Statement: Let $Z_{2t}$ and $Z_t$ be two geometric random variables of mean $2t$ and $t$ respectively, then by the coupling of $Z_{2t}$ and $Z_{t}$, it follows that $Z_{2t}-Z_{t}$ can be expressed as:
$$Z_{2t}-Z_{t} = \xi G_{2t}$$ where $\xi\sim$ Bernoulli$(1/2)$ and $G_{2t}\sim $ Geometric random variable of mean $2t$ and independent of $\xi$.
I know the definition of coupling but may I know how this follows? Thanks.
As pointed out above, the coupling isn't exactly obvious, and the one we want definitely isn't the independent one.
The usual way of generating a geometric random variable $Z$ with parameter $p$ (and hence, mean $\frac{1}{1-p}$) would be to let $(X_n)_{n\in \mathbb{N}}$ be iid Ber($p$) variables and define $Z$ to be the first time $n$ such that $X_n=0$.
However, if the $(X_n)_{n\in\mathbb{N}}$ are uniformly distributed on $[0,1]$ instead, we can define $Z$ to be the first time such that $X_n\geq p$ (since $1_{(X_n< p)}$ is Bernoulli with parameter $p$). This yields a very natural coupling between geometric variables of different parameters.
Note that $Z_{2t}$ corresponds to parameter $p=\frac{2t-1}{2t}$ and $Z_t$ corresponds to parameter $q=\frac{t-1}{t}$. Under the above coupling, it's clear that $Z_{2t}\geq Z_t$. Furthermore, we see that, by independence of the $X_n$, $$ \mathbb{P}(Z_{2t}=Z_t|Z_t=n)=\mathbb{P}\left(X_n\geq \frac{2t-1}{2t} \big|X_n\geq \frac{t-1}{t}\right)=\frac{1-\frac{2t-1}{2t}}{1-\frac{t-1}{t}}=\frac{1}{2} $$ By the law of total probability, we get that $\mathbb{P}(Z_{2t}>Z_t)=\frac{1}{2}$. Furthermore, we see that for $k\geq 1$, we have, again by independence, that
\begin{align} \mathbb{P}(Z_{2t}-Z_t=k) &=\sum_{n=0}^{\infty}\mathbb{P}(Z_{2t}=n+k|Z_t=n)\mathbb{P}(Z_t=n) \\ &=\sum_{n=0}^{\infty} \mathbb{P}\left(X_n\leq \frac{2t-1}{2t}|X_n\geq \frac{t-1}{t}\right) \mathbb{P}\left(X_{n+k}\geq \frac{2t-1}{2t}\right)\prod_{j=1}^{k-1} \mathbb{P}\left( X_{n+j}\leq \frac{2t-1}{2t} \right) \mathbb{P}(Z_t=n) \\ &=\frac{1}{2}(1-p)p^{k-1} \sum_{n=0}^{\infty} \mathbb{P}(Z_t=n)\\ &= \frac{1}{2}(1-p)p^{k-1} \end{align} Which is exactly equal to $\mathbb{P}(\xi G_{2t}=k)$. This shows the desired.