= Solution
An <exponential distribution> of rate $\lambda>0$ has density $f_X(x)=\lambda e^{-\lambda x}$ for $x\geq0$ and zero for $x<0$, with survival probability $\mathbb P(X>x)=e^{-\lambda x}$. For $s,t\geq0$, its <memoryless property> follows from
$$
\boxed{\mathbb P(X>s+t\mid X>s)=\frac{e^{-\lambda(s+t)}}{e^{-\lambda s}}=e^{-\lambda t}=\mathbb P(X>t).}
$$
For the minimum, <independence> gives $\mathbb P(Z>z)=\mathbb P(X>z)\mathbb P(Y>z)=e^{-2\lambda z}$. Hence
$$
\boxed{f_Z(z)=2\lambda e^{-2\lambda z}\quad(z\geq0),\qquad \mathbb P(X>Y)=\frac12.}
$$
The density is zero for negative $z$. The comparison probability follows from identical continuous laws and zero tie probability; directly it is $\int_0^\infty\lambda e^{-2\lambda y}dy=1/2$. These are <competing exponential clocks>.
For the last density, substitute $y=s^2$ into the normalizing integral to obtain $C=2\int_0^\infty e^{-s^2}ds=\sqrt\pi$, the <Gaussian integral>. Equivalently, each variable has a <gamma distribution> with shape $1/2$ and rate one. <Independence> permits <convolution> of the densities. For $z>0$,
$$
f_{G_1+G_2}(z)=\frac{e^{-z}}{C^2}\int_0^z\frac{dy}{\sqrt{y(z-y)}}.
$$
Putting $y=z\sin^2\theta$ turns the integral into $2\int_0^{\pi/2}d\theta=\pi$. Consequently
$$
\boxed{f_{G_1+G_2}(z)=e^{-z}\quad(z\geq0),\qquad G_1+G_2\sim\operatorname{Exp}(1).}
$$
This is also the <additivity of independent gamma distributions with a common rate>: their shapes add to one. Density values at the single endpoint zero are immaterial.
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