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Let X and Y be geometrically distributed iid r.v.s. Find the pmf of M = min(X, Y), and the pmf and D = X - Y.

I thought $$ P(M = m) = P(X = x) \cdot P(Y > x) + P(Y = y) \cdot P(X > y) + P(D = 0)$$ $$ = 2 \cdot P(X = x) \cdot P(Y > x) + P(D = 0)\\$$ $$= 2 \cdot (1-p)^m \: p \cdot (1-p)^{m} + (1-p)^{2m}$$ $$(1-p)^{2m}(2p + 1)$$

But the answer is

$$p^{2m}(1-p^2)$$

Pie_day_1239191919
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2 Answers2

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The pmf of $min(X,Y)$

Given: $X \sim Geometric(p)$ for $x \in \{0,1,2, \dots\}$ with pmf $f(x)$:


(source: tri.org.au)

There are a number of ways to find the pmf of $min(X,Y)$ where $X$ and $Y$ are iid. Since $X$ and $Y$ are iid, the method I prefer is ... think of the problem as generating a sample of size 2 from a single random variable $X$, and then picking the smaller outcome. In other words, we are seeking the pmf of the $1^{st}$ order statistic, in a sample of size 2. Using mathStatica to automate, the pmf is:


(source: tri.org.au)

for $x \in \{0,1,2, \dots\}$.

Notes

  1. Why is your textbook answer different? It seems that they are using slightly different notation for the Geometric pmf than that which you used. Substituting $p = 1-q$, our answer becomes $q^{2 x} (1 - q^2)$ which matches the 'known' answer the OP stated.

  2. As disclosure, I should perhaps note that I am one of the authors of the software used.

Glorfindel
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wolfies
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If $\mathbb P(X=m)=(1-p)^mp$ for $m\geqslant 0$ then $$\mathbb P(X\geqslant m)=\sum_{k=m}^\infty (1-p)^kp = p(1-p)^m\sum_{k=0}^\infty (1-p)^k=(1-p)^m. $$ Moreover, we have $$ \{X\wedge Y=m\} = \{X=m,Y\geqslant m\}\cup\{Y=m,X\geqslant m\}, $$ with each event on the right-hand side sharing the member $\{X=m,Y=m\}$. Subtracting this off, we have \begin{align} \mathbb P(M=m) &= \mathbb P(X\wedge Y=m)\\ &= \mathbb P(X=m)\mathbb P(Y\geqslant m) + \mathbb P(Y=m)\mathbb P(X\geqslant m) - \mathbb P(X=m)\mathbb P(Y=m)\\ &= (1-p)^mp + (1-p)^mp - (1-p)^mp(1-p)^mp\\ &= 2(1-p)^{2m}p - (1-p)^{2m}p^2\\ &= (1-p)^{2m}p(2-p). \end{align} (Substituting $q=1-p$ yields $q^{2m}(1-q^2)$, as noted in @wolfies's answer.)

For $D=X-Y$, note that for $m\geqslant 0$, $$\{X-Y=m\} = \{X=Y+m\} = \bigcup_{k=0}^\infty \{X=k,Y=m+k\},$$ and so \begin{align} \mathbb P(D=m) &= \mathbb P(X-Y=m)\\ &= \mathbb P\left(\bigcup_{k=0}^\infty \{X=k,Y=m+k\}\right)\\ &= \sum_{k=0}^\infty (1-p)^k(1-p)^{m+k}p^2\\ &= (1-p)^mp^2\sum_{k=0}^\infty (1-p)^{2k}\\ &= \frac{(1-p)^mp^2}{1-(1-p)^2}\\ &= \frac{(1-p)^mp}{2-p}. \end{align} For $m<0$ we have $$\{X-Y=m\} = \{Y-X=-m\} = \{Y-X=|m|\}, $$ so by symmetry $$ \mathbb P(D=m) = \frac{(1-p)^{|m|}p}{2-p}. $$

Math1000
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