Given two geometric random variables: $\; X_{j}\;$ for $\, j=1,2 \;$ with probability mass function $P(X_{1}=m)=P(X_{2}=m)=(1-q)^{m-1} \cdot q \,$ where $m=1,2,\dots \,$ and $0<q<1.$ Find the probability mass function $\, \mbox{max}(X_{1},X_{2}).$
First, can we assume that these random variables are independent? If so, is the probability mass function $\, \mbox{max}(X_{1},X_{2})$ equal to the following product $\mbox{max} \bigg((1-q)^{2m-2} \cdot q^{2} \bigg)?$
I'd go this way : say $Z = \mathrm{max}(X_1,X_2)$
Then for $m\in\mathbb{N}^{*}$, $$P(Z=m) = $$ $$ P((X_1 = m \quad \mathrm{and} \quad X_2< m)\quad \mathrm{or}\quad (X_2=m\quad\mathrm{and}\quad X_1< m)\quad \mathrm{or} \quad (X_1 = m \quad \mathrm{and} \quad X_2 =m))$$
These three events are disjoints, so
$$P(Z=m) = $$ $$ P((X_1 = m \quad \mathrm{and} \quad X_2< m))+P((X_2=m\quad\mathrm{and}\quad X_1< m))+P((X_1 = m \quad \mathrm{and} \quad X_2 =m))$$
If $X_1$ and $X_2$ are independant, we have
$$P(Z=m) = 2P(X = m)P(X< m)+P(X = m)^2$$
The term $P(X<m)$ is easy to compute (sum of the terms of a geometric sequence)
