Trying to work out $\operatorname{Var}(\bar{X})$ of Pareto distribution Pareto distribution with pdf $$f(x;\lambda)=\lambda x^{-(\lambda-1)} \;  $$
Using the standard equation for the variance I have that:
$$\operatorname{Var}(\bar{X})=E[\bar{X^2}]-(E[\bar{X}])^2$$
I have $E[\bar{X}]=\frac{\lambda}{\lambda-1}$ but I'm having trouble with working out $E[\bar{X^2}]$
So far I've got:
$$E[\bar{X^2}]=E\left[\left(\frac{1}{n}\sum X_i\right)^2\right] = \frac{1}{n^2} E\left[\left(\sum X_i\right)^2\right]$$
but I don't know how to go on from there.
Thanks very much for you help!
 A: The question is phrased in a confused way.  I'm going to assume the density was intended to be
$$
f(x) = \begin{cases} \lambda x^{-(\lambda+1)} & \text{for }x>1, \\[6pt] 0 & \text{for }x<1. \end{cases}
$$
(One could write either $-\lambda-1$ or $-(\lambda+1)$.)
I'm also going to surmise that, despite the way the question was phrased, what is sought is the expected value and variance of this distribution.  The expected value is
$$
\int_1^\infty xf(x)\,dx = \int_1^\infty x \lambda x^{-(\lambda+1)} \,dx = \lambda\int_1^\infty x^{-\lambda}\,dx = \lambda\left[\frac{x^{-\lambda+1}}{-\lambda+1}\right]_{x=1}^\infty = \frac{\lambda}{\lambda-1}.
$$
Calling the random variable (capital) $X$, we have the variance:
$$
\operatorname{var}(X) = \mathbb E(X^2) - (\mathbb E(X))^2 = \mathbb E(X^2) -\left(\frac{\lambda}{\lambda-1}\right)^2.
$$
So now we need $\mathbb E(X^2)$.
$$
\mathbb E(X^2) = \int_1^\infty x^2 \lambda x^{-(\lambda+1)} \, dx = \lambda\int_1^\infty x^{-\lambda+1} \,dx = \lambda\left[\frac{x^{-\lambda+2}}{-\lambda+2}\right]_{x=1}^\infty = \frac{\lambda}{\lambda-2}.
$$
Then some trivial algebraic simplifications can be done.
The stuff in the question about sums of $X_i$, with $i$ running from $1$ to $n$, has no relevance to what is done above.  If you want to get into sample means, that is in a way a separate problem, which would rely on the results found above.
A: You started all right and you already know that $E[X_i]=a$ for every $i$, with $a=\frac{\lambda}{\lambda-1}$, and that $E[\bar X]=a$. To carry on, note that
$$
\left(\sum_i X_i\right)^2=\sum_iX_i^2+\sum_{i\ne j}X_iX_j.
$$
For every $i$, $E[X_i^2]=b$ with $b=\frac{\lambda}{\lambda-2}$ (if $\lambda\gt2$, but if $\lambda\leqslant2$ there is no variance anyway), and there are $n$ such terms in the sum. For every $i\ne j$, $E[X_iX_j]=a^2$, and there are $n(n-1)$ such terms in the sum. Thus,
$$
E[\bar X^2]=\frac1{n^2}(nb+n(n-1)a^2),
$$
hence
$$
\mathrm{var}(\bar X^2)=\frac1{n^2}(nb+n(n-1)a^2)-a^2=\frac1n(b-a^2)=\frac1n\mathrm{var}(X_1).
$$
A more direct approach would be to note that $n\bar X_n$ is the sum of $n$ i.i.d. centered random variables distributed like $X$, hence
$$
n^2\mathrm{var}(\bar X_n)=\mathrm{var}(n\bar X_n)=\mathrm{var}(X_1+\cdots+X_n)=\mathrm{var}(X_1)+\cdots+\mathrm{var}(X_n)=n\mathrm{var}(X).
$$
To be complete, note that
$$
\mathrm{var}(X)=b-a^2=\frac{\lambda}{(\lambda-2)(\lambda-1)^2}.
$$
A: Because the $X_i$'s are independent you have that
$$ \text{var} \left( \sum_{i=1}^n X_i \right) = \sum_{i=1}^n \text{var}(X_i).  $$
