Calculate the asymptotic dystribution Let $X_1,...,X_n$ be an i.i.d. random sample from a continuous distribution with density given by: $f(x;\theta)=(\theta-x)\frac{2}{\theta^2}$ if $0<=x<=\theta$ and 0 otherwise.
We have the following distribution $Q=\frac{1}{\sqrt n} \sum_{i=1}^{n}({\frac{X_i}{\theta}-\frac{1}{3}}) $  this is the pivotal quantity for $\theta$.


*

*Compute the asymptotic distribution of Q, as n increases. 

*Use the previous question to construct an approximate 95% c.i. for $\theta$. You can use: $z_{0.95}=1.645, z_{0.975}=1.96$ and the mean of $x=13.7$.


If you haven't answered question 1, assume that Q follows a normal distribution with mean 0 and variance 1/10.
Thank you for your help! :)
 A: One of the most important result you will learn in an elementary probability/statistics course is the following:

Central Limit Theorem. Let $X_1,X_2,X_3,\ldots$ be a sequence of i.i.d. random variables with mean $\mu$ and variance $\sigma^2$. Then, as $n\to\infty$,
  \begin{align}
\frac1{\sqrt n}\sum_{i=1}^n(X_i-\mu)\tag{1}
\end{align}
  converges in distribution to $N(0,\sigma^2)$.

The quantity $Q$ in your question looks a lot like $(1)$ in the statement of the central limit theorem I have written above.
If you can figure out what is the expected value and variance of the $X_i/\theta$ using $f(x,\theta)$,
then you should be able to determine the distribution to which $Q$ converges.
Hint: Given a random variable $X$ with density $f(x)$,
one has
$$E[X]=\int x\cdot f(x)~dx,$$
and
$$Var[X]=E[X^2]-E[X]^2=\int x^2\cdot f(x)~dx-\left(\int x\cdot f(x)~dx\right)^2,$$
and don't forget that for a constant $k$,
one has $E[kX]=kE[X]$, and $Var[kX]=k^2Var[X]$.
For the second part,
you know the asymptotic distribution of $Q$,
so you should be able to find a number $z>0$ such that
$$\Pr[-z\leq Q\leq z]=0.95$$
for $n$ large.
This is a $95%$ confidence interval for $Q$.
But this is not what you want,
you want a confidence interval for $\theta$.
However,
you can isolate $\theta$ in $Q$.
