Constructing a confidence interval using the asymptotic approach Consider a random pair of scalars $(x; y)$ for which a random sample $\{(x_i; y_i)\}^n_{i=1}$ is available. Denote $g(x) = E [y|x]$ the conditional mean. Suppose we are interested in constructing a confidence interval of confidence level $99%$ for $g (1)-g(-1)$.
We have to show how to do it using the asymptotic approach within the parametric model
$g(x)= \gamma e^{\delta x}$, where $\gamma$ and $\delta$ are unknown.
Firstly, we have to create confidence interval for $\gamma (e^{\delta}-e^{-\delta})$. In general case, we can use the CLT or Delta method.
Could you please help me with making it corectly?
 A: First of all you have to assume a distribution for $Y_i$. Suppose
$$Y_i \sim N(g(x_i),\sigma^2)$$
$$g(x_i) = \gamma e^{\delta x_i},\quad i=1,\ldots,n$$
and $Y_i$'s are independent.
Let $\tau = (\gamma, \delta)$ and let the parameter be $\theta = (\tau,\sigma)$ and let $\hat\theta$ be the maximum likelihood estimator for $\theta$. Under mild regularity conditions
$$
\hat\tau \overset{a}{\sim}\,N_2(\tau, I(\hat\theta)^{\tau\tau}),
$$
where $\overset{a}{\sim}$ means asymptotically as $n\to \infty$, and $I(\hat\tau)^{\tau\tau}$ is the asymptotic variance of $\hat\tau$, the MLE of $\tau$.
Let $t(z_1,z_2):\mathbb{R}^2\to \mathbb{R}$ with $t(z_1,z_2) = z_1(e^{z_2}-e^{-z_2})$ and let
$$
B(z_1,z_2) = (\partial t(z_1,z_2)/\partial z_1,\partial t(z_1,z_2)/\partial z_2),
$$
be the $1\times 2$ vector of partial derivatives which we assume do not vanish in a neighbour of $\tau$. Let $B_\tau = B(\gamma, \delta).$ Then by the delta method, for large $n$,
$$
\hat\gamma(e^{\hat\delta}-e^{-\hat\delta}) \overset{a}{\sim} N(\gamma(e^{\delta}-e^{-\delta}),B_\tau I(\hat\theta)^{\tau\tau}B_\tau^T).
$$
This formula is not yet usable since $B_\tau$ is unknown, but we can replace it with its consistent estimate $B_\hat\tau$. An approximate confidence interval for $\gamma(e^{\delta}-e^{-\delta})$ of confidence level $1-\alpha$ is thus
$$
\hat\gamma(e^{\hat\delta}-e^{-\hat\delta}) \pm z_{1-\alpha/2} \hat{\text{se}},
$$
where $\hat{\text{se}} = (B_\tau I(\hat\theta)^{\tau\tau}B_\tau^T)^{1/2}$ and $z_{1-\alpha/2}$ is the $(1-\alpha/2)$th quantile of the standard normal distribution.
Note that the result is asymptotically valid even if you use the observed information matrix in place of $I(\hat\theta)$.
