# Probability of type I error

## 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
Homework Helper
Gold Member
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.

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
Homework Helper
Gold Member
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

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

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
Homework Helper
Gold Member
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

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

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

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

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

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

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