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If $X$ and $Y$ are independent random variables with the same geometric distribution with parameter $p$, find:

$P(X = Y)$ and $P(X \geq Y)$

I have done the joint distribution table of both variables and found, with some sums, these results, but I don't find in books some confirmation of my procedures:

1) $\frac{p}{2-p}$

2) $\frac{1}{2-p}$

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  • $\begingroup$ Which parameterization of the geomtric distribution are you using? Supported on $\{0,1,2,\dots\}$ or on $\{1,2,\dots\}$? $\endgroup$ – Clement C. May 19 '18 at 20:46
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    $\begingroup$ It's {0,1,2,...}. $\endgroup$ – EduardoGM May 19 '18 at 20:53
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  • For the first: you have $$ \mathbb{P}\{X=Y\} = \mathbb{P}\bigcup_{k=1}^\infty \{X=Y=k\} = \sum_{k=1}^\infty \mathbb{P}\{X=Y=k\} = \sum_{k=1}^\infty \mathbb{P}\{X=k\} \cdot \mathbb{P}\{Y=k\} $$ the second equality since the events are disjoint; the last equality by independence of $X,Y$. Using the pmf of the Geometric distribution, we get $$ \mathbb{P}\{X=Y\} = \sum_{k=1}^\infty p(1-p)^{k-1}\cdot p(1-p)^{k-1} = p^2 \sum_{k=0}^\infty (1-p)^{2k} = \frac{p^2}{1-(1-p)^2} = \boxed{\frac{p}{2-p}} $$ where to compute the sum we used the expression of a geometric series.

  • For the second, we can do something a bit more "fun": note that $$ \mathbb{P}\{X\geq Y\} = \mathbb{P}\{X = Y\} + \mathbb{P}\{X > Y\} \tag{1} $$ and that $\mathbb{P}\{X > Y\} = \mathbb{P}\{X < Y\}$ by symmetry (Why?). Since $$ 1 = \mathbb{P}\{X = Y\} + \mathbb{P}\{X > Y\} + \mathbb{P}\{X < Y\} \tag{2} $$ (Why?) we thus get $$ \mathbb{P}\{X > Y\} = \frac{1-\mathbb{P}\{X = Y\}}{2} $$ and therefore by (1) $$ \mathbb{P}\{X\geq Y\} = \mathbb{P}\{X = Y\}+\frac{1-\mathbb{P}\{X = Y\}}{2} = \frac{1+\mathbb{P}\{X = Y\}}{2} $$ which you can compute given the first part: $$ \frac{1+\frac{p}{2-p}}{2} = \boxed{\frac{1}{2-p}}\,. $$

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  • $\begingroup$ I had done the second by a double sum and I didn't trust in it, but your method it's beautiful ! $\endgroup$ – EduardoGM May 19 '18 at 21:05
  • $\begingroup$ @EduardoGM Symmetry is a beautiful thing :) $\endgroup$ – Clement C. May 19 '18 at 21:13
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Write $q=1-p$

a)

$$P(X=Y) = \sum _{n=1}^{\infty}P((X=n)\cap (Y=n)) $$ $$ =\sum _{n=1}^{\infty}P(X=n)P(Y=n) $$ $$ =\sum _{n=1}^{\infty}q^{n-1}p q^{n-1}p =p^2\sum _{n=0}^{\infty}q^{2n} $$

$$ = p^2{1\over 1-q^2}= {p\over 1+q} = {p\over 2-p}$$


b)

$$P(X\geq Y) = \sum _{n=1}^{\infty}\sum _{k=1}^nP((X=n)\cap (Y=k)) $$ $$ =\sum _{n=1}^{\infty}P(X=n)\sum _{k=1}^nP(Y=k) $$ $$ =\sum _{n=1}^{\infty}q^{n-1}p \sum _{k=1}^nq^{k-1}p =p^2\sum _{n=0}^{\infty}q^{n-1}\sum _{k=0}^{n-1}q^{k} $$

$$ =p^2\sum _{n=0}^{\infty}q^{n}{1-q^{n}\over 1-q} $$ $$ =p\sum _{n=0}^{\infty}q^{n}(1-q^{n}) $$ $$ =p\Big(\sum _{n=0}^{\infty}q^{n}-\sum _{n=0}^{\infty}q^{2n}\Big) $$ $$ = p({1\over 1-q}-{1\over 1-q^2}) = 1 - {1\over 1+q} = {q\over 1+q}$$

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