When to reject the null hypothesis given a uniformly most powerful test?
Let $X$ be a random variable that takes values in ${-1,0,1}$. We want to test
$H_{0}:Pr(X = -1|Theta = 0) = Pr(X = 0|Theta = 0) = Pr(X = 1| Theta = 0) = 1/3$
versus
$H_{1}=Pr(X = -1 |Theta = 1) = Pr(X = 1|Theta = 1) = 1/4,,Pr(X = 0|Theta = 1) = 1/2.$
Exercise: Calculate the power of this test and show that the UMP test of size $alpha=1/3$ based on $X$ leads to rejecting the null hypothesis when $X=0$.
My attempt: I'm new to this subject and am not sure what a general approach is to these types of questions so any help regarding that would also be much appreciated. Anyway, I figured it would be good to use the Neyman-Pearson Lemma about uniformly most powerful tests but that was about as far as I got.
Thanks!
probability probability-theory statistics statistical-inference hypothesis-testing
asked Dec 2 '18 at 19:42
S. Crim
12710
add a comment |
Let $X$ be a random variable that takes values in ${-1,0,1}$. We want to test
$H_{0}:Pr(X = -1|Theta = 0) = Pr(X = 0|Theta = 0) = Pr(X = 1| Theta = 0) = 1/3$
versus
$H_{1}=Pr(X = -1 |Theta = 1) = Pr(X = 1|Theta = 1) = 1/4,,Pr(X = 0|Theta = 1) = 1/2.$
Exercise: Calculate the power of this test and show that the UMP test of size $alpha=1/3$ based on $X$ leads to rejecting the null hypothesis when $X=0$.
My attempt: I'm new to this subject and am not sure what a general approach is to these types of questions so any help regarding that would also be much appreciated. Anyway, I figured it would be good to use the Neyman-Pearson Lemma about uniformly most powerful tests but that was about as far as I got.
Thanks!
probability probability-theory statistics statistical-inference hypothesis-testing
asked Dec 2 '18 at 19:42
S. Crim
12710
You wrote "Calculate the power of this test" but there is no test provided.
– NCh
Dec 3 '18 at 4:39
add a comment |
Let $X$ be a random variable that takes values in ${-1,0,1}$. We want to test
$H_{0}:Pr(X = -1|Theta = 0) = Pr(X = 0|Theta = 0) = Pr(X = 1| Theta = 0) = 1/3$
versus
$H_{1}=Pr(X = -1 |Theta = 1) = Pr(X = 1|Theta = 1) = 1/4,,Pr(X = 0|Theta = 1) = 1/2.$
Exercise: Calculate the power of this test and show that the UMP test of size $alpha=1/3$ based on $X$ leads to rejecting the null hypothesis when $X=0$.
My attempt: I'm new to this subject and am not sure what a general approach is to these types of questions so any help regarding that would also be much appreciated. Anyway, I figured it would be good to use the Neyman-Pearson Lemma about uniformly most powerful tests but that was about as far as I got.
Thanks!
probability probability-theory statistics statistical-inference hypothesis-testing
asked Dec 2 '18 at 19:42
S. Crim
12710
Let $X$ be a random variable that takes values in ${-1,0,1}$. We want to test
$H_{0}:Pr(X = -1|Theta = 0) = Pr(X = 0|Theta = 0) = Pr(X = 1| Theta = 0) = 1/3$
versus
$H_{1}=Pr(X = -1 |Theta = 1) = Pr(X = 1|Theta = 1) = 1/4,,Pr(X = 0|Theta = 1) = 1/2.$
Exercise: Calculate the power of this test and show that the UMP test of size $alpha=1/3$ based on $X$ leads to rejecting the null hypothesis when $X=0$.
My attempt: I'm new to this subject and am not sure what a general approach is to these types of questions so any help regarding that would also be much appreciated. Anyway, I figured it would be good to use the Neyman-Pearson Lemma about uniformly most powerful tests but that was about as far as I got.
