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Bias of MLE of simple PDF
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Given a sample $x_1, x_2, cdots x_n$ from the pdf:
$$
f(x ; theta) = (theta + 1) x^theta
$$
where $0 < x < 1$ and $theta > -1$ is unknown. What is the bias of the MLE of $theta$?
I've found the MLE to be
$$
hattheta = frac-nsum_i=1^n log(x_i) - 1
$$
but I'm stuck on finding the bias of this estimator. The sum in the denominator makes it hard to take the expected value. I think there is something simple here that I am missing...
self-study mathematical-statistics maximum-likelihood bias
asked 1 hour ago
bill_e
1,48211326
add a comment |Â
up vote
1
down vote
favorite
Given a sample $x_1, x_2, cdots x_n$ from the pdf:
$$
f(x ; theta) = (theta + 1) x^theta
$$
where $0 < x < 1$ and $theta > -1$ is unknown. What is the bias of the MLE of $theta$?
I've found the MLE to be
$$
hattheta = frac-nsum_i=1^n log(x_i) - 1
$$
but I'm stuck on finding the bias of this estimator. The sum in the denominator makes it hard to take the expected value. I think there is something simple here that I am missing...
self-study mathematical-statistics maximum-likelihood bias
asked 1 hour ago
bill_e
1,48211326
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given a sample $x_1, x_2, cdots x_n$ from the pdf:
$$
f(x ; theta) = (theta + 1) x^theta
$$
where $0 < x < 1$ and $theta > -1$ is unknown. What is the bias of the MLE of $theta$?
I've found the MLE to be
$$
hattheta = frac-nsum_i=1^n log(x_i) - 1
$$
but I'm stuck on finding the bias of this estimator. The sum in the denominator makes it hard to take the expected value. I think there is something simple here that I am missing...
self-study mathematical-statistics maximum-likelihood bias
asked 1 hour ago
bill_e
1,48211326
Given a sample $x_1, x_2, cdots x_n$ from the pdf:
$$
f(x ; theta) = (theta + 1) x^theta
$$
where $0 < x < 1$ and $theta > -1$ is unknown. What is the bias of the MLE of $theta$?
I've found the MLE to be
$$
hattheta = frac-nsum_i=1^n log(x_i) - 1
$$
but I'm stuck on finding the bias of this estimator. The sum in the denominator makes it hard to take the expected value. I think there is something simple here that I am missing...
self-study mathematical-statistics maximum-likelihood bias
self-study mathematical-statistics maximum-likelihood bias
asked 1 hour ago
bill_e
1,48211326
asked 1 hour ago
bill_e
1,48211326
asked 1 hour ago
bill_e
1,48211326
asked 1 hour ago
bill_e
1,48211326
asked 1 hour ago
bill_e
1,48211326
1,48211326
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Make the substitution
$$
Y_i = -log X_i.
$$
It is easy to show that $Y_i$ has density
$$
f(y ; theta) = (theta + 1) e^-(theta + 1) y I(y > 0).
$$
so $Y_i sim operatornameExponential(theta + 1)$. It follows that $sum_i -log (X_i) sim operatornameGamma(n, theta + 1)$. Recall that if $Z sim operatornameGamma(alpha, beta)$ then $E Z^-1 = fracbetaalpha - 1$. Therefore
$$
E_theta(widehat theta) = ... = fracntheta + 1n - 1
$$
and the bias is then easily shown to be $(theta + 1) / (n - 1)$.
answered 1 hour ago
guy
3,95811336
"it's easy to show..." Could you elaborate here a little bit? What was your thought process here? Is the original pdf of the x's some sort of standard distribution (with a known identity for Y = -logX) I ought to know? Or did you intuit trying the Y=-logX transformation and then worked out the distribution for Y from scratch?
– bill_e
1 hour ago
2
@bill_e My logic was that, because this looks like a homework problem, it must be the case that $sum_i -log (X_i)$ has a simple distribution. Hence $-log (X_i)$ probably has some kind of gamma distribution, because if it didn't then the question would be difficult (intro Math Stat courses learn that sums of iid gammas are gammas, and that is the only type of identity of this sort that seems like it might be useful). Anyway, the original pdf is a $operatornameBeta(theta+1, 1)$, but that fact didn't motivate the approach I took.
– guy
59 mins ago
1
@bill_e Also, I worked out the density by doing a change of variables (i.e., substitue $e^-y$ for $x$ and multiply by $|frac d dy e^-y|$).
– guy
55 mins ago
Ahh that makes sense, thanks!
