Probability of type I error

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Homework Statement


Given that X is a uniform random variable on the interval [tex](0, \theta)[/tex], we might test [tex]Ho: \theta = 1[/tex] versus the alternative [tex]H_{1}: \theta = 2[/tex] by taking a sample of 2 observations of X and rejecting Ho if [tex]\bar{X} > 0.99[/tex]. Compute [tex]\alpha[/tex]


2. The attempt at a solution
[tex]\alpha[/tex] = P[type I error]
= P[rejecting Ho| Ho is true]
= P[[tex]\bar{X} > 0.99 given that \theta = 1][/tex]

I just know that if X is a uniform random variable, it has a pdf:
[tex]f\left(x; a, b\right) = \frac{1}{b - a}I_{[a, b]}(x)[/tex]

Kindly help me what to do next.
 

Answers and Replies

  • #2
LCKurtz
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So, under H0, what is the joint pdf of X1 and X2? You didn't state it but I'm guessing the two observations are independent. Once you have that you can calculate α directly.
 
  • #3
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So, under H0, what is the joint pdf of X1 and X2? You didn't state it but I'm guessing the two observations are independent. Once you have that you can calculate α directly.

the joint pdf of X[tex]_{1}[/tex] and X[tex]_{2}[/tex] is [tex]\frac{1}{\theta}[/tex] [tex]\frac{1}{\theta}[/tex] = [tex]\frac{1}{\theta^{2}}[/tex]
But I still don't get how is it related in getting [tex]\alpha[/tex]
 
  • #4
LCKurtz
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the joint pdf of X[tex]_{1}[/tex] and X[tex]_{2}[/tex] is [tex]\frac{1}{\theta}[/tex] [tex]\frac{1}{\theta}[/tex] = [tex]\frac{1}{\theta^{2}}[/tex]
But I still don't get how is it related in getting [tex]\alpha[/tex]

But under H0, θ = 1, so your joint probability is 1 on the unit square. Do you know how to calculate a probability when you know the joint pdf? You just need to calculate the probability

[tex]P\left(\frac {X_1+X_2}{2} >.99\right)[/tex]
 
  • #5
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But under H0, θ = 1, so your joint probability is 1 on the unit square. Do you know how to calculate a probability when you know the joint pdf? You just need to calculate the probability

[tex]P\left(\frac {X_1+X_2}{2} >.99\right)[/tex]

I really am confused how to calculate this probability.
Also I think there is no table for uniform distribution.
 
  • #6
LCKurtz
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But under H0, θ = 1, so your joint probability is 1 on the unit square. Do you know how to calculate a probability when you know the joint pdf? You just need to calculate the probability

[tex]P\left(\frac {X_1+X_2}{2} >.99\right)[/tex]

I really am confused how to calculate this probability.
Also I think there is no table for uniform distribution.

You don't need a table. This problem can be done without calculus, but the normal way to calculate a probability is with an integral. Have you had calculus?

Generally, if X and Y have joint pdf f(x,y) and A is a subset of the sample space

[tex]P(A) = \iint_A f(x,y)\,dxdy[/tex]

In your case

[tex]A =\left\{ (x_1,x_2):\frac{x_1+x_2}{2} > .99 \right\}[/tex]
 
  • #7
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You don't need a table. This problem can be done without calculus, but the normal way to calculate a probability is with an integral. Have you had calculus?

Generally, if X and Y have joint pdf f(x,y) and A is a subset of the sample space

[tex]P(A) = \iint_A f(x,y)\,dxdy[/tex]

In your case

[tex]A =\left\{ (x_1,x_2):\frac{x_1+x_2}{2} > .99 \right\}[/tex]

[tex]P\left[\frac{X_{1}+X_{2}}{2} > 0.99\right][/tex] = [tex]\int^{1}_{0}\int^{0.98}_{0} 1 dx_{1}dx_{2}[/tex] = 0.98

Is this right?
If I double integrate the joint pdf with when X1 goes from 0 to 1 and X2 goes from 0 to 1, I got [tex]P\left[\frac{X_{1}+X_{2}}{2} > 0.99\right][/tex] = 1.

I think it is not right because it is too high to be a value of [tex]\alpha[/tex]
 
  • #8
LCKurtz
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[tex]P\left[\frac{X_{1}+X_{2}}{2} > 0.99\right][/tex] = [tex]\int^{1}_{0}\int^{0.98}_{0} 1 dx_{1}dx_{2}[/tex] = 0.98

Is this right?
If I double integrate the joint pdf with when X1 goes from 0 to 1 and X2 goes from 0 to 1, I got [tex]P\left[\frac{X_{1}+X_{2}}{2} > 0.99\right][/tex] = 1.

I think it is not right because it is too high to be a value of [tex]\alpha[/tex]

You are correct thinking that isn't right. You need to draw a picture of the unit square and shade the portion of it where (x1+x2)/2 > .99. Then integrate over that region. Hint: It isn't a rectangle, which your attempt is, having constant limits.
 

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