Thanks!
probability probability-theory statistics statistical-inference hypothesis-testing
probability probability-theory statistics statistical-inference hypothesis-testing
asked Dec 2 '18 at 19:42
S. Crim
12710
asked Dec 2 '18 at 19:42
S. Crim
12710
asked Dec 2 '18 at 19:42
S. Crim
12710
asked Dec 2 '18 at 19:42
S. Crim
12710
asked Dec 2 '18 at 19:42
S. Crim
12710
12710
You wrote "Calculate the power of this test" but there is no test provided.
– NCh
Dec 3 '18 at 4:39
add a comment |
You wrote "Calculate the power of this test" but there is no test provided.
– NCh
Dec 3 '18 at 4:39
You wrote "Calculate the power of this test" but there is no test provided.
– NCh
Dec 3 '18 at 4:39
You wrote "Calculate the power of this test" but there is no test provided.
– NCh
Dec 3 '18 at 4:39
add a comment |
1 Answer
1
active
oldest
votes
Calculate likelihood ratio $frac{L_0(X)}{L_1(X)}$ for $X=-1,0,1$ separately:
$$
dfrac{L_0(-1)}{L_1(-1)}=dfrac{L_0(1)}{L_1(1)}=dfrac{1/3}{1/4}=frac43,
$$
$$
dfrac{L_0(0)}{L_1(0)}=dfrac{1/3}{1/2}=frac23.
$$
Likelihood ratio for $X=0$ is smaller than for $X=pm 1$. UMP test rejects null hypothesis for small values of likelihood ratio, so we can simply try to find size of a test
$$
delta(X) = begin{cases}H_0, & frac{L_0(X)}{L_1(X)} >frac23,cr H_1, & frac{L_0(X)}{L_1(X)} leq frac23 end{cases} = begin{cases}H_0, & X = pm 1,cr H_1, & X=0end{cases}
$$
If it is equal to $frac13$, we construct the UMP test of size $alpha=frac13$.
$$alpha=mathbb P_{H_0}(delta=H_1)=mathbb P_{H_0}(X=0)=frac13$$ as required.
answered Dec 3 '18 at 4:54
NCh
6,2682723
Thank you for the answer! I did write ''give the power of this test'' in the wrong place, apologies. It is asked to compute the power of the UMP test of size $alpha=1/3$. I'm not quite getting a step in your answer though so maybe you could help me out. You give the size of the test and then say ''if it is equal to $frac{1}{3}$, we can construct the UMP test of size $alpha=1/3$. I'm not getting what you mean by this and how the earlier representation of $delta(X)$ represents the size of the test. Could you elaborate? I'm new to the subject so maybe I'm missing easy terms.
– S. Crim
Dec 3 '18 at 7:45
1
@S.Crim I mean that if the size of the decribed test is $1/3$, this is exactly the test we needed: it is UMP since it is constructed by NP Lemma, and it has required size.
– NCh
Dec 3 '18 at 9:28
What would the power of the test you just gave be in this case?
– S. Crim
Dec 3 '18 at 11:11
Do you know the definition of test power? The answer follows immediately from the definition.
– NCh
Dec 3 '18 at 11:14
1
The test $delta(X)$ is given. Simply find $beta=mathbb P_{H_1}(delta=H_1)$.
