9

Suppose $X_1$ and $X_2$ are independent geometric random variables with parameter $p$. What is the probability that $X_1 \geq X_2$?

I am confused about this question because we aren't told anything about $X_1$ and $X_2$ other than they are geometric. Wouldn't this be $50\%$ because $X_1$ and $X_2$ can be anything in the range?

EDIT: New attempt

$P(X1 ≥ X2) = P(X1 > X2) + P(X1 = X2)$

$P(X1 = X2)$ = $\sum_{x}$ $(1-p)^{x-1}p(1-p)^{x-1}p$ = $\frac{p}{2-p}$

$P(X1 > X2)$ = $P(X1 < X2)$ and $P(X1 < X2) + P(X1 > X2) + P(X1 = X2) = 1$

Therefore, $P(X1 > X2)$ = $\frac{1-P(X1 = X2)}{2}$ = $\frac{1-p}{2-p}$
Adding $P(X1 = X2)=\frac{p}{2-p}$ to that , I get $P(X1 ≥ X2)$ = $\frac{1}{2-p}$

Is this correct?

StatsStudent
  • 10,205
  • 4
  • 37
  • 68
IrCa
  • 191
  • 7

2 Answers2

13

It can't be $50\%$ because $P(X_1=X_2)>0$

One approach:

Consider the three events $P(X_1>X_2), P(X_2>X_1)$ and $P(X_1=X_2)$, which partition the sample space.

There's an obvious connection between the first two. Write an expression for the third and simplify. Hence solve the question.

Glen_b
  • 257,508
  • 32
  • 553
  • 939
  • I edited, my post with my new answer. Could you take a look and see if it's correct? – IrCa Feb 24 '19 at 03:02
  • 1
    Yes, your answers look correct. An alternative method (using a similar idea) would be to note that $P(X_1\geq X_2)=\frac12 + \frac12 P(X_1=X_2)$ (again, exploiting the symmetry/exchangeability of $X_1$ and $X_2$). – Glen_b Feb 24 '19 at 04:26
6

Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:

\begin{align} \Pr\{X_1\geq X_2\} &= \sum_{k=0}^\infty \Pr\{X_1\geq X_2\mid X_2=k\} \Pr\{X_2=k\} \\ &= \sum_{k=0}^\infty \sum_{\ell=k}^\infty \Pr\{X_1=\ell\}\Pr\{X_2=k\}. \end{align}

This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.

Zen
  • 21,786
  • 3
  • 72
  • 114
  • 4
    note - your way is better for applying to new problems I think. Because it is based on first principles. The trick/intuiton from glen_b's answer usually comes *after* the problem has been solved your way – probabilityislogic Feb 24 '19 at 04:28
  • 3
    @probabilityislogic I share your enthusiasm for derivations from "first principles." However, to a modern mathematician, looking for and exploiting symmetry is even more fundamental than the first principles (definitions) to which you refer: we might call it a *metaprinciple* of mathematics. It's much more than a mere "trick." – whuber Feb 24 '19 at 17:13