If $f (ymid theta)=(theta + 1)y^theta$, find an estimator for $θ$ by the method of moments.
Let $Y_1, Y_2, . . . , Y_n$ denote a random sample from the probability density function
$$f (y mid theta)=begin{cases} (theta + 1)y^theta, & 0 < y < 1; theta > −1,\ 0 ,& text{ elsewhere }end{cases}$$
Find an estimator for $theta$ by the method of moments.
I am told that that $μ = frac{theta + 1}{theta + 2} $. I am wondering how do they show this ?
probability statistics probability-theory probability-distributions
edited Dec 10 '18 at 18:43
StubbornAtom
5,32411138
asked Apr 11 '14 at 4:48
afsdf dfsaf
59611235
add a comment |
Let $Y_1, Y_2, . . . , Y_n$ denote a random sample from the probability density function
$$f (y mid theta)=begin{cases} (theta + 1)y^theta, & 0 < y < 1; theta > −1,\ 0 ,& text{ elsewhere }end{cases}$$
Find an estimator for $theta$ by the method of moments.
I am told that that $μ = frac{theta + 1}{theta + 2} $. I am wondering how do they show this ?
probability statistics probability-theory probability-distributions
edited Dec 10 '18 at 18:43
StubbornAtom
5,32411138
asked Apr 11 '14 at 4:48
afsdf dfsaf
59611235
add a comment |
Let $Y_1, Y_2, . . . , Y_n$ denote a random sample from the probability density function
$$f (y mid theta)=begin{cases} (theta + 1)y^theta, & 0 < y < 1; theta > −1,\ 0 ,& text{ elsewhere }end{cases}$$
Find an estimator for $theta$ by the method of moments.
I am told that that $μ = frac{theta + 1}{theta + 2} $. I am wondering how do they show this ?
probability statistics probability-theory probability-distributions
edited Dec 10 '18 at 18:43
StubbornAtom
5,32411138
asked Apr 11 '14 at 4:48
afsdf dfsaf
59611235
Let $Y_1, Y_2, . . . , Y_n$ denote a random sample from the probability density function
$$f (y mid theta)=begin{cases} (theta + 1)y^theta, & 0 < y < 1; theta > −1,\ 0 ,& text{ elsewhere }end{cases}$$
Find an estimator for $theta$ by the method of moments.
I am told that that $μ = frac{theta + 1}{theta + 2} $. I am wondering how do they show this ?
probability statistics probability-theory probability-distributions
probability statistics probability-theory probability-distributions
edited Dec 10 '18 at 18:43
StubbornAtom
5,32411138
asked Apr 11 '14 at 4:48
afsdf dfsaf
59611235
edited Dec 10 '18 at 18:43
StubbornAtom
5,32411138
asked Apr 11 '14 at 4:48
afsdf dfsaf
59611235
edited Dec 10 '18 at 18:43
StubbornAtom
5,32411138
edited Dec 10 '18 at 18:43
StubbornAtom
5,32411138
edited Dec 10 '18 at 18:43
StubbornAtom
5,32411138
5,32411138
asked Apr 11 '14 at 4:48
afsdf dfsaf
59611235
asked Apr 11 '14 at 4:48
afsdf dfsaf
59611235
asked Apr 11 '14 at 4:48
afsdf dfsaf
59611235
59611235
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
The motivation of the method of moments estimate is that it produces a model that has the same first $n$ raw moments as the data (as represented by the empirical distribution). Let $Y_1, Y_2, cdots, Y_n$ be a random sample, therefore
$$
text{E}left[Y^nright]=frac{1}{n}sum_{i=1}^n y_i^n.tag1
$$
Let us obtain the first raw moment of the data.
$$
overline{y}=frac{1}{n}sum_{i=1}^n y_i.tag2
$$
Now, we obtain the first raw moment of the sample distribution.
