Mixing
Find the MLE of $p$ where $f(y;p)=2p^2y^{-3}$
$begingroup$
Find the MLE of $p$ where $f(y;p)=2p^2y^{-3}$.
Attempt:
Method: find the likelihood function, differentiate with respect to $p$ then set to zero and solve for $p$.
$L(p;y)=prodlimits_{i=1}^n [2p^2y^{-3}_i] =prodlimits_{i=1}^n[2] prodlimits_{i=1}^n[p^2] prodlimits_{i=1}^n [y^{-3}_i] = 2^n p^{2n} prodlimits_{i=1}^n [y^{-3}_i]$
Differentiating the log likelihood gives $2n/p$ (i think) and the $y_i$ all disappear. So what happens when you set this to zero? Because if $0=2n/p$ then surely $0=2n$ and $n=0$.
Have I misstated $L(p;y)$? Or is something going wrong with my differentiation? (Or both?)
statistics statistical-inference maximum-likelihood
edited Jan 6 at 12:40
Did
247k23223459
asked Jan 6 at 12:07
Maths BarryMaths Barry
458
$endgroup$
add a comment |
$begingroup$
Find the MLE of $p$ where $f(y;p)=2p^2y^{-3}$.
Attempt:
Method: find the likelihood function, differentiate with respect to $p$ then set to zero and solve for $p$.
$L(p;y)=prodlimits_{i=1}^n [2p^2y^{-3}_i] =prodlimits_{i=1}^n[2] prodlimits_{i=1}^n[p^2] prodlimits_{i=1}^n [y^{-3}_i] = 2^n p^{2n} prodlimits_{i=1}^n [y^{-3}_i]$
Differentiating the log likelihood gives $2n/p$ (i think) and the $y_i$ all disappear. So what happens when you set this to zero? Because if $0=2n/p$ then surely $0=2n$ and $n=0$.
Have I misstated $L(p;y)$? Or is something going wrong with my differentiation? (Or both?)
statistics statistical-inference maximum-likelihood
edited Jan 6 at 12:40
Did
247k23223459
asked Jan 6 at 12:07
Maths BarryMaths Barry
458
$endgroup$
1
$begingroup$
Previously asked: math.stackexchange.com/questions/446142/pareto-distribution-mle. Without mentioning support of the distribution, your $f(y;p)$ is incomplete. And when you mention the support, say by using indicator functions, you will see that differentiation is not valid to derive the MLE because the likelihood is not differentiable everywhere in the first place. See the linked threads here.
$endgroup$
– StubbornAtom
Jan 6 at 14:16
$begingroup$
THANK YOU! I didn't realise this was a Pareto distribution so I am extremely grateful for this.
$endgroup$
– Maths Barry
Jan 6 at 14:36
add a comment |
$begingroup$
Find the MLE of $p$ where $f(y;p)=2p^2y^{-3}$.
Attempt:
Method: find the likelihood function, differentiate with respect to $p$ then set to zero and solve for $p$.
$L(p;y)=prodlimits_{i=1}^n [2p^2y^{-3}_i] =prodlimits_{i=1}^n[2] prodlimits_{i=1}^n[p^2] prodlimits_{i=1}^n [y^{-3}_i] = 2^n p^{2n} prodlimits_{i=1}^n [y^{-3}_i]$
Differentiating the log likelihood gives $2n/p$ (i think) and the $y_i$ all disappear. So what happens when you set this to zero? Because if $0=2n/p$ then surely $0=2n$ and $n=0$.
Have I misstated $L(p;y)$? Or is something going wrong with my differentiation? (Or both?)
statistics statistical-inference maximum-likelihood
edited Jan 6 at 12:40
Did
247k23223459
asked Jan 6 at 12:07
Maths BarryMaths Barry
458
$endgroup$
Find the MLE of $p$ where $f(y;p)=2p^2y^{-3}$.
Attempt:
Method: find the likelihood function, differentiate with respect to $p$ then set to zero and solve for $p$.
