$Z_1, Z_2, Z_3,...$ are independent and identically distributed R>V.s s.t. $E(Z)_i^- < \infty$ and $E(Z)_i^+ = \infty$. Prove that $$\frac {Z_1+Z_2+Z_3+...+Z_n} n \to \infty$$ almost surely.
What does $E(Z)_i^+$ $E(Z)_i^-$ mean? i dont understand the +,- sign. But If E(Z)=$\infty$, then according to law of large numbers, the average of $\sum Zi$ is of course $\infty$. Or do i need to use other theorems to prove this?
The sequence $\{Z_n^{+}\}$ and $\{Z_n^{-}\}$ are also i.i.d.. Suppose we prove that $\frac {Z_1^{+}+...+Z_n^{+}} n \to \infty$ almost surely. Then we can apply Strong Law to $\{Z_n^{-}\}$ and use $Z_n=Z_n^{+}-Z_n^{-}$ to complete the proof. Fix a positive integer $N$ and let $Y_n =\min (Z_n^{+},N)$. Then $\{Y_n\}$ is i.i.d. and Strong Law gives $\frac {Y_1+Y_2+...+Y_n} n \to EY_1$ almost surely. Since $Y_n \leq Z_N^{+}$ this gives $\liminf \frac {Z_1^{+}+Z_2^{+}+...+Z_n^{+}} n \geq EY_1=E\min (Z_1^{+},N)$. This is true for each $N$. Now note that (by Monotone Convergence Theorem) $E\min (Z_1^{+},N) \to EZ_1^{+}=\infty$. This proves that $\frac {Z_1^{+}+...+Z_n^{+}} n \to \infty$ almost surely.