Proving independence of a sequence of events

Question

Let $(X_i)_{i \geq 1}$ be a sequence of i.i.d. random variables with $P(X_i = 1) = P(X_i = −1) = 1/2$. Define $Z_k = \prod_{i=1}^{k} X_i$. Prove that $(Z_k)_{k≥1}$ is a sequence of independent random variables.

My solution:

For any sequence $Z_k$ the probability of an event $\omega=(\omega_1\cdot\omega_2\ldots\omega_k)$ is given by the product $\frac{1}{2^k}$. Therefore for any two $Z_k$'s (suppose $Z_k$ and $Z_m$) in the sequence $(Z_k)_{k \geq 1}$ is going to be the product $\frac{1}{2^k}\cdot\frac{1}{2^{k-m}}$ (if $k \geq m$) since the first $m$ events would be common to both. However I don't understand how this makes them independent. If the answer was $\frac{1}{2^k}\cdot\frac{1}{2^m}$ instead, that would be a proof of independence, no?

Answer 1

For any $k$, $P(Z_k=1)=P(Z_k=-1)=1/2$. See here for a combinatorial justification.

Even more generally, $P\left(\prod_{i=m+1}^k X_i = 1\right)=1/2$ for any $m \le k$.

So for example, $$P(Z_k=1,Z_m=-1) = P(Z_m=-1) P\left(\prod_{i=m+1}^k X_i = -1\right) = P(Z_m=-1)\frac{1}{2}= P(Z_m=-1) P(Z_k=1).$$

Similarly, one can show $P(Z_k=z_k, Z_m=z_m) = P(Z_k=z_k) P(Z_m=z_m)$ for any $z_k, z_m \in \{\pm 1\}$.

Finally, this can be extended to show $P \bigcap_{j=1}^n \{Z_{k_j} = z_{k_j}\} = \prod_{j=1}^n P(Z_{k_j} = z_{k_j})$ for any $n$ and $k_1,\ldots, k_n$, which is what you need to show to verify independence.