# Finding the distributions of the max and min of random variables from the geometric distribution

Let $$X$$ and $$Y$$ be independent random variables following the same geometric distribution, that is $$P(X=k)=P(Y=k) = (1-p)p^k, k=0,1,\ldots,$$. Let $$U=min\{X,Y\}$$, $$V=max\{X,Y\}$$,and $$W=V-U$$. How do I compute $$P(U=i)$$ and $$P(W=j)$$?

The problem gives me a hint for each, but I don't really understand them. It gives me $$P(U=i) = \sum_{j=1}^\infty P(X=i,Y=j) + \sum_{j=i+i}^\infty P(X=j, Y=i)=(1-p^2)p^{2i}$$ but I am unsure where this comes from.

It also gives me that $$j=1,2,...$$ $$P(W=j) = \sum_{j=0}^\infty P(X=i,Y=i+j) + \sum_{i=0}^\infty P(X=i+j, Y=i)=\frac{2(1-p)p^j}{1+p}$$ and for $$j=0$$ $$P(W=0)=P(W=Y)=\sum_{i=0}^\infty P(X=i,Y=i)=\frac{1-p}{1+p}$$ but again I am unsure where this comes from.

If anyone has any information on how I would be able to set these equations up and how to simplify them to the answers it would be greatly appreciated.

• There must be a typo in the first summation in your expression for $P(U=i)$. It will not give the desired formula. The first summation should go from $j=i$ to $\infty$. Thus we see that the event $U=i$ can be split into the mutually exclusive events $X=i, Y \ge i$ and $X > i, Y = i$. Jan 21, 2022 at 4:51

Although there are closed-form solutions for the distribution of the order statistics of discrete random variables, the problem can also can be answered by transforming events of $$U$$ or $$W$$ into events of $$X$$ and $$Y$$. First note that $$X$$ and $$Y$$ are i.i.d. and $$\mbox{Pr}[X \ge i] = p^i$$.
For instance, if we observe the event $$U = i$$, then there are three cases: (1) $$\left(X=i, Y=i\right)$$, (2) $$\left(X=i, Y>i\right)$$, (3) $$\left(X>i, Y=i\right)$$. Now we can combine case (1) with either case (2) or case (3). Combining it with case (2), we have two sets of events: (1*) $$\left(X=i, Y \ge i\right)$$ and (2*) $$\left(X>i, Y=i\right)$$. Since the events (1*) and (2*) are mutually exclusive and equivalent to the event $$U=i$$, we have that $$\begin{eqnarray*} \mbox{Pr} \left[U=i\right] &=& \mbox{Pr} \left[X=i, Y \ge i\right] + \mbox{Pr} \left[X>i, Y=i\right] \\ &=& \sum_{j=i}^\infty \mbox{Pr} \left[X=i, Y =j \right] + \sum_{j=i+i}^\infty \mbox{Pr} \left[X=j, Y =i \right]. \end{eqnarray*}$$ Using the fact that $$X$$ and $$Y$$ are i.i.d., we can simplify the above to $$\begin{eqnarray*} \mbox{Pr} \left[U=i\right] &=& \mbox{Pr} \left[X=i\right] \left(\mbox{Pr} \left[Y \ge i\right]+ \mbox{Pr} \left[X>i\right] \right) \\ &=& \mbox{Pr} \left[X=i\right] \left(\mbox{Pr} \left[Y \ge i\right]+ \mbox{Pr} \left[X \ge i+1\right] \right) \\ &=& (1-p)p^i \left(p^i + p^{i+1}\right) \\ &=& (1-p)(1+p)p^{2i} \\ &=& (1-p^2)p^{2i}. \end{eqnarray*}$$
Now to calculate the distribution of the range of a discrete random variable, we generally split it up into the case $$W=0$$ and $$W \ne 0$$. If $$W=0$$, then clearly the maximum and minimum of $$(X,Y)$$ must be equal; hence, $$X=Y=i$$ for all $$i \in \{0, 1, \cdots \}$$. Therefore, $$\begin{eqnarray*} \mbox{Pr} \left[W=0\right] &=& \sum_{i=0}^\infty \mbox{Pr} \left[X=i, Y = i \right] \\ &=& (1-p)^2 \sum_{i=0}^\infty p^{2i} \\ &=& \frac{(1-p)^2}{1-p^2} \\ &=& \frac{1-p}{1+p}. \end{eqnarray*}$$
Finally, consider the event $$W=j$$ for $$j \in \{1, 2, \cdots \}$$. Clearly for $$j \ne 0$$, $$X \ne Y$$; hence, if $$X>Y$$ then the event $$W=j$$ is equivalent to the event $$X-Y=j$$ and if $$Y>X$$ then the event $$W=j$$ is equivalent to the event $$Y-X=j$$. Thus the event $$W=j$$ is equivalent to the event that the difference of $$X$$ and $$Y$$ and the the difference of $$Y$$ and $$X$$ are both $$j$$. If $$X-Y=j$$, then this is equivalent to the event that $$(X=i+j,Y=i)$$ for $$i \in \{0, 1, \cdots \}$$. Likewise if $$Y-X=j$$, then this is equivalent to the event that $$(X=i,Y=i+j)$$ for $$i \in \{0, 1, \cdots \}$$. Coupling these results, we have that for $$j \ne 0$$ $$\begin{eqnarray*} \mbox{Pr} \left[W=j\right] &=& \sum_{i=0}^\infty \mbox{Pr} \left[X=i+j, Y = i \right] + \sum_{i=0}^\infty \mbox{Pr} \left[X=i, Y = i+j \right] \\ &=& 2(1-p)^2p^j \sum_{i=0}^\infty p^{2i} \\ &=& 2\frac{1-p}{1+p}p^j. \end{eqnarray*}$$