$L(p;y)=prodlimits_{i=1}^n [2p^2y^{-3}_i] =prodlimits_{i=1}^n[2] prodlimits_{i=1}^n[p^2] prodlimits_{i=1}^n [y^{-3}_i] = 2^n p^{2n} prodlimits_{i=1}^n [y^{-3}_i]$
Differentiating the log likelihood gives $2n/p$ (i think) and the $y_i$ all disappear. So what happens when you set this to zero? Because if $0=2n/p$ then surely $0=2n$ and $n=0$.
Have I misstated $L(p;y)$? Or is something going wrong with my differentiation? (Or both?)
statistics statistical-inference maximum-likelihood
statistics statistical-inference maximum-likelihood
edited Jan 6 at 12:40
Did
247k23223459
asked Jan 6 at 12:07
Maths BarryMaths Barry
458
edited Jan 6 at 12:40
Did
247k23223459
asked Jan 6 at 12:07
Maths BarryMaths Barry
458
edited Jan 6 at 12:40
Did
247k23223459
edited Jan 6 at 12:40
Did
247k23223459
edited Jan 6 at 12:40
Did
247k23223459
247k23223459
asked Jan 6 at 12:07
Maths BarryMaths Barry
458
asked Jan 6 at 12:07
Maths BarryMaths Barry
458
asked Jan 6 at 12:07
Maths BarryMaths Barry
458
458
1
$begingroup$
Previously asked: math.stackexchange.com/questions/446142/pareto-distribution-mle. Without mentioning support of the distribution, your $f(y;p)$ is incomplete. And when you mention the support, say by using indicator functions, you will see that differentiation is not valid to derive the MLE because the likelihood is not differentiable everywhere in the first place. See the linked threads here.
$endgroup$
– StubbornAtom
Jan 6 at 14:16
$begingroup$
THANK YOU! I didn't realise this was a Pareto distribution so I am extremely grateful for this.
$endgroup$
– Maths Barry
Jan 6 at 14:36
add a comment |
1
$begingroup$
Previously asked: math.stackexchange.com/questions/446142/pareto-distribution-mle. Without mentioning support of the distribution, your $f(y;p)$ is incomplete. And when you mention the support, say by using indicator functions, you will see that differentiation is not valid to derive the MLE because the likelihood is not differentiable everywhere in the first place. See the linked threads here.
$endgroup$
– StubbornAtom
Jan 6 at 14:16
$begingroup$
THANK YOU! I didn't realise this was a Pareto distribution so I am extremely grateful for this.
$endgroup$
– Maths Barry
Jan 6 at 14:36
1
1
$begingroup$
Previously asked: math.stackexchange.com/questions/446142/pareto-distribution-mle. Without mentioning support of the distribution, your $f(y;p)$ is incomplete. And when you mention the support, say by using indicator functions, you will see that differentiation is not valid to derive the MLE because the likelihood is not differentiable everywhere in the first place. See the linked threads here.
$endgroup$
– StubbornAtom
Jan 6 at 14:16
$begingroup$
Previously asked: math.stackexchange.com/questions/446142/pareto-distribution-mle. Without mentioning support of the distribution, your $f(y;p)$ is incomplete. And when you mention the support, say by using indicator functions, you will see that differentiation is not valid to derive the MLE because the likelihood is not differentiable everywhere in the first place. See the linked threads here.
$endgroup$
– StubbornAtom
Jan 6 at 14:16
$begingroup$
THANK YOU! I didn't realise this was a Pareto distribution so I am extremely grateful for this.
$endgroup$
– Maths Barry
Jan 6 at 14:36
$begingroup$
THANK YOU! I didn't realise this was a Pareto distribution so I am extremely grateful for this.
$endgroup$
– Maths Barry
Jan 6 at 14:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Both the PDF and the likelihood in your post are incorrect, due to the fact that you forget to include indicator functions.
