The 2-parameter exponential cumulative density function is defined as $1-e^{\lambda (x-\gamma)}\quad$ (see e.g. this page )
Question:
what does $\quad1-\left(1-e^{\lambda (x-\gamma)}\right)^2\quad$ measure?
Any information about that distribution would be appreciated.
$1-e^{\lambda (x-\gamma)}$ is $\mathbb P(X\le x)$ for a two parameter exponentially distributed random variable with rate $\lambda$ and minimum possible value $\gamma$ where $x\ge \gamma$.
Suppose you have two independent random variables $X_1$ and $X_2$ with this distribution. Then $\left(1-e^{\lambda (x-\gamma)}\right)^2$ is $\mathbb P(X_1\le x, X_2\le x) = \mathbb P(\max(X_1, X_2)\le x)$. So $1-\left(1-e^{\lambda (x-\gamma)}\right)^2$ is $\mathbb P(\max(X_1, X_2)\gt x)$
Alternatively, if $X_1$ and $X_2$ have random variables with independent one parameter exponential distributions and rates of $\lambda$, then $1-\left(1-e^{\lambda (x-\gamma)}\right)^2$ is $\mathbb P(\max(X_1, X_2)\gt x \mid \min(X_1, X_2)\gt \gamma)$ so long as $0 \le \gamma \le x$