– bill_e
39 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Make the substitution
$$
Y_i = -log X_i.
$$
It is easy to show that $Y_i$ has density
$$
f(y ; theta) = (theta + 1) e^-(theta + 1) y I(y > 0).
$$
so $Y_i sim operatornameExponential(theta + 1)$. It follows that $sum_i -log (X_i) sim operatornameGamma(n, theta + 1)$. Recall that if $Z sim operatornameGamma(alpha, beta)$ then $E Z^-1 = fracbetaalpha - 1$. Therefore
$$
E_theta(widehat theta) = ... = fracntheta + 1n - 1
$$
and the bias is then easily shown to be $(theta + 1) / (n - 1)$.
answered 1 hour ago
guy
3,95811336
"it's easy to show..." Could you elaborate here a little bit? What was your thought process here? Is the original pdf of the x's some sort of standard distribution (with a known identity for Y = -logX) I ought to know? Or did you intuit trying the Y=-logX transformation and then worked out the distribution for Y from scratch?
– bill_e
1 hour ago
2
@bill_e My logic was that, because this looks like a homework problem, it must be the case that $sum_i -log (X_i)$ has a simple distribution. Hence $-log (X_i)$ probably has some kind of gamma distribution, because if it didn't then the question would be difficult (intro Math Stat courses learn that sums of iid gammas are gammas, and that is the only type of identity of this sort that seems like it might be useful). Anyway, the original pdf is a $operatornameBeta(theta+1, 1)$, but that fact didn't motivate the approach I took.
– guy
59 mins ago
1
@bill_e Also, I worked out the density by doing a change of variables (i.e., substitue $e^-y$ for $x$ and multiply by $|frac d dy e^-y|$).
– guy
55 mins ago
Ahh that makes sense, thanks!
– bill_e
39 mins ago
add a comment |Â
up vote
3
down vote
accepted
Make the substitution
$$
Y_i = -log X_i.
$$
It is easy to show that $Y_i$ has density
$$
f(y ; theta) = (theta + 1) e^-(theta + 1) y I(y > 0).
$$
so $Y_i sim operatornameExponential(theta + 1)$. It follows that $sum_i -log (X_i) sim operatornameGamma(n, theta + 1)$. Recall that if $Z sim operatornameGamma(alpha, beta)$ then $E Z^-1 = fracbetaalpha - 1$. Therefore
$$
E_theta(widehat theta) = ... = fracntheta + 1n - 1
$$
and the bias is then easily shown to be $(theta + 1) / (n - 1)$.
answered 1 hour ago
guy
3,95811336
"it's easy to show..." Could you elaborate here a little bit? What was your thought process here? Is the original pdf of the x's some sort of standard distribution (with a known identity for Y = -logX) I ought to know? Or did you intuit trying the Y=-logX transformation and then worked out the distribution for Y from scratch?
– bill_e
1 hour ago
2
@bill_e My logic was that, because this looks like a homework problem, it must be the case that $sum_i -log (X_i)$ has a simple distribution. Hence $-log (X_i)$ probably has some kind of gamma distribution, because if it didn't then the question would be difficult (intro Math Stat courses learn that sums of iid gammas are gammas, and that is the only type of identity of this sort that seems like it might be useful). Anyway, the original pdf is a $operatornameBeta(theta+1, 1)$, but that fact didn't motivate the approach I took.
– guy
59 mins ago
1
@bill_e Also, I worked out the density by doing a change of variables (i.e., substitue $e^-y$ for $x$ and multiply by $|frac d dy e^-y|$).
– guy
55 mins ago
Ahh that makes sense, thanks!
– bill_e
39 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Make the substitution
$$
Y_i = -log X_i.
$$
It is easy to show that $Y_i$ has density
$$
f(y ; theta) = (theta + 1) e^-(theta + 1) y I(y > 0).
$$
so $Y_i sim operatornameExponential(theta + 1)$. It follows that $sum_i -log (X_i) sim operatornameGamma(n, theta + 1)$. Recall that if $Z sim operatornameGamma(alpha, beta)$ then $E Z^-1 = fracbetaalpha - 1$. Therefore
$$
E_theta(widehat theta) = ... = fracntheta + 1n - 1
$$
and the bias is then easily shown to be $(theta + 1) / (n - 1)$.
answered 1 hour ago
guy
3,95811336
Make the substitution
$$
Y_i = -log X_i.
$$
It is easy to show that $Y_i$ has density
$$
f(y ; theta) = (theta + 1) e^-(theta + 1) y I(y > 0).
$$
so $Y_i sim operatornameExponential(theta + 1)$. It follows that $sum_i -log (X_i) sim operatornameGamma(n, theta + 1)$. Recall that if $Z sim operatornameGamma(alpha, beta)$ then $E Z^-1 = fracbetaalpha - 1$. Therefore
$$
E_theta(widehat theta) = ... = fracntheta + 1n - 1
$$
and the bias is then easily shown to be $(theta + 1) / (n - 1)$.
answered 1 hour ago
guy
3,95811336
answered 1 hour ago
guy
3,95811336
answered 1 hour ago
guy
3,95811336
answered 1 hour ago
guy
3,95811336
3,95811336
"it's easy to show..." Could you elaborate here a little bit? What was your thought process here? Is the original pdf of the x's some sort of standard distribution (with a known identity for Y = -logX) I ought to know? Or did you intuit trying the Y=-logX transformation and then worked out the distribution for Y from scratch?
