KindiakMath

The Geometric Distribution

Definition 1. For K \subseteq \mathbb R, a random variable X : \Omega \to K satisfies the memoryless property if the following holds: for any s, t \in K,

\displaystyle \mathbb P(X > s + t \mid X > t) = \mathbb P(X > s).

Problem 1. If K = \mathbb N^+ and X satisfies the memoryless property, compute an expression for \mathbb P(X = x) in terms of p := \mathbb P(X = 1).

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Solution. Define the function G by G(x):= \mathbb P(X > x). By the definition of conditional probability,

\begin{aligned} G(s+t) = \mathbb P(X > s+t) &= \mathbb P(X > s +t, X>t) \\ &= \mathbb P(X > s +t \mid X>t) \cdot \mathbb P(X > t) \\ &= \mathbb P(X > s) