Neyman-Pearson lemma for the sample of size one
$begingroup$
Let $X$ be a continuous random variable on $[0, 1]$ with a density function
$$f(theta |x ) = frac{x^{theta}}{theta + 1}$$
Let's take a sample that consists on the only one element from the given distribution, say $x_{1}$ and test the following hypothesis:
$$H_{0}: theta = 0 text{vs. } H_{1}: theta = 1$$
Neyman - Pearson lemma gives
$$lambda(x) = frac{L(theta = 0 | x)}{L(theta = 1 | x)} = frac{2}{x}$$
Thus, the rejection regions consists of presicely those $x$ that satsify $frac{2}{x} leq a$.
In order to find $a$, we calculate
$$mathbb{P}(lambda(x) leq a | H_{0}) = mathbb{P}(x geq frac{2}{a} | H_{0}) = int_{frac{2}{a}}^{1} {1_{[0, 1]} dx} = 1 - frac{2}{a}$$
Thus $frac{2}{a} = 1 - alpha$ and we reject the hypothesis if $x leq 1 - alpha$, where $alpha$ is the significance level.
As for me that sounds quite counter-intuitive, since for the large values of $alpha$ we have the critical value concentrated near the end on an interval, which i strongly doubt. Am i missing something crucial here?
statistics statistical-inference hypothesis-testing
asked Dec 14 '18 at 15:35
hyperkahlerhyperkahler
1,483714
$endgroup$
add a comment |
$begingroup$
Let $X$ be a continuous random variable on $[0, 1]$ with a density function
$$f(theta |x ) = frac{x^{theta}}{theta + 1}$$
Let's take a sample that consists on the only one element from the given distribution, say $x_{1}$ and test the following hypothesis:
$$H_{0}: theta = 0 text{vs. } H_{1}: theta = 1$$
Neyman - Pearson lemma gives
$$lambda(x) = frac{L(theta = 0 | x)}{L(theta = 1 | x)} = frac{2}{x}$$
Thus, the rejection regions consists of presicely those $x$ that satsify $frac{2}{x} leq a$.
In order to find $a$, we calculate
$$mathbb{P}(lambda(x) leq a | H_{0}) = mathbb{P}(x geq frac{2}{a} | H_{0}) = int_{frac{2}{a}}^{1} {1_{[0, 1]} dx} = 1 - frac{2}{a}$$
Thus $frac{2}{a} = 1 - alpha$ and we reject the hypothesis if $x leq 1 - alpha$, where $alpha$ is the significance level.
As for me that sounds quite counter-intuitive, since for the large values of $alpha$ we have the critical value concentrated near the end on an interval, which i strongly doubt. Am i missing something crucial here?
statistics statistical-inference hypothesis-testing
asked Dec 14 '18 at 15:35
hyperkahlerhyperkahler
1,483714
$endgroup$
$begingroup$
Your $f$ is not a pdf in the first place; $f(xmidtheta)=(theta+1)x^{theta}mathbf1_{[0,1]}(x)$ for $theta>-1$ makes more sense.
$endgroup$
– StubbornAtom
Dec 14 '18 at 20:03
$begingroup$
@StubbornAtom Ouh, indeed, thanks. In fact, this is a distribution in case $theta = 0$, but in this case the problem boils down to somewhat very strange and trivial
$endgroup$
– hyperkahler
Dec 14 '18 at 20:10
add a comment |
$begingroup$
Let $X$ be a continuous random variable on $[0, 1]$ with a density function
$$f(theta |x ) = frac{x^{theta}}{theta + 1}$$
Let's take a sample that consists on the only one element from the given distribution, say $x_{1}$ and test the following hypothesis:
$$H_{0}: theta = 0 text{vs. } H_{1}: theta = 1$$
Neyman - Pearson lemma gives
$$lambda(x) = frac{L(theta = 0 | x)}{L(theta = 1 | x)} = frac{2}{x}$$
Thus, the rejection regions consists of presicely those $x$ that satsify $frac{2}{x} leq a$.
In order to find $a$, we calculate
$$mathbb{P}(lambda(x) leq a | H_{0}) = mathbb{P}(x geq frac{2}{a} | H_{0}) = int_{frac{2}{a}}^{1} {1_{[0, 1]} dx} = 1 - frac{2}{a}$$
Thus $frac{2}{a} = 1 - alpha$ and we reject the hypothesis if $x leq 1 - alpha$, where $alpha$ is the significance level.
As for me that sounds quite counter-intuitive, since for the large values of $alpha$ we have the critical value concentrated near the end on an interval, which i strongly doubt. Am i missing something crucial here?
statistics statistical-inference hypothesis-testing
asked Dec 14 '18 at 15:35
hyperkahlerhyperkahler
1,483714
$endgroup$
Let $X$ be a continuous random variable on $[0, 1]$ with a density function
$$f(theta |x ) = frac{x^{theta}}{theta + 1}$$
Let's take a sample that consists on the only one element from the given distribution, say $x_{1}$ and test the following hypothesis:
$$H_{0}: theta = 0 text{vs. } H_{1}: theta = 1$$
Neyman - Pearson lemma gives
$$lambda(x) = frac{L(theta = 0 | x)}{L(theta = 1 | x)} = frac{2}{x}$$
Thus, the rejection regions consists of presicely those $x$ that satsify $frac{2}{x} leq a$.
In order to find $a$, we calculate
$$mathbb{P}(lambda(x) leq a | H_{0}) = mathbb{P}(x geq frac{2}{a} | H_{0}) = int_{frac{2}{a}}^{1} {1_{[0, 1]} dx} = 1 - frac{2}{a}$$
Thus $frac{2}{a} = 1 - alpha$ and we reject the hypothesis if $x leq 1 - alpha$, where $alpha$ is the significance level.
As for me that sounds quite counter-intuitive, since for the large values of $alpha$ we have the critical value concentrated near the end on an interval, which i strongly doubt. Am i missing something crucial here?
statistics statistical-inference hypothesis-testing
statistics statistical-inference hypothesis-testing
asked Dec 14 '18 at 15:35
hyperkahlerhyperkahler
1,483714
asked Dec 14 '18 at 15:35
hyperkahlerhyperkahler
1,483714
asked Dec 14 '18 at 15:35
hyperkahlerhyperkahler
1,483714
asked Dec 14 '18 at 15:35
hyperkahlerhyperkahler
1,483714
asked Dec 14 '18 at 15:35
hyperkahlerhyperkahler
1,483714
1,483714
$begingroup$
Your $f$ is not a pdf in the first place; $f(xmidtheta)=(theta+1)x^{theta}mathbf1_{[0,1]}(x)$ for $theta>-1$ makes more sense.
$endgroup$
– StubbornAtom
Dec 14 '18 at 20:03
$begingroup$
@StubbornAtom Ouh, indeed, thanks. In fact, this is a distribution in case $theta = 0$, but in this case the problem boils down to somewhat very strange and trivial
$endgroup$
– hyperkahler
Dec 14 '18 at 20:10
add a comment |
$begingroup$
Your $f$ is not a pdf in the first place; $f(xmidtheta)=(theta+1)x^{theta}mathbf1_{[0,1]}(x)$ for $theta>-1$ makes more sense.
$endgroup$
– StubbornAtom
Dec 14 '18 at 20:03
$begingroup$
@StubbornAtom Ouh, indeed, thanks. In fact, this is a distribution in case $theta = 0$, but in this case the problem boils down to somewhat very strange and trivial
$endgroup$
– hyperkahler
Dec 14 '18 at 20:10
$begingroup$
Your $f$ is not a pdf in the first place; $f(xmidtheta)=(theta+1)x^{theta}mathbf1_{[0,1]}(x)$ for $theta>-1$ makes more sense.
$endgroup$
– StubbornAtom
Dec 14 '18 at 20:03
$begingroup$
Your $f$ is not a pdf in the first place; $f(xmidtheta)=(theta+1)x^{theta}mathbf1_{[0,1]}(x)$ for $theta>-1$ makes more sense.
$endgroup$
– StubbornAtom
Dec 14 '18 at 20:03
$begingroup$
@StubbornAtom Ouh, indeed, thanks. In fact, this is a distribution in case $theta = 0$, but in this case the problem boils down to somewhat very strange and trivial
$endgroup$
– hyperkahler
Dec 14 '18 at 20:10
$begingroup$
@StubbornAtom Ouh, indeed, thanks. In fact, this is a distribution in case $theta = 0$, but in this case the problem boils down to somewhat very strange and trivial
$endgroup$
– hyperkahler
Dec 14 '18 at 20:10
add a comment |
1 Answer
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$begingroup$
It seems there is nothing wrong with that. For example, let $alpha = 0.99$. It means that we allow large probability of type 1 error (almost don't care about it) and we may end up with large critical region. Actually, corresponding critical region is
$$
xgeq 1-alpha = 0.01,
$$ and almost any value of $x$ leads to rejection of $H_0$ as expected. As a result, we get a very powerful test at the expense of significance level.
answered Dec 14 '18 at 15:47
SongSong
16.7k1941
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It seems there is nothing wrong with that. For example, let $alpha = 0.99$. It means that we allow large probability of type 1 error (almost don't care about it) and we may end up with large critical region. Actually, corresponding critical region is
$$
xgeq 1-alpha = 0.01,
$$ and almost any value of $x$ leads to rejection of $H_0$ as expected. As a result, we get a very powerful test at the expense of significance level.
answered Dec 14 '18 at 15:47
SongSong
16.7k1941
$endgroup$
add a comment |
$begingroup$
It seems there is nothing wrong with that. For example, let $alpha = 0.99$. It means that we allow large probability of type 1 error (almost don't care about it) and we may end up with large critical region. Actually, corresponding critical region is
$$
xgeq 1-alpha = 0.01,
$$ and almost any value of $x$ leads to rejection of $H_0$ as expected. As a result, we get a very powerful test at the expense of significance level.
answered Dec 14 '18 at 15:47
SongSong
16.7k1941
$endgroup$
add a comment |
$begingroup$
It seems there is nothing wrong with that. For example, let $alpha = 0.99$. It means that we allow large probability of type 1 error (almost don't care about it) and we may end up with large critical region. Actually, corresponding critical region is
$$
xgeq 1-alpha = 0.01,
$$ and almost any value of $x$ leads to rejection of $H_0$ as expected. As a result, we get a very powerful test at the expense of significance level.
answered Dec 14 '18 at 15:47
SongSong
16.7k1941
$endgroup$
It seems there is nothing wrong with that. For example, let $alpha = 0.99$. It means that we allow large probability of type 1 error (almost don't care about it) and we may end up with large critical region. Actually, corresponding critical region is
$$
xgeq 1-alpha = 0.01,
$$ and almost any value of $x$ leads to rejection of $H_0$ as expected. As a result, we get a very powerful test at the expense of significance level.
answered Dec 14 '18 at 15:47
SongSong
16.7k1941
edited Dec 14 '18 at 15:52
answered Dec 14 '18 at 15:47
SongSong
16.7k1941
answered Dec 14 '18 at 15:47
SongSong
16.7k1941
answered Dec 14 '18 at 15:47
SongSong
16.7k1941
16.7k1941
add a comment |
add a comment |
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$begingroup$
Your $f$ is not a pdf in the first place; $f(xmidtheta)=(theta+1)x^{theta}mathbf1_{[0,1]}(x)$ for $theta>-1$ makes more sense.
$endgroup$
– StubbornAtom
Dec 14 '18 at 20:03
$begingroup$
@StubbornAtom Ouh, indeed, thanks. In fact, this is a distribution in case $theta = 0$, but in this case the problem boils down to somewhat very strange and trivial
$endgroup$
– hyperkahler
Dec 14 '18 at 20:10