What kind of distribution is $f_X(x) = 2 \lambda \pi x e^{-\lambda \pi x ^2}$? What kind of function is:
$f_X(x) = 2 \lambda \pi x e^{-\lambda \pi x ^2}$
Is this a common distribution? I am trying to find a confidence interval of $\lambda$ using the estimator $\hat{\lambda}=\frac{n}{\pi \sum^n_{i=1} X^2_i}$ and I am struggling to prove if this estimator has Asymptotic Normality.
Thanks
 A: It is a square root of exponential distribution with rate $\pi\lambda$. This means that if $Y\sim\exp(\pi\lambda)$, then $\sqrt{Y}\sim f_X$. 
Since your estimate is maximum likelihood estimate it should be asymptotically normal. This follows immediately from the properties of maximum likelihood estimates. In this particular case:
$$\sqrt{n}(\hat\lambda-\lambda)\to N(0,\lambda^2)$$
since 
$$E\frac{\partial^2}{\partial \lambda^2}\log f_X(X)=-\frac{1}{\lambda^2}.$$
A: Why do you care about asymptotics when the exact answer is just as simple (and exact)?  I am assuming that you want asymptotic normality so that you can use the $\mathrm{Est}\pm z_{\alpha}\mathrm{StdErr}$ type of confidence interval
If you make the probability transformation $Y_{i}=X_{i}^{2}$ then you have an exponential sampling distribution (as @mpiktas has mentioned):
$$\newcommand{\Gamma}{\mathrm{Gamma}}
\newcommand{\MLE}{\mathrm{MLE}}
\newcommand{\Pr}{\mathrm{Pr}}
f_{Y_{i}}(y_{i})=f_{X_{i}}(\sqrt{y_{i}})|\frac{\partial\sqrt{y_{i}}}{\partial y_{i}}|=2 \lambda \pi \sqrt{y_{i}} \exp(-\lambda \pi \sqrt{y_{i}} ^2)\frac{1}{2\sqrt{y_{i}}}=\lambda\pi\exp(-\lambda\pi y_{i})$$
So the joint log-likelihood in terms of $D\equiv\{y_{1},\dots,y_{N}\}$ becomes:
$$\log[f(D|\lambda)]=N\log(\pi)+N\log(\lambda)-\lambda\pi\sum_{i=1}^{N}y_{i}$$
Now the only way the data enters the analysis is through the total $T_{N}=\sum_{i=1}^{N}y_{i}$ (and the sample size $N$).  Now it is an elementary sampling theory calculation to show that $T_{N}\sim \Gamma(N,\pi\lambda)$, and further that $\pi N^{-1}T_{N}\sim \Gamma(N,N\lambda)$.  We can further make this a "pivotal" quantity by taking $\lambda$ out of the equations (via the same way that I just put $N$ into them).  And we have:
$$\lambda\pi N^{-1}T_{N}=\frac{\lambda}{\hat{\lambda}_{\MLE}}\sim \Gamma(N,N)$$
Note that thus we now have a distribution which involves the MLE and whose sampling distribution is independent of the parameter $\lambda$.  Now your MLE is equal to $\frac{1}{\pi N^{-1}T_{N}}$ And so writing quantities $L_{\alpha}$ and $U_{\alpha}$ such that the following holds:
$$\Pr(L_{\alpha} < G < U_{\alpha})=1-\alpha\;\;\;\;\;\;\;G\sim \Gamma(N,N)$$
And we then have:
$$\Pr(L_{\alpha} < \frac{\lambda}{\hat{\lambda}_{\MLE}} < U_{\alpha})=\Pr(L_{\alpha}\hat{\lambda}_{\MLE} > \lambda > U_{\alpha}\hat{\lambda}_{\MLE})=1-\alpha$$
And you have an exact $1-\alpha$ confidence interval for $\lambda$.
NOTE: The Gamma distribution I am using is the "precision" style, so that a $\Gamma(N,N)$ density looks like:
$$f_{\Gamma(N,N)}(g)=\frac{N^{N}}{\Gamma(N)}g^{N-1}\exp(-Ng)$$
