Determining the distribution of Q

Question

Let $X_1, X_2,..., X_9$ denote a sample. Assume that $X_i \sim N(4\theta, \theta^2)$ for $i = 1,...,9$ with an unknown $\theta > 0$. We want to find the confidence interval for $\theta$.

a) Find the distribution of $$ Q = \frac{3\bar{X}}{\theta} - 12 $$ where $\bar{X} = \displaystyle{\frac{X_1 + ... + X_n}{n}}$, and explain why Q is a pivot.

It's been quite sometime i had my probability and statistics course, so i'm really lost and don't know where to begin with this.

Answer 1

Since $X_i$ are normal, so are $\bar{X}_n$ and $Q$. We then just have to characterize their means and variances. We have ($n = 9$):

$$ \mathbb{E}(\bar{X}_n) = \frac{1}{9} \sum_i \mathbb{E}(X_i) = 4 \theta, \quad \text{Var}(\bar{X}_n) = \frac{1}{81}\text{Var}(X_i) = \frac{\theta^2}{9} $$

Hence we get

$$ \mathbb{E}(Q) = \frac{3 \; \mathbb{E}(\bar{X}_n)}{\theta} -12 = 0, \quad \text{Var}(Q) = \frac{9}{\theta^2}\text{Var}(\bar{X}_n) = 1 $$

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Proof of Variance

• Fact1: $\text{Var}(aX+b) = a^2 \text{Var}(X)$, i.e. the constants can be omitted from the computation and the variance is a quadratic operator
• Fact2: from Fact1 we then have $\text{Var}(3/\theta\bar{X}_n-12) = (3/\theta)^2 \text{Var}(\bar{X}_n)=(9/\theta^2) (\theta^2/9) = 1$

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Therefore $Q \sim \mathcal{N}(0,1)$. Since its distribution does not depend on unknown parameters, it is a pivotal quantity