– bill_e
1 hour ago
2
@bill_e My logic was that, because this looks like a homework problem, it must be the case that $sum_i -log (X_i)$ has a simple distribution. Hence $-log (X_i)$ probably has some kind of gamma distribution, because if it didn't then the question would be difficult (intro Math Stat courses learn that sums of iid gammas are gammas, and that is the only type of identity of this sort that seems like it might be useful). Anyway, the original pdf is a $operatornameBeta(theta+1, 1)$, but that fact didn't motivate the approach I took.
– guy
59 mins ago
1
@bill_e Also, I worked out the density by doing a change of variables (i.e., substitue $e^-y$ for $x$ and multiply by $|frac d dy e^-y|$).
– guy
55 mins ago
Ahh that makes sense, thanks!
– bill_e
39 mins ago
add a comment |Â
"it's easy to show..." Could you elaborate here a little bit? What was your thought process here? Is the original pdf of the x's some sort of standard distribution (with a known identity for Y = -logX) I ought to know? Or did you intuit trying the Y=-logX transformation and then worked out the distribution for Y from scratch?
– bill_e
1 hour ago
2
@bill_e My logic was that, because this looks like a homework problem, it must be the case that $sum_i -log (X_i)$ has a simple distribution. Hence $-log (X_i)$ probably has some kind of gamma distribution, because if it didn't then the question would be difficult (intro Math Stat courses learn that sums of iid gammas are gammas, and that is the only type of identity of this sort that seems like it might be useful). Anyway, the original pdf is a $operatornameBeta(theta+1, 1)$, but that fact didn't motivate the approach I took.
– guy
59 mins ago
1
@bill_e Also, I worked out the density by doing a change of variables (i.e., substitue $e^-y$ for $x$ and multiply by $|frac d dy e^-y|$).
– guy
55 mins ago
Ahh that makes sense, thanks!
– bill_e
39 mins ago
"it's easy to show..." Could you elaborate here a little bit? What was your thought process here? Is the original pdf of the x's some sort of standard distribution (with a known identity for Y = -logX) I ought to know? Or did you intuit trying the Y=-logX transformation and then worked out the distribution for Y from scratch?
– bill_e
1 hour ago
"it's easy to show..." Could you elaborate here a little bit? What was your thought process here? Is the original pdf of the x's some sort of standard distribution (with a known identity for Y = -logX) I ought to know? Or did you intuit trying the Y=-logX transformation and then worked out the distribution for Y from scratch?
– bill_e
1 hour ago
2
2
@bill_e My logic was that, because this looks like a homework problem, it must be the case that $sum_i -log (X_i)$ has a simple distribution. Hence $-log (X_i)$ probably has some kind of gamma distribution, because if it didn't then the question would be difficult (intro Math Stat courses learn that sums of iid gammas are gammas, and that is the only type of identity of this sort that seems like it might be useful). Anyway, the original pdf is a $operatornameBeta(theta+1, 1)$, but that fact didn't motivate the approach I took.
– guy
59 mins ago
@bill_e My logic was that, because this looks like a homework problem, it must be the case that $sum_i -log (X_i)$ has a simple distribution. Hence $-log (X_i)$ probably has some kind of gamma distribution, because if it didn't then the question would be difficult (intro Math Stat courses learn that sums of iid gammas are gammas, and that is the only type of identity of this sort that seems like it might be useful). Anyway, the original pdf is a $operatornameBeta(theta+1, 1)$, but that fact didn't motivate the approach I took.
– guy
59 mins ago
1
1
@bill_e Also, I worked out the density by doing a change of variables (i.e., substitue $e^-y$ for $x$ and multiply by $|frac d dy e^-y|$).
– guy
55 mins ago
@bill_e Also, I worked out the density by doing a change of variables (i.e., substitue $e^-y$ for $x$ and multiply by $|frac d dy e^-y|$).
– guy
55 mins ago
Ahh that makes sense, thanks!
– bill_e
39 mins ago
Ahh that makes sense, thanks!
– bill_e
39 mins ago
add a comment |Â
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