$$X,Y\sim \exp(\lambda)$$
$$S=\min(X,Y)$$ $$T=\mathbf1_{X\le Y}=\begin{Bmatrix} 1 & \text{ if } X\le Y \\ 0 & \text{ if } X\gt Y \end{Bmatrix}$$
I have that $P(S>s)=e^{-2\lambda s}$ (so $S\sim \exp(2\lambda)$) and that $P(T=1)=\frac{1}{2}$
I am now struggling to find out how to show their independence. Using $P(S>s,T=1)=P(S>s)P(T=1)$, I attempted to work out the left hand side of the equation to show it was equal to $\frac{1}{2}e^{-2\lambda s}$ as follows:
$$\begin{align} P(S>s,T=1)&=P(X>s,Y>s,X\le Y) \\ &=P(X>s|Y>s,X\le Y)+P(Y>s|X\le Y)\\ &=P(X>s|X\le Y)+P(Y>s|X\le Y)\\ &=\frac{P(X>s,X\le Y)+P(Y>s,X\le Y)}{P(X\le Y)} \end{align}$$
However, I could not figure out how to work out $P(X>s,X\le Y)$ and $P(Y>s,X\le Y)$
I was thinking something along the lines of:
$$P(X>s,X\le Y)=\int_{s}^{\infty}\int_{x}^{\infty}2\lambda e^{-2\lambda s}\mathbf1_{X\le Y}dydx$$ but this diverges so it has left me a bit stumped about where to go from here. Any clues to help me get over this hurdle would be much appreciated! Or if I'm completely barking up the wrong tree it would be useful if I could be pointed in the right direction :)
You should have \begin{align} P(X>s,X\le Y) &= \int_0^\infty dx \int_0^\infty dy\, \rho_X(x)\rho_Y(y) {\bf 1}_{x>s} {\bf 1}_{y\ge x} = \\ &= \int_s^\infty dx \int_x^\infty dy\, \rho_X(x)\rho_Y(y) = \\ &= \int_s^\infty dx\, \rho_X(x) P(Y\ge x) = \\ &= \int_s^\infty dx\,\lambda e^{-\lambda x} \cdot e^{-\lambda x}= \\ &= \lambda \int_s^\infty dx\,e^{-2\lambda x} = \\ &= \lambda \frac{e^{-2\lambda s}}{2\lambda} = \\ &= \frac12 e^{-2\lambda s} \end{align}