Say $X=2^Z$ and $Z$ is a geometric random variable with $p=1/2$.
It follows that, $E[X] = \infty$
So setting the upper bound by the markov inequality, $$P(X \geq t) \leq \frac{E[X]}{t} = \frac{\infty}{t}$$ for some value $t$.
However, this doesn't make much sense since we know the maximum probability is $1$.
Any advice on how to estimate the upper bound probability $P(X \geq t)$ ?
Thanks in advance.
The Markov inequality still holds, it just isn't useful here. For your specific example though, $Z$ takes values in $1,2,3,\ldots$ so $X$ takes values in $2,4,8,\ldots$, and $$\mathbb P(X=k)=\mathbb P(2^Z = k)=\mathbb P(Z=\lg k)=\left(\frac12\right)^{\lg k}, k=2,4,8,\ldots. $$ (where $\lg$ denotes base-$2$ logarithm). If $t>0$, let $n=\min\{k\geqslant 1 : \lg k\geqslant t \}.$ Then $$\mathbb P(X\geqslant t)=\sum_{k=n}^\infty\mathbb P(X=2^k)=\sum_{k=n}^\infty \left(\frac12\right)^k=\left(\frac12\right)^{n-1}. $$