What is the Probability of Type I Error for a Uniform Random Variable Test?

[safina]

Homework Statement

Given that X is a uniform random variable on the interval (0, \theta), we might test Ho: \theta = 1 versus the alternative H_{1}: \theta = 2 by taking a sample of 2 observations of X and rejecting Ho if \bar{X} > 0.99. Compute \alpha2. The attempt at a solution
\alpha = P[type I error]
= P[rejecting Ho| Ho is true]
= P[\bar{X} > 0.99 given that \theta = 1]

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

Kindly help me what to do next.

 

[LCKurtz]

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.

 

[safina]

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_{1} and X_{2} is \frac{1}{\theta} \frac{1}{\theta} = \frac{1}{\theta^{2}}
But I still don't get how is it related in getting \alpha

 

[LCKurtz]

the joint pdf of X_{1} and X_{2} is \frac{1}{\theta} \frac{1}{\theta} = \frac{1}{\theta^{2}}
But I still don't get how is it related in getting \alpha

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

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

 

[safina]

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

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

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

 

[LCKurtz]

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

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

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

P(A) = \iint_A f(x,y)\,dxdy

In your case

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

 

[safina]

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

P(A) = \iint_A f(x,y)\,dxdy

In your case

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

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

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

 

[LCKurtz]

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

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

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.