# Find $P(X=Y)$ if $X$ and $Y$ are independent random variables with same geometric distribution

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

• Which parameterization of the geomtric distribution are you using? Supported on $\{0,1,2,\dots\}$ or on $\{1,2,\dots\}$? – Clement C. May 19 '18 at 20:46
• It's {0,1,2,...}. – EduardoGM May 19 '18 at 20:53

• 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: $$\frac{1+\frac{p}{2-p}}{2} = \boxed{\frac{1}{2-p}}\,.$$

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

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)

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

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