-x^alpha} = e^{-u}quadtext{for }u>0.
end{align}
Next, we have $ln U=alphaln X$, so
$$
frac{ln X_1}{ln X_2} = frac{ln U_1}{ln U_2}.
$$
Since the distribution of neither $U_1$ nor $U_2$ depends on $alpha$, the distribution of this random variable does not depend on $alpha$.
edited Dec 28 '18 at 20:30
Community♦
1
answered Mar 12 '15 at 4:22
Michael HardyMichael Hardy
1
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have $f(x),dx = e^{-x^alpha}Big( alpha x^{alpha-1},dxBig) = e^{-u},du$. This suggests we consider $U=X^alpha$, so that
begin{align}
f_U(u) & = frac d {du} Pr(Ule u) \[10pt]
& = left(frac d{dx} Pr(Ule u)right) cdot frac{dx}{du} \[10pt]
& = left(frac d{dx} Pr(X le x)right) cdot frac 1 {alpha x^{alpha-1}} \[10pt]
& = alpha x^{alpha-1} e^{-x^alpha}cdotfrac 1 {alpha x^{alpha-1}} \[10pt]
& = e^{-x^alpha} = e^{-u}quadtext{for }u>0.
end{align}
Next, we have $ln U=alphaln X$, so
$$
frac{ln X_1}{ln X_2} = frac{ln U_1}{ln U_2}.
$$
Since the distribution of neither $U_1$ nor $U_2$ depends on $alpha$, the distribution of this random variable does not depend on $alpha$.
edited Dec 28 '18 at 20:30
Community♦
1
answered Mar 12 '15 at 4:22
Michael HardyMichael Hardy
1
$endgroup$
add a comment |
$begingroup$
You have $f(x),dx = e^{-x^alpha}Big( alpha x^{alpha-1},dxBig) = e^{-u},du$. This suggests we consider $U=X^alpha$, so that
begin{align}
f_U(u) & = frac d {du} Pr(Ule u) \[10pt]
& = left(frac d{dx} Pr(Ule u)right) cdot frac{dx}{du} \[10pt]
& = left(frac d{dx} Pr(X le x)right) cdot frac 1 {alpha x^{alpha-1}} \[10pt]
& = alpha x^{alpha-1} e^{-x^alpha}cdotfrac 1 {alpha x^{alpha-1}} \[10pt]
& = e^{-x^alpha} = e^{-u}quadtext{for }u>0.
end{align}
Next, we have $ln U=alphaln X$, so
$$
frac{ln X_1}{ln X_2} = frac{ln U_1}{ln U_2}.
$$
Since the distribution of neither $U_1$ nor $U_2$ depends on $alpha$, the distribution of this random variable does not depend on $alpha$.
edited Dec 28 '18 at 20:30
Community♦
1
answered Mar 12 '15 at 4:22
Michael HardyMichael Hardy
1
$endgroup$
add a comment |
$begingroup$
You have $f(x),dx = e^{-x^alpha}Big( alpha x^{alpha-1},dxBig) = e^{-u},du$. This suggests we consider $U=X^alpha$, so that
begin{align}
f_U(u) & = frac d {du} Pr(Ule u) \[10pt]
& = left(frac d{dx} Pr(Ule u)right) cdot frac{dx}{du} \[10pt]
& = left(frac d{dx} Pr(X le x)right) cdot frac 1 {alpha x^{alpha-1}} \[10pt]
& = alpha x^{alpha-1} e^{-x^alpha}cdotfrac 1 {alpha x^{alpha-1}} \[10pt]
& = e^{-x^alpha} = e^{-u}quadtext{for }u>0.
end{align}
Next, we have $ln U=alphaln X$, so
$$
frac{ln X_1}{ln X_2} = frac{ln U_1}{ln U_2}.
$$
Since the distribution of neither $U_1$ nor $U_2$ depends on $alpha$, the distribution of this random variable does not depend on $alpha$.
edited Dec 28 '18 at 20:30
Community♦
1
answered Mar 12 '15 at 4:22
Michael HardyMichael Hardy
1
$endgroup$
You have $f(x),dx = e^{-x^alpha}Big( alpha x^{alpha-1},dxBig) = e^{-u},du$. This suggests we consider $U=X^alpha$, so that
begin{align}
f_U(u) & = frac d {du} Pr(Ule u) \[10pt]
& = left(frac d{dx} Pr(Ule u)right) cdot frac{dx}{du} \[10pt]
& = left(frac d{dx} Pr(X le x)right) cdot frac 1 {alpha x^{alpha-1}} \[10pt]
& = alpha x^{alpha-1} e^{-x^alpha}cdotfrac 1 {alpha x^{alpha-1}} \[10pt]
& = e^{-x^alpha} = e^{-u}quadtext{for }u>0.
end{align}
Next, we have $ln U=alphaln X$, so
$$
frac{ln X_1}{ln X_2} = frac{ln U_1}{ln U_2}.
$$
Since the distribution of neither $U_1$ nor $U_2$ depends on $alpha$, the distribution of this random variable does not depend on $alpha$.
edited Dec 28 '18 at 20:30
Community♦
1
answered Mar 12 '15 at 4:22
Michael HardyMichael Hardy
1
edited Dec 28 '18 at 20:30
Community♦
1
edited Dec 28 '18 at 20:30
Community♦
1
edited Dec 28 '18 at 20:30
Community♦
1
1
answered Mar 12 '15 at 4:22
Michael HardyMichael Hardy
1
answered Mar 12 '15 at 4:22
Michael HardyMichael Hardy
1
answered Mar 12 '15 at 4:22
Michael HardyMichael Hardy
1
1
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