In fact, the PDF is $$f(y;p)=2p^2y^{-3}mathbf 1_{ygeqslant p}$$ hence, the likelihood of some i.i.d. sample $mathbf y=(y_i)_{1leqslant ileqslant n}$ for the PDF $f( ;p)$ is $$L(p;mathbf y)=2^np^{2n}left(prod_iy_i^{-3}right)mathbf 1_{m(mathbf y)geqslant p}$$ where $$m(mathbf y)=min_{1leqslant ileqslant n} y_i$$ One sees readily that $$L(p;mathbf y)=c(mathbf y)p^{2n}mathbf 1_{m(mathbf y)geqslant p}$$ for some positive constant $c(mathbf y)$ independent of $p$, hence $L( ;mathbf y)$ is maximum at $$hat p=m(mathbf y)$$ No differentiation is involved, rather a precise understanding of the situation.
answered Jan 6 at 12:36
DidDid
247k23223459
$endgroup$
$begingroup$
What if we assume $Y geq p$?
$endgroup$
– Maths Barry
Jan 6 at 13:14
$begingroup$
Sorry, but what if what?
$endgroup$
– Did
Jan 6 at 13:22
$begingroup$
What if the range is $0<pleq y<infty$. The indicator function is not still needed then is it? And differentation would be the only way to get the MLE?
$endgroup$
– Maths Barry
Jan 6 at 13:33
$begingroup$
Yes the range of each $f( ;p)$ is $[p,infty)$. This is exactly why the indicator function is (much) needed. "And different[i]ation would be the only way to get the MLE?" Sorry but what are you talking about? If you can present a proof based on solving $$frac{partial L(p;mathbf y)}{partial p}=0$$ please do so (there are none...).
$endgroup$
– Did
Jan 6 at 14:13
add a comment |
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1 Answer
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active
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1 Answer
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$begingroup$
Both the PDF and the likelihood in your post are incorrect, due to the fact that you forget to include indicator functions.
In fact, the PDF is $$f(y;p)=2p^2y^{-3}mathbf 1_{ygeqslant p}$$ hence, the likelihood of some i.i.d. sample $mathbf y=(y_i)_{1leqslant ileqslant n}$ for the PDF $f( ;p)$ is $$L(p;mathbf y)=2^np^{2n}left(prod_iy_i^{-3}right)mathbf 1_{m(mathbf y)geqslant p}$$ where $$m(mathbf y)=min_{1leqslant ileqslant n} y_i$$ One sees readily that $$L(p;mathbf y)=c(mathbf y)p^{2n}mathbf 1_{m(mathbf y)geqslant p}$$ for some positive constant $c(mathbf y)$ independent of $p$, hence $L( ;mathbf y)$ is maximum at $$hat p=m(mathbf y)$$ No differentiation is involved, rather a precise understanding of the situation.
answered Jan 6 at 12:36
DidDid
247k23223459
$endgroup$
$begingroup$
What if we assume $Y geq p$?
$endgroup$
– Maths Barry
Jan 6 at 13:14
$begingroup$
Sorry, but what if what?
$endgroup$
– Did
Jan 6 at 13:22
$begingroup$
What if the range is $0<pleq y<infty$. The indicator function is not still needed then is it? And differentation would be the only way to get the MLE?
$endgroup$
– Maths Barry
Jan 6 at 13:33
$begingroup$
Yes the range of each $f( ;p)$ is $[p,infty)$. This is exactly why the indicator function is (much) needed. "And different[i]ation would be the only way to get the MLE?" Sorry but what are you talking about? If you can present a proof based on solving $$frac{partial L(p;mathbf y)}{partial p}=0$$ please do so (there are none...).
$endgroup$
– Did
Jan 6 at 14:13
add a comment |
$begingroup$
Both the PDF and the likelihood in your post are incorrect, due to the fact that you forget to include indicator functions.