– NCh
Dec 3 '18 at 14:37
|
show 1 more comment
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1 Answer
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1 Answer
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Calculate likelihood ratio $frac{L_0(X)}{L_1(X)}$ for $X=-1,0,1$ separately:
$$
dfrac{L_0(-1)}{L_1(-1)}=dfrac{L_0(1)}{L_1(1)}=dfrac{1/3}{1/4}=frac43,
$$
$$
dfrac{L_0(0)}{L_1(0)}=dfrac{1/3}{1/2}=frac23.
$$
Likelihood ratio for $X=0$ is smaller than for $X=pm 1$. UMP test rejects null hypothesis for small values of likelihood ratio, so we can simply try to find size of a test
$$
delta(X) = begin{cases}H_0, & frac{L_0(X)}{L_1(X)} >frac23,cr H_1, & frac{L_0(X)}{L_1(X)} leq frac23 end{cases} = begin{cases}H_0, & X = pm 1,cr H_1, & X=0end{cases}
$$
If it is equal to $frac13$, we construct the UMP test of size $alpha=frac13$.
$$alpha=mathbb P_{H_0}(delta=H_1)=mathbb P_{H_0}(X=0)=frac13$$ as required.
answered Dec 3 '18 at 4:54
NCh
6,2682723
Thank you for the answer! I did write ''give the power of this test'' in the wrong place, apologies. It is asked to compute the power of the UMP test of size $alpha=1/3$. I'm not quite getting a step in your answer though so maybe you could help me out. You give the size of the test and then say ''if it is equal to $frac{1}{3}$, we can construct the UMP test of size $alpha=1/3$. I'm not getting what you mean by this and how the earlier representation of $delta(X)$ represents the size of the test. Could you elaborate? I'm new to the subject so maybe I'm missing easy terms.
– S. Crim
Dec 3 '18 at 7:45
1
@S.Crim I mean that if the size of the decribed test is $1/3$, this is exactly the test we needed: it is UMP since it is constructed by NP Lemma, and it has required size.
– NCh
Dec 3 '18 at 9:28
What would the power of the test you just gave be in this case?
– S. Crim
Dec 3 '18 at 11:11
Do you know the definition of test power? The answer follows immediately from the definition.
– NCh
Dec 3 '18 at 11:14
1
The test $delta(X)$ is given. Simply find $beta=mathbb P_{H_1}(delta=H_1)$.
– NCh
Dec 3 '18 at 14:37
|
show 1 more comment
Calculate likelihood ratio $frac{L_0(X)}{L_1(X)}$ for $X=-1,0,1$ separately:
$$
dfrac{L_0(-1)}{L_1(-1)}=dfrac{L_0(1)}{L_1(1)}=dfrac{1/3}{1/4}=frac43,
$$
$$
dfrac{L_0(0)}{L_1(0)}=dfrac{1/3}{1/2}=frac23.
$$
Likelihood ratio for $X=0$ is smaller than for $X=pm 1$. UMP test rejects null hypothesis for small values of likelihood ratio, so we can simply try to find size of a test
$$
delta(X) = begin{cases}H_0, & frac{L_0(X)}{L_1(X)} >frac23,cr H_1, & frac{L_0(X)}{L_1(X)} leq frac23 end{cases} = begin{cases}H_0, & X = pm 1,cr H_1, & X=0end{cases}
$$
If it is equal to $frac13$, we construct the UMP test of size $alpha=frac13$.
$$alpha=mathbb P_{H_0}(delta=H_1)=mathbb P_{H_0}(X=0)=frac13$$ as required.
answered Dec 3 '18 at 4:54
NCh
6,2682723
Thank you for the answer! I did write ''give the power of this test'' in the wrong place, apologies. It is asked to compute the power of the UMP test of size $alpha=1/3$. I'm not quite getting a step in your answer though so maybe you could help me out. You give the size of the test and then say ''if it is equal to $frac{1}{3}$, we can construct the UMP test of size $alpha=1/3$. I'm not getting what you mean by this and how the earlier representation of $delta(X)$ represents the size of the test. Could you elaborate? I'm new to the subject so maybe I'm missing easy terms.
– S. Crim
Dec 3 '18 at 7:45
1
@S.Crim I mean that if the size of the decribed test is $1/3$, this is exactly the test we needed: it is UMP since it is constructed by NP Lemma, and it has required size.
