Confidence interval for $\theta$ when $X_i$'s are i.i.d $N(\theta,\theta)$ Let $X_i$ be i.i.d. r.v. with $N(\theta,\theta)$
I calculated $$E[\bar{X_n}] = \theta$$ $$Var[\bar{X_n}] = \theta/n$$
And want to construct a confidence interval $I_{\theta}$ that is centered around $\bar{X_n}$ such that $\theta \in P(I_{\theta}) = 0.9$ for all n, where $q_{\alpha/2}=q_{0.05}=1.6448$
I am using the general formula: $$\bar{X} \pm Z \frac{s}{\sqrt(n)} = \bar{X_n} \pm 1.6448 \frac{\sqrt{Var[\bar{X_n}]}}{\sqrt(n)} = \bar{X} \pm 1.6448 \frac{\sqrt{\theta}/\sqrt(n)}{\sqrt(n)} = \bar{X} \pm 1.6448 \frac{\sqrt{\theta}}{n}$$
But it seems I am getting it wrong here. Any input would be appreciated. It seems as there is a simple solution to this.
 A: The statistic to be used is, as usual, 
$\bar{X}_n\sim N(\theta;\frac{\theta}{n})$
The pivotal quantity is the following
$$ \bbox[5px,border:2px solid red]
{
\frac{\bar{X}_n-\theta}{\sqrt{\theta}}\sqrt{n}\sim \Phi
\qquad (1) 
}
$$
So you can find your Confidence interval solving this double inequality with respect to $\theta$
$$-1.64 \leq \frac{\bar{X}_n-\theta}{\sqrt{\theta}}\sqrt{n} \leq 1.64$$
this lead to the exact solution:
$$ \bbox[5px,border:2px solid black]
{
\bar{X}_n+\frac{1.64^2}{2n} \pm \sqrt{(\frac{1.64^2}{2n})^2+ \bar{X}_n \frac{1.64^2}{n} }
\qquad  (2)
}
$$
If you want to find an approx solution, in a very easy way, you can substitute $\sqrt{\theta}$ with its estimation, $\sqrt{\bar{X}_n}$ and you immediately find the following CI from (1)
$$ \bbox[5px,border:2px solid black]
{
\bar{X}_n \pm 1.64 \sqrt{ \frac{\bar{X}_n}{n}}
\qquad  (3)
}
$$
So my conclusion is the following:
If you are interested in finding a CI of $\theta$ centered around $\bar{X}_n$ the only alternative is to calculate an approximate Interval.
Calculations from (1) to  (2)
$-z<\frac{ \bar{X}-\theta}{\sqrt{\theta}}\sqrt{n}<z$
$( \bar{X}-\theta)^2<\frac{z^2\theta}{n}$
$\bar{X}^2-2\bar{X}\theta+\theta^2-\frac{z^2\theta}{n}<0$
$\theta^2 -\theta(2\bar{X}+\frac{z^2}{n})+\bar{X}^2<0   $
$\theta=\frac{2\bar{X}+\frac{z^2}{n}\pm\sqrt{4\bar{X}^2+\frac{z^4}{n^2}+\frac{4\bar{X}z^2   } {n}    -4\bar{X}^2 }}{2}  $
$\theta=\bar{X}+\frac{z^2}{2n}\pm\sqrt{(\frac{z^2}{2n})^2+\frac{\bar{X}z^2   } {n} } $
1) the notation $\theta=a\pm b$ is a standard notation used in Confidence Intervals. See here for an example. Anyway it is just a problem of notation.
2) On the countrary this, in effect, it is a problem. At a first sight I would have solved the exercise with my formula 2) that is a correct 
$$ \bbox[5px,border:2px solid black]
{
\bar{X}_n+\frac{z^2}{2n} - \sqrt{(\frac{z^2}{2n})^2+ \bar{X}_n \frac{z^2}{n} }< \theta<\bar{X}_n+\frac{z^2}{2n} + \sqrt{(\frac{z^2}{2n})^2+ \bar{X}_n \frac{z^2}{n} }
\qquad  (2)
}
$$
Note that, for your happiness, I avoided to use the previous notation and I substituted 1.64 with $z$, indicating a general percentile of the Gaussian. (let me avoid to indicate $z_{\frac{\alpha}{2}}$ and $z_{(1-\frac{\alpha}{2})}$ to simplify the notation)
As formula 3) is concerned, I tried to find a CI that was symmetric around $\bar{X}_n$ as requested in the exercise but your observation is absolutely correct: it is a standard way to proceed in other situation but with a $N(\theta;\theta)$ we must pay more attention. Anyway, before answering, I did some simulations and the two formulas (2) and (3) give more or less the same numerical result when $n \rightarrow \infty$
In conclusion: my formula (2) is absolutely correct; (3) works, I do not know why but it works... :)
