Show $\lim\limits_{i\to\infty} \mathbb P[e^{\sum_{i=0}^n Z_i}>1]=\frac 1 2$ for $(Z_i)_{i\in \mathbb{N}}$ $iid$ $\mathbb E[Z_i]=0, Var[Z_i]=4$

Question

Show $\lim\limits_{i\to\infty} \mathbb P[\mathrm{e}^{\sum_{i=0}^n Z_i}>1]=\frac 1 2$ for $(Z_i)_{i\in \mathbb{N}}$ $iid$ $\mathbb E[Z_i]=0, Var[Z_i]=4$

I think it is intuitively clear but how am I able to show this rigorously? Of course $\mathbb P[\mathrm{e}^{\sum_{i=0}^n Z_i}>1]=\mathbb P[\sum_{i=0}^n Z_i>0]$ and tried to use Chebyshev's and Markov's inequality but it didn't work because of the "$0$" and I would have only gotten an inequality either way.

$Var[\sum_{i=0}^n Z_i]=4n$ and $\mathbb E[\sum_{i=0}^n Z_i]=0$ because of independence.

So what would be the correct approach?

Answer 1

From Central Limit Theorem, it follows that $$ \lim_{n\to \infty} P \left (\frac{\sum_{i=0}^n Z_i}{2\sqrt{n}} > 0 \right) = 1 - \Phi (0) = \frac{1}{2}$$