# Probability Distribution

Can some one give the technique of this question? $X$ and $Y$ are two independent geometric random variables. $U = \min(X,Y)$ and $V = \max(X,Y)$, $W = V -U$, Find $P(U =i)$ and $P(W =j)$, Also Show that $U$ and $W$ are independent.

GP

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We have $U=\min(X,Y)$; $V=\max(X,Y)$. Also, we have been told, $X$ and $Y$ are geometrically distributed independent random variables! Consider the parameter of the distribution of $X$, equivalently that of $Y$, to be $p$, with $0<p<1$.

Note that the range of $X$ and $Y$ are the set of all whole numbers. This means that, the range of $U$ will be the set of whole numbers as well.

The following quantities are often useful:

$$P(X=i)=p(1-p)^i$$ $$P(X>i)=\sum_{k=i+1}^\infty {p(1-p)^k}=p \cdot\dfrac{(1-p)^{i+1}}{1-(1-p)}=(1-p)^{i+1}$$

Now, use the above to compute the following:

$$P(U=i)=P(\min(X,Y)=i)=P(X>i,Y=i)+P(X=i,Y>i)+P(X=i,Y=i)$$

Leaving all the details of actual computation, [You'll need to use independence and the results I've pointed out previously!], you'll get,

$$P(U=i)=(2p-p^2)\cdot(1-p)^{2i}$$ It is also clear from here that $\min(X,Y)$ follows a geometric distribution with $2p-p^2$ as its parameter. [Note that this parameter is indeed legal as $p^2<2p$.]

Now the second part of the question is interesting: We are interested in the density of $W=V-U$.

$$P(W=j)=P(V-U=j)=\sum_{k=0}^\infty{P(U=k,V=j+k)}$$

Now, if minimum and maximum of two numbers are fixed, essentially both the numbers are known. So, this precisely what is happening here. So, you'll have,

\begin{align*}P(W=j)&=\sum_{k=0}^\infty\{P(X=k;Y=j+k)+P(X=j+k;Y=k)\}\\ &=\sum_{k=0}^\infty{2p^2(1-p)^{j+2k}}\\ &=\dfrac{2p^2(1-p)^j}{1-(1-p)^2}\\&=\dfrac{2p^2(1-p)^j}{2p-p^2}\end{align*}

Now for the independence, look at the following recipe,

\begin{align*}P(U=i;W=j)&=P(U=i;V=i+j)\\&=P(X=i;Y=i+j)+P(X=i+j;Y=i)\end{align*}

From here, I assume you'll take through!

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