Law of large numbers? 
Given random variables $Z_1,Z_2,Z_3,\ldots$, which are uniformly distributed for $[8,10]$:

*

*If $X_k =\min\{Z_1, Z_2,Z_3,\ldots,Z_k\}$, prove convergence in probability and find the constant.


*Prove and find that the mean of the first $k$ of these variables converge in probably to some constant.

I think this involves the law of large numbers and possible central limit theorem, but I am not really sure.
 A: Hints: The sequence $(X_k)$ is nonincreasing pointwise and lower bounded, hence its pointwise limit $X$ is well defined and finite. A simple argument shows that, if the random variables $(Z_k)$ are independent then $X=8$ almost surely (can you write it down?).
The rest follows: $X_k\to X$ in probability since almost sure convergence implies convergence in probability, and $E[X_k]\to E[X]$ since the random variables $X_k$ are uniformly bounded.
Neither a law of large numbers nor a central limit theorem are useful here. The common distribution of the random variables $Z_k$ is not needed either, simply the fact that $Z_k\geqslant8$ almost surely and that, for every $x\gt8$, $P[Z_k\leqslant x]\ne0$. But the independence property (which is unfortunately omitted in the question) is crucial.
Edit: What remains without the independence hypothesis is that $X_k\to X$ almost surely, hence in probability, and that $E[X_k]\to E[X]$, for some random variable $X$ with values in $[8,10]$ such that $E[X]\leqslant E[Z_1]=9$... and this is it. Note as an extreme example that one could define $Z_k=Z_1$ for every $k$, then the preceding results hold with $X=Z_1$. In particular, the limit may not be a constant, and there is still no law of large numbers nor central limit theorem in the picture.
A: Hint:
1.)
Step 1: $\Pr(X_k > a) = \Pr(Z_1 > a, Z_2 > a, \cdots, Z_k > a) = (1 - F_Z(a))^k$
where $F_Z(.)$ is the c.d.f of the random variables.
Step 2: Use the above and let $k \to \infty$ to show that the sequence 'converges in distribution' to a constant (can you guess the constant?).
Step 3: Now use the fact that convergence in distribution to a constant implies 'convergence in probability'
2.)
The mean of the $X_k$ can be calculated using $\mathbb{E}(X_k) = \int_{0}^{\infty} (1 - F_{X_k} (x)) dx $, since $X_k$ is a positive random variable.
Use the Step 1 of the previous part to calculate the mean.