– NCh
Dec 3 '18 at 9:28
What would the power of the test you just gave be in this case?
– S. Crim
Dec 3 '18 at 11:11
Do you know the definition of test power? The answer follows immediately from the definition.
– NCh
Dec 3 '18 at 11:14
1
The test $delta(X)$ is given. Simply find $beta=mathbb P_{H_1}(delta=H_1)$.
– NCh
Dec 3 '18 at 14:37
|
show 1 more comment
Calculate likelihood ratio $frac{L_0(X)}{L_1(X)}$ for $X=-1,0,1$ separately:
$$
dfrac{L_0(-1)}{L_1(-1)}=dfrac{L_0(1)}{L_1(1)}=dfrac{1/3}{1/4}=frac43,
$$
$$
dfrac{L_0(0)}{L_1(0)}=dfrac{1/3}{1/2}=frac23.
$$
Likelihood ratio for $X=0$ is smaller than for $X=pm 1$. UMP test rejects null hypothesis for small values of likelihood ratio, so we can simply try to find size of a test
$$
delta(X) = begin{cases}H_0, & frac{L_0(X)}{L_1(X)} >frac23,cr H_1, & frac{L_0(X)}{L_1(X)} leq frac23 end{cases} = begin{cases}H_0, & X = pm 1,cr H_1, & X=0end{cases}
$$
If it is equal to $frac13$, we construct the UMP test of size $alpha=frac13$.
$$alpha=mathbb P_{H_0}(delta=H_1)=mathbb P_{H_0}(X=0)=frac13$$ as required.
answered Dec 3 '18 at 4:54
NCh
6,2682723
Calculate likelihood ratio $frac{L_0(X)}{L_1(X)}$ for $X=-1,0,1$ separately:
$$
dfrac{L_0(-1)}{L_1(-1)}=dfrac{L_0(1)}{L_1(1)}=dfrac{1/3}{1/4}=frac43,
$$
$$
dfrac{L_0(0)}{L_1(0)}=dfrac{1/3}{1/2}=frac23.
$$
Likelihood ratio for $X=0$ is smaller than for $X=pm 1$. UMP test rejects null hypothesis for small values of likelihood ratio, so we can simply try to find size of a test
$$
delta(X) = begin{cases}H_0, & frac{L_0(X)}{L_1(X)} >frac23,cr H_1, & frac{L_0(X)}{L_1(X)} leq frac23 end{cases} = begin{cases}H_0, & X = pm 1,cr H_1, & X=0end{cases}
$$
If it is equal to $frac13$, we construct the UMP test of size $alpha=frac13$.
$$alpha=mathbb P_{H_0}(delta=H_1)=mathbb P_{H_0}(X=0)=frac13$$ as required.
answered Dec 3 '18 at 4:54
NCh
6,2682723
answered Dec 3 '18 at 4:54
NCh
6,2682723
answered Dec 3 '18 at 4:54
NCh
6,2682723
answered Dec 3 '18 at 4:54
NCh
6,2682723
6,2682723
Thank you for the answer! I did write ''give the power of this test'' in the wrong place, apologies. It is asked to compute the power of the UMP test of size $alpha=1/3$. I'm not quite getting a step in your answer though so maybe you could help me out. You give the size of the test and then say ''if it is equal to $frac{1}{3}$, we can construct the UMP test of size $alpha=1/3$. I'm not getting what you mean by this and how the earlier representation of $delta(X)$ represents the size of the test. Could you elaborate? I'm new to the subject so maybe I'm missing easy terms.
– S. Crim
Dec 3 '18 at 7:45
1
@S.Crim I mean that if the size of the decribed test is $1/3$, this is exactly the test we needed: it is UMP since it is constructed by NP Lemma, and it has required size.
– NCh
Dec 3 '18 at 9:28
What would the power of the test you just gave be in this case?
– S. Crim
Dec 3 '18 at 11:11
Do you know the definition of test power? The answer follows immediately from the definition.
