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 $$\alpha$$2. 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.