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In this question, we will further investigate the geometric distribution. Let X, Y be i.i.d. geometric random variables with parameter p. Let U = min{X,Y} and V = max{X,Y} − min{X,Y}. Compute the joint distribution of (U,V) and prove that U and V are independent. [Hint: If X ∼ Geometric( p) and Y ∼ Geometric(q) are independent, then min{X , Y } ∼ Geometric( p + q − pq).]

  • How do I approach this question? I'm very conceptually confused and not quite sure how to solve it. Thanks –  Jul 15 '17 at 23:08
  • Here's a straightforward solution for those who have studied other distributions: notice that the floor function commutes with $\max$ and $\min$. Because the Geometric distribution is the floor of the Exponential distribution and for the Exponential distribution the range is independent of the minimum (that's its "memoryless" property, easily proven), the same result holds for the Geometric distribution, QED. – whuber Jul 19 '17 at 12:32

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I'm not sure how to use the hint, but here's a direct approach. Let $U = min(X,Y)$ and $V = max(X,Y)-min(X,Y)$. Let's split up the cases where $X > Y$, $X < Y$ and $X = Y$.

Case 1: $X > Y$

Then $U = Y$ and $V = X - Y$. Then, $P(U = u, V = v) = P(Y = u, X = u+v, v>0)$.

Case 2: $X < Y$

Then $U = X$ and $V = Y - X$. Then, $P(U = u, V = v) = P(X = u, Y = u+v, v>0)$.

Case 3: $X = Y$

Then, $U = X = Y$ and $V = 0$. Then, $P(U = u, V = 0) = P(X = Y = u, v = 0)$.

So it looks like we can write the joint pmf in the form,

$$ P(U = u, V = v) = \begin{cases} p^2(1-p)^{2u-2} & v = 0, u = 1,2,3,..\\ 2p^2(1-p)^{2u+v-2} & v > 0, u=1,2,3,... \end{cases}$$

From here, you should be able to deduce independence by finding the marginals of $U$ and $V$.

Michael R. Chernick
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