constructing a 95% confidence interval - manipulating inequalities

Question

given the asymptotic distribution of $\hat{\theta_1}$ construct a 95% confidence interval for $\theta$ for large samples:

$\hat{\theta_1} = \frac{\hat{\theta_1}-\theta}{\frac{\theta}{\sqrt{n}}}$

I know that the confidence interval will be :

$-1.96 \leq \frac{\hat{\theta_1}-\theta}{\frac{\theta}{\sqrt{n}}} \leq 1.96$

and hence:

$\frac{\hat{\theta_1}}{1+\frac{1.96}{\sqrt{n}}} \leq \theta \leq \frac{\hat{\theta_1}}{1-\frac{1.96}{\sqrt{n}}}$

Question:

How to get from $-1.96 \leq \frac{\hat{\theta_1}-\theta}{\frac{\theta}{\sqrt{n}}} \leq 1.96$ to $\frac{\hat{\theta_1}}{1+\frac{1.96}{\sqrt{n}}} \leq \theta \leq \frac{\hat{\theta_1}}{1-\frac{1.96}{\sqrt{n}}}$ ? I am aware it is just inequality manipulation, however i am not able to solve it. Please provide detailed steps as my mathematical background is rather weak.

Answer 1

For ease, let $\hat{\theta_1}=x, \theta=y$.

Divide by $\sqrt{n}$: $$-\frac{1.96}{\sqrt{n}}\le \frac{x}{y}-1 \le \frac{1.96}{\sqrt{n}}.$$ Add $1$: $$1-\frac{1.96}{\sqrt{n}}\le \frac{x}{y} \le 1+\frac{1.96}{\sqrt{n}}.$$ Raise to power $-1$: $$\frac{1}{1-\frac{1.96}{\sqrt{n}}}\ge \frac{y}{x} \ge \frac{1}{1+\frac{1.96}{\sqrt{n}}}.$$ Note: $2<3 \iff \frac12>\frac13$.

Now multiply by $x$ to get the final result.

Answer 2

Treat each of the inequalities like you'd treat an equation: Multiply both sides by the denominator; bring mulitples of $\theta$ to one side and everything else to the other; divide by the coefficient of $\theta$. The only thing specific to inequalities is to make sure you reverse the inequality when you multiply or divide by a negative quantity. In the present case, assuming that $\theta\gt0$ and $n\ge4$, no reversal occurs. In the end, you can put the two inequalities back together.