A question regarding independence of $\min{\{X,Y\}}$ and $X-Y$ when $X,Y$ follows iid geometric distribution

Question

THE PROBLEM :

Let $X$ and $Y$ be independent and geometrically distributed with parameter $p$. Prove that $U=\min\{X,Y\}$ and $V=X-Y$ are independent. (This property characterizes geometric distributions.)

What I did was a brute-force computation of $P(U \leq u) \times P(V\leq v)$ which came out to be $$(1-q^u) \times \left(1-\frac{q^{v+1}}{1+q}\right)$$

However, computation of $P(U \leq u, V\leq v)$ is messier with lot of cases depending on the values of $u$ and $v$. I cannot get to the end. Is there any clever, nice way to solve the problem (without getting into messy calculations)? Thanks in advance.

Answer 1

I take the pmf of $X$ to be of the form $$P(X=k)=pq^k\mathbf 1_{k\in\{0,1,2,\cdots\}}\quad, 0<p=1-q<1$$

Note that support of $U=\min(X,Y)$ is same as the support of $X$ or $Y$, i.e. the set of non-negative integers $\{0,1,\ldots\}$, whereas support of $V=X-Y$ is the set of all integers $\mathbb Z$.

Now,

\begin{align} P(U=u,V=v)&=P(U=u,V=v,X\ge Y)+P(U=u,V=v,X<Y) \\\\&=P(X-Y=v,Y=u,X\ge Y)+P(X=u,X-Y=v,X<Y) \\\\&=P(X=u+v)P(Y=u)\mathbf1_{v\ge 0}+P(X=u)P(Y=u-v)\mathbf1_{v<0} \\\\&=pq^{u+v}\mathbf1_{u+v\ge 0}\cdot pq^u\mathbf1_{u\ge 0}\mathbf1_{v\ge 0}+pq^u\mathbf1_{u\ge 0}\cdot pq^{u-v}\mathbf1_{u-v\ge 0}\mathbf1_{v<0} \\\\&=p^2q^{2u+v}\mathbf1_{u\ge 0,v\ge 0}+p^2q^{2u-v}\mathbf1_{u\ge 0,v<0} \end{align}

That is,

\begin{align} P(U=u,V=v)&=p^2q^{2u+|v|}\mathbf1_{u\in\{0,1,\ldots\},v\in\mathbb{Z}} \\&=\underbrace{pq^{2u}(1+q)\mathbf1_{u\in\{0,1,\ldots\}}}_{P(U=u)}\cdot\underbrace{\frac{pq^{|v|}}{1+q}\mathbf1_{v\in\mathbb{Z}}}_{P(V=v)} \end{align}