$$
begin{align}
mu&=text{E}[Y]\
&=int_{y=0}^1 yf(y|theta) dy\
&=int_{y=0}^1 y(theta+1)y^theta dy\
&=(theta+1)int_{y=0}^1 y^{theta+1} dy\
&=(theta+1)cdotleft.frac{y^{theta+2}}{theta+2}right|_{y=0}^1\
&=frac{theta+1}{theta+2}.tag3
end{align}
$$
Thus, based on $(1)$, from $(2)$ and $(3)$ we obtain
$$
begin{align}
frac{hat{theta}+1}{hat{theta}+2}&=overline{y}\
hat{theta}+1&=overline{y}(hat{theta}+2)\
hat{theta}+1&=overline{y}hat{theta}+2overline{y}\
hat{theta}-overline{y}hat{theta}&=2overline{y}-1\
hat{theta}(1-overline{y})&=2overline{y}-1\
hat{theta}&=frac{2overline{y}-1}{1-overline{y}}\
&=-frac{2overline{y}-1}{overline{y}-1}.
end{align}
$$
$$\$$
$$Largecolor{blue}{text{# }mathbb{Q.E.D.}text{ #}}$$
answered Apr 11 '14 at 10:41
Tunk-Fey
23k969100
add a comment |
For the method of moments, you want to estimate the parameter $theta$ as a function of the sample mean, $bar{y}$.
Begin by finding the expected value $mu = int_0^1y*(theta +1)*y^{theta}dy = int_0^1(theta +1)*y^{theta+1}dy = frac{theta +1}{theta +2}*(y^{theta+2}|_0^1 = frac{theta+1}{theta +2}$.
Now replace $mu$ with the sample mean, $bar{y} = frac{1}{n}sum_{i=1}^nY_i$: $bar{y} = frac{theta+1}{theta +2}$.
Now solve $theta$ in terms of $bar{y}$: $hat{theta} = frac{1-2bar{y}}{bar{y}-1}$
answered Apr 11 '14 at 5:11
Erroldactyl
605
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
The motivation of the method of moments estimate is that it produces a model that has the same first $n$ raw moments as the data (as represented by the empirical distribution). Let $Y_1, Y_2, cdots, Y_n$ be a random sample, therefore
$$
text{E}left[Y^nright]=frac{1}{n}sum_{i=1}^n y_i^n.tag1
$$
Let us obtain the first raw moment of the data.
$$
overline{y}=frac{1}{n}sum_{i=1}^n y_i.tag2
$$
Now, we obtain the first raw moment of the sample distribution.
$$
begin{align}
mu&=text{E}[Y]\
&=int_{y=0}^1 yf(y|theta) dy\
&=int_{y=0}^1 y(theta+1)y^theta dy\
&=(theta+1)int_{y=0}^1 y^{theta+1} dy\
&=(theta+1)cdotleft.frac{y^{theta+2}}{theta+2}right|_{y=0}^1\
&=frac{theta+1}{theta+2}.tag3
end{align}
$$
Thus, based on $(1)$, from $(2)$ and $(3)$ we obtain
$$
begin{align}
frac{hat{theta}+1}{hat{theta}+2}&=overline{y}\
hat{theta}+1&=overline{y}(hat{theta}+2)\
hat{theta}+1&=overline{y}hat{theta}+2overline{y}\
hat{theta}-overline{y}hat{theta}&=2overline{y}-1\
hat{theta}(1-overline{y})&=2overline{y}-1\
hat{theta}&=frac{2overline{y}-1}{1-overline{y}}\
&=-frac{2overline{y}-1}{overline{y}-1}.
end{align}
$$
$$\$$
$$Largecolor{blue}{text{# }mathbb{Q.E.D.}text{ #}}$$
answered Apr 11 '14 at 10:41
Tunk-Fey
23k969100
add a comment |
The motivation of the method of moments estimate is that it produces a model that has the same first $n$ raw moments as the data (as represented by the empirical distribution). Let $Y_1, Y_2, cdots, Y_n$ be a random sample, therefore
$$
text{E}left[Y^nright]=frac{1}{n}sum_{i=1}^n y_i^n.tag1
$$
Let us obtain the first raw moment of the data.
$$
overline{y}=frac{1}{n}sum_{i=1}^n y_i.tag2
$$
Now, we obtain the first raw moment of the sample distribution.