– NCh
Dec 3 '18 at 11:14
1
The test $delta(X)$ is given. Simply find $beta=mathbb P_{H_1}(delta=H_1)$.
– NCh
Dec 3 '18 at 14:37
|
show 1 more comment
Thank you for the answer! I did write ''give the power of this test'' in the wrong place, apologies. It is asked to compute the power of the UMP test of size $alpha=1/3$. I'm not quite getting a step in your answer though so maybe you could help me out. You give the size of the test and then say ''if it is equal to $frac{1}{3}$, we can construct the UMP test of size $alpha=1/3$. I'm not getting what you mean by this and how the earlier representation of $delta(X)$ represents the size of the test. Could you elaborate? I'm new to the subject so maybe I'm missing easy terms.
– S. Crim
Dec 3 '18 at 7:45
1
@S.Crim I mean that if the size of the decribed test is $1/3$, this is exactly the test we needed: it is UMP since it is constructed by NP Lemma, and it has required size.
– NCh
Dec 3 '18 at 9:28
What would the power of the test you just gave be in this case?
– S. Crim
Dec 3 '18 at 11:11
Do you know the definition of test power? The answer follows immediately from the definition.
– NCh
Dec 3 '18 at 11:14
1
The test $delta(X)$ is given. Simply find $beta=mathbb P_{H_1}(delta=H_1)$.
– NCh
Dec 3 '18 at 14:37
Thank you for the answer! I did write ''give the power of this test'' in the wrong place, apologies. It is asked to compute the power of the UMP test of size $alpha=1/3$. I'm not quite getting a step in your answer though so maybe you could help me out. You give the size of the test and then say ''if it is equal to $frac{1}{3}$, we can construct the UMP test of size $alpha=1/3$. I'm not getting what you mean by this and how the earlier representation of $delta(X)$ represents the size of the test. Could you elaborate? I'm new to the subject so maybe I'm missing easy terms.
– S. Crim
Dec 3 '18 at 7:45
Thank you for the answer! I did write ''give the power of this test'' in the wrong place, apologies. It is asked to compute the power of the UMP test of size $alpha=1/3$. I'm not quite getting a step in your answer though so maybe you could help me out. You give the size of the test and then say ''if it is equal to $frac{1}{3}$, we can construct the UMP test of size $alpha=1/3$. I'm not getting what you mean by this and how the earlier representation of $delta(X)$ represents the size of the test. Could you elaborate? I'm new to the subject so maybe I'm missing easy terms.
– S. Crim
Dec 3 '18 at 7:45
1
1
@S.Crim I mean that if the size of the decribed test is $1/3$, this is exactly the test we needed: it is UMP since it is constructed by NP Lemma, and it has required size.
– NCh
Dec 3 '18 at 9:28
@S.Crim I mean that if the size of the decribed test is $1/3$, this is exactly the test we needed: it is UMP since it is constructed by NP Lemma, and it has required size.
– NCh
Dec 3 '18 at 9:28
What would the power of the test you just gave be in this case?
– S. Crim
Dec 3 '18 at 11:11
What would the power of the test you just gave be in this case?
– S. Crim
Dec 3 '18 at 11:11
Do you know the definition of test power? The answer follows immediately from the definition.
– NCh
Dec 3 '18 at 11:14
Do you know the definition of test power? The answer follows immediately from the definition.
– NCh
Dec 3 '18 at 11:14
1
1
The test $delta(X)$ is given. Simply find $beta=mathbb P_{H_1}(delta=H_1)$.
– NCh
Dec 3 '18 at 14:37
The test $delta(X)$ is given. Simply find $beta=mathbb P_{H_1}(delta=H_1)$.
– NCh
Dec 3 '18 at 14:37
|
show 1 more comment
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You wrote "Calculate the power of this test" but there is no test provided.
– NCh
Dec 3 '18 at 4:39