In fact, the PDF is $$f(y;p)=2p^2y^{-3}mathbf 1_{ygeqslant p}$$ hence, the likelihood of some i.i.d. sample $mathbf y=(y_i)_{1leqslant ileqslant n}$ for the PDF $f( ;p)$ is $$L(p;mathbf y)=2^np^{2n}left(prod_iy_i^{-3}right)mathbf 1_{m(mathbf y)geqslant p}$$ where $$m(mathbf y)=min_{1leqslant ileqslant n} y_i$$ One sees readily that $$L(p;mathbf y)=c(mathbf y)p^{2n}mathbf 1_{m(mathbf y)geqslant p}$$ for some positive constant $c(mathbf y)$ independent of $p$, hence $L( ;mathbf y)$ is maximum at $$hat p=m(mathbf y)$$ No differentiation is involved, rather a precise understanding of the situation.
answered Jan 6 at 12:36
DidDid
247k23223459
$endgroup$
$begingroup$
What if we assume $Y geq p$?
$endgroup$
– Maths Barry
Jan 6 at 13:14
$begingroup$
Sorry, but what if what?
$endgroup$
– Did
Jan 6 at 13:22
$begingroup$
What if the range is $0<pleq y<infty$. The indicator function is not still needed then is it? And differentation would be the only way to get the MLE?
$endgroup$
– Maths Barry
Jan 6 at 13:33
$begingroup$
Yes the range of each $f( ;p)$ is $[p,infty)$. This is exactly why the indicator function is (much) needed. "And different[i]ation would be the only way to get the MLE?" Sorry but what are you talking about? If you can present a proof based on solving $$frac{partial L(p;mathbf y)}{partial p}=0$$ please do so (there are none...).
$endgroup$
– Did
Jan 6 at 14:13
add a comment |
$begingroup$
Both the PDF and the likelihood in your post are incorrect, due to the fact that you forget to include indicator functions.
In fact, the PDF is $$f(y;p)=2p^2y^{-3}mathbf 1_{ygeqslant p}$$ hence, the likelihood of some i.i.d. sample $mathbf y=(y_i)_{1leqslant ileqslant n}$ for the PDF $f( ;p)$ is $$L(p;mathbf y)=2^np^{2n}left(prod_iy_i^{-3}right)mathbf 1_{m(mathbf y)geqslant p}$$ where $$m(mathbf y)=min_{1leqslant ileqslant n} y_i$$ One sees readily that $$L(p;mathbf y)=c(mathbf y)p^{2n}mathbf 1_{m(mathbf y)geqslant p}$$ for some positive constant $c(mathbf y)$ independent of $p$, hence $L( ;mathbf y)$ is maximum at $$hat p=m(mathbf y)$$ No differentiation is involved, rather a precise understanding of the situation.
answered Jan 6 at 12:36
DidDid
247k23223459
$endgroup$
Both the PDF and the likelihood in your post are incorrect, due to the fact that you forget to include indicator functions.
In fact, the PDF is $$f(y;p)=2p^2y^{-3}mathbf 1_{ygeqslant p}$$ hence, the likelihood of some i.i.d. sample $mathbf y=(y_i)_{1leqslant ileqslant n}$ for the PDF $f( ;p)$ is $$L(p;mathbf y)=2^np^{2n}left(prod_iy_i^{-3}right)mathbf 1_{m(mathbf y)geqslant p}$$ where $$m(mathbf y)=min_{1leqslant ileqslant n} y_i$$ One sees readily that $$L(p;mathbf y)=c(mathbf y)p^{2n}mathbf 1_{m(mathbf y)geqslant p}$$ for some positive constant $c(mathbf y)$ independent of $p$, hence $L( ;mathbf y)$ is maximum at $$hat p=m(mathbf y)$$ No differentiation is involved, rather a precise understanding of the situation.
answered Jan 6 at 12:36
DidDid
247k23223459
edited Jan 6 at 14:46
answered Jan 6 at 12:36
DidDid
247k23223459
answered Jan 6 at 12:36
DidDid
247k23223459
answered Jan 6 at 12:36
DidDid
247k23223459
247k23223459
$begingroup$
What if we assume $Y geq p$?
