Asymptotically confidence interval of level $1-2\alpha$ For fixed $k$, the variables $(2k+1)\hat{f}_k(\lambda)$ are asymptotically distributed according the Gamma distribution with shape parameter $2k+1$ and mean $(2k+1)f_X(\lambda)$. It turns out that this suggests a confidence interval of the form
\begin{align}
\bigg(\frac{(4k+2)\hat{f}_k(\lambda)}{\chi^2_{4k+2,1-\alpha}}, \frac{(4k+2)\hat{f}_k(\lambda)}{\chi^2_{4k+2,\alpha}} \bigg), \qquad (*)
\end{align} 
where $\chi^2_{k,\alpha}$ the $\alpha$-quantile of the chi-square distribution with $k$ degrees of freedom.
I want to show that this interval is asymptotically of level $1-2\alpha$.
I know that a $\Gamma(m/2,1/2)$-distribution is identical to a $\chi^2_m$-dsitribution. So in $(*)$ we have a fraction of two object which are Gamma-distributed. However, I do not see how to make ends meet? Any help is appreciated! 
 A: 
For fixed $k$, the variables $(2k+1)\hat{f}_k(\lambda)$ are asymptotically distributed according the Gamma distribution with shape parameter $2k+1$ and mean $(2k+1)f_X(\lambda)$.

The mean of $\Gamma(\alpha,\beta)$ equals to 
$$\frac{\alpha}{\beta}=\frac{2k+1}{\beta}=(2k+1)f_X(\lambda) \quad\Rightarrow\quad\beta=\frac{1}{f_X(\lambda)}.
$$ 
Rewrite your statement as
$$
(2k+1)\,\hat{f}_k(\lambda) \xrightarrow{d} \Gamma\left(2k+1,\frac{1}{f_X(\lambda)}\right).
$$
Use the PDF of $\Gamma(\alpha,\beta)$:
$$
f_Y(x)=\dfrac{\beta^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x},\,x>0
$$ 
and check that the multiplying $Y\sim\Gamma(\alpha,\beta)$ by $c>0$ results in changing $\beta$ to $\beta/c$ only: $cY\sim \Gamma(\alpha, \frac{\beta}{c})$. 
So, in order to get asymptotically chi-squared distribution $\chi^2_{4k+2}=\Gamma(2k+1,\frac12)$, divide original estimator $(2k+1)\hat{f}_k(\lambda)$ by $f_X(\lambda)$ and multiply by $2$:
$$
\frac{2(2k+1)\,\hat{f}_k(\lambda)}{f_X(\lambda)} \xrightarrow{d} \Gamma\left(2k+1,\frac{1}{2}\right)=\chi^2_{4k+2}.
$$
$$\begin{multline}
\mathbb P\bigg(\frac{(4k+2)\hat{f}_k(\lambda)}{\chi^2_{4k+2,1-\alpha}}\leq f_X(\lambda) \leq \frac{(4k+2)\hat{f}_k(\lambda)}{\chi^2_{4k+2,\alpha}} \bigg)=\cr =\mathbb P\bigg(\chi^2_{4k+2,\alpha}\leq\underbrace{\frac{(4k+2)\hat{f}_k(\lambda)}{f_X(\lambda)}}_{\begin{array}{c}\displaystyle\downarrow{\tiny{d}}\cr\displaystyle\chi^2_{4k+2} \end{array}} \leq \chi^2_{4k+2,1-\alpha} \bigg)\to 1-2\alpha.
\end{multline}
$$