$$
begin{align}
mu&=text{E}[Y]\
&=int_{y=0}^1 yf(y|theta) dy\
&=int_{y=0}^1 y(theta+1)y^theta dy\
&=(theta+1)int_{y=0}^1 y^{theta+1} dy\
&=(theta+1)cdotleft.frac{y^{theta+2}}{theta+2}right|_{y=0}^1\
&=frac{theta+1}{theta+2}.tag3
end{align}
$$
Thus, based on $(1)$, from $(2)$ and $(3)$ we obtain
$$
begin{align}
frac{hat{theta}+1}{hat{theta}+2}&=overline{y}\
hat{theta}+1&=overline{y}(hat{theta}+2)\
hat{theta}+1&=overline{y}hat{theta}+2overline{y}\
hat{theta}-overline{y}hat{theta}&=2overline{y}-1\
hat{theta}(1-overline{y})&=2overline{y}-1\
hat{theta}&=frac{2overline{y}-1}{1-overline{y}}\
&=-frac{2overline{y}-1}{overline{y}-1}.
end{align}
$$
$$\$$
$$Largecolor{blue}{text{# }mathbb{Q.E.D.}text{ #}}$$
answered Apr 11 '14 at 10:41
Tunk-Fey
23k969100
add a comment |
The motivation of the method of moments estimate is that it produces a model that has the same first $n$ raw moments as the data (as represented by the empirical distribution). Let $Y_1, Y_2, cdots, Y_n$ be a random sample, therefore
$$
text{E}left[Y^nright]=frac{1}{n}sum_{i=1}^n y_i^n.tag1
$$
Let us obtain the first raw moment of the data.
$$
overline{y}=frac{1}{n}sum_{i=1}^n y_i.tag2
$$
Now, we obtain the first raw moment of the sample distribution.
$$
begin{align}
mu&=text{E}[Y]\
&=int_{y=0}^1 yf(y|theta) dy\
&=int_{y=0}^1 y(theta+1)y^theta dy\
&=(theta+1)int_{y=0}^1 y^{theta+1} dy\
&=(theta+1)cdotleft.frac{y^{theta+2}}{theta+2}right|_{y=0}^1\
&=frac{theta+1}{theta+2}.tag3
end{align}
$$
Thus, based on $(1)$, from $(2)$ and $(3)$ we obtain
$$
begin{align}
frac{hat{theta}+1}{hat{theta}+2}&=overline{y}\
hat{theta}+1&=overline{y}(hat{theta}+2)\
hat{theta}+1&=overline{y}hat{theta}+2overline{y}\
hat{theta}-overline{y}hat{theta}&=2overline{y}-1\
hat{theta}(1-overline{y})&=2overline{y}-1\
hat{theta}&=frac{2overline{y}-1}{1-overline{y}}\
&=-frac{2overline{y}-1}{overline{y}-1}.
end{align}
$$
$$\$$
$$Largecolor{blue}{text{# }mathbb{Q.E.D.}text{ #}}$$
answered Apr 11 '14 at 10:41
Tunk-Fey
23k969100
The motivation of the method of moments estimate is that it produces a model that has the same first $n$ raw moments as the data (as represented by the empirical distribution). Let $Y_1, Y_2, cdots, Y_n$ be a random sample, therefore
$$
text{E}left[Y^nright]=frac{1}{n}sum_{i=1}^n y_i^n.tag1
$$
Let us obtain the first raw moment of the data.
$$
overline{y}=frac{1}{n}sum_{i=1}^n y_i.tag2
$$
Now, we obtain the first raw moment of the sample distribution.
$$
begin{align}
mu&=text{E}[Y]\
&=int_{y=0}^1 yf(y|theta) dy\
&=int_{y=0}^1 y(theta+1)y^theta dy\
&=(theta+1)int_{y=0}^1 y^{theta+1} dy\
&=(theta+1)cdotleft.frac{y^{theta+2}}{theta+2}right|_{y=0}^1\
&=frac{theta+1}{theta+2}.tag3
end{align}
$$
Thus, based on $(1)$, from $(2)$ and $(3)$ we obtain
$$
begin{align}
frac{hat{theta}+1}{hat{theta}+2}&=overline{y}\
hat{theta}+1&=overline{y}(hat{theta}+2)\
hat{theta}+1&=overline{y}hat{theta}+2overline{y}\
hat{theta}-overline{y}hat{theta}&=2overline{y}-1\
hat{theta}(1-overline{y})&=2overline{y}-1\
hat{theta}&=frac{2overline{y}-1}{1-overline{y}}\
&=-frac{2overline{y}-1}{overline{y}-1}.