$endgroup$
– Maths Barry
Jan 6 at 13:14
$begingroup$
Sorry, but what if what?
$endgroup$
– Did
Jan 6 at 13:22
$begingroup$
What if the range is $0<pleq y<infty$. The indicator function is not still needed then is it? And differentation would be the only way to get the MLE?
$endgroup$
– Maths Barry
Jan 6 at 13:33
$begingroup$
Yes the range of each $f( ;p)$ is $[p,infty)$. This is exactly why the indicator function is (much) needed. "And different[i]ation would be the only way to get the MLE?" Sorry but what are you talking about? If you can present a proof based on solving $$frac{partial L(p;mathbf y)}{partial p}=0$$ please do so (there are none...).
$endgroup$
– Did
Jan 6 at 14:13
add a comment |
$begingroup$
What if we assume $Y geq p$?
$endgroup$
– Maths Barry
Jan 6 at 13:14
$begingroup$
Sorry, but what if what?
$endgroup$
– Did
Jan 6 at 13:22
$begingroup$
What if the range is $0<pleq y<infty$. The indicator function is not still needed then is it? And differentation would be the only way to get the MLE?
$endgroup$
– Maths Barry
Jan 6 at 13:33
$begingroup$
Yes the range of each $f( ;p)$ is $[p,infty)$. This is exactly why the indicator function is (much) needed. "And different[i]ation would be the only way to get the MLE?" Sorry but what are you talking about? If you can present a proof based on solving $$frac{partial L(p;mathbf y)}{partial p}=0$$ please do so (there are none...).
$endgroup$
– Did
Jan 6 at 14:13
$begingroup$
What if we assume $Y geq p$?
$endgroup$
– Maths Barry
Jan 6 at 13:14
$begingroup$
What if we assume $Y geq p$?
$endgroup$
– Maths Barry
Jan 6 at 13:14
$begingroup$
Sorry, but what if what?
$endgroup$
– Did
Jan 6 at 13:22
$begingroup$
Sorry, but what if what?
$endgroup$
– Did
Jan 6 at 13:22
$begingroup$
What if the range is $0<pleq y<infty$. The indicator function is not still needed then is it? And differentation would be the only way to get the MLE?
$endgroup$
– Maths Barry
Jan 6 at 13:33
$begingroup$
What if the range is $0<pleq y<infty$. The indicator function is not still needed then is it? And differentation would be the only way to get the MLE?
$endgroup$
– Maths Barry
Jan 6 at 13:33
$begingroup$
Yes the range of each $f( ;p)$ is $[p,infty)$. This is exactly why the indicator function is (much) needed. "And different[i]ation would be the only way to get the MLE?" Sorry but what are you talking about? If you can present a proof based on solving $$frac{partial L(p;mathbf y)}{partial p}=0$$ please do so (there are none...).
$endgroup$
– Did
Jan 6 at 14:13
$begingroup$
Yes the range of each $f( ;p)$ is $[p,infty)$. This is exactly why the indicator function is (much) needed. "And different[i]ation would be the only way to get the MLE?" Sorry but what are you talking about? If you can present a proof based on solving $$frac{partial L(p;mathbf y)}{partial p}=0$$ please do so (there are none...).
$endgroup$
– Did
Jan 6 at 14:13
add a comment |
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$begingroup$
Previously asked: math.stackexchange.com/questions/446142/pareto-distribution-mle. Without mentioning support of the distribution, your $f(y;p)$ is incomplete. And when you mention the support, say by using indicator functions, you will see that differentiation is not valid to derive the MLE because the likelihood is not differentiable everywhere in the first place. See the linked threads here.
$endgroup$
– StubbornAtom
Jan 6 at 14:16
$begingroup$
THANK YOU! I didn't realise this was a Pareto distribution so I am extremely grateful for this.
$endgroup$
– Maths Barry
Jan 6 at 14:36