end{align}
$$
$$\$$
$$Largecolor{blue}{text{# }mathbb{Q.E.D.}text{ #}}$$
answered Apr 11 '14 at 10:41
Tunk-Fey
23k969100
answered Apr 11 '14 at 10:41
Tunk-Fey
23k969100
answered Apr 11 '14 at 10:41
Tunk-Fey
23k969100
answered Apr 11 '14 at 10:41
Tunk-Fey
23k969100
23k969100
add a comment |
add a comment |
For the method of moments, you want to estimate the parameter $theta$ as a function of the sample mean, $bar{y}$.
Begin by finding the expected value $mu = int_0^1y*(theta +1)*y^{theta}dy = int_0^1(theta +1)*y^{theta+1}dy = frac{theta +1}{theta +2}*(y^{theta+2}|_0^1 = frac{theta+1}{theta +2}$.
Now replace $mu$ with the sample mean, $bar{y} = frac{1}{n}sum_{i=1}^nY_i$: $bar{y} = frac{theta+1}{theta +2}$.
Now solve $theta$ in terms of $bar{y}$: $hat{theta} = frac{1-2bar{y}}{bar{y}-1}$
answered Apr 11 '14 at 5:11
Erroldactyl
605
add a comment |
For the method of moments, you want to estimate the parameter $theta$ as a function of the sample mean, $bar{y}$.
Begin by finding the expected value $mu = int_0^1y*(theta +1)*y^{theta}dy = int_0^1(theta +1)*y^{theta+1}dy = frac{theta +1}{theta +2}*(y^{theta+2}|_0^1 = frac{theta+1}{theta +2}$.
Now replace $mu$ with the sample mean, $bar{y} = frac{1}{n}sum_{i=1}^nY_i$: $bar{y} = frac{theta+1}{theta +2}$.
Now solve $theta$ in terms of $bar{y}$: $hat{theta} = frac{1-2bar{y}}{bar{y}-1}$
answered Apr 11 '14 at 5:11
Erroldactyl
605
add a comment |
For the method of moments, you want to estimate the parameter $theta$ as a function of the sample mean, $bar{y}$.
Begin by finding the expected value $mu = int_0^1y*(theta +1)*y^{theta}dy = int_0^1(theta +1)*y^{theta+1}dy = frac{theta +1}{theta +2}*(y^{theta+2}|_0^1 = frac{theta+1}{theta +2}$.
Now replace $mu$ with the sample mean, $bar{y} = frac{1}{n}sum_{i=1}^nY_i$: $bar{y} = frac{theta+1}{theta +2}$.
Now solve $theta$ in terms of $bar{y}$: $hat{theta} = frac{1-2bar{y}}{bar{y}-1}$
answered Apr 11 '14 at 5:11
Erroldactyl
605
For the method of moments, you want to estimate the parameter $theta$ as a function of the sample mean, $bar{y}$.
Begin by finding the expected value $mu = int_0^1y*(theta +1)*y^{theta}dy = int_0^1(theta +1)*y^{theta+1}dy = frac{theta +1}{theta +2}*(y^{theta+2}|_0^1 = frac{theta+1}{theta +2}$.
Now replace $mu$ with the sample mean, $bar{y} = frac{1}{n}sum_{i=1}^nY_i$: $bar{y} = frac{theta+1}{theta +2}$.
Now solve $theta$ in terms of $bar{y}$: $hat{theta} = frac{1-2bar{y}}{bar{y}-1}$
answered Apr 11 '14 at 5:11
Erroldactyl
605
answered Apr 11 '14 at 5:11
Erroldactyl
605
answered Apr 11 '14 at 5:11
Erroldactyl
605
answered Apr 11 '14 at 5:11
Erroldactyl
605
605
add a comment |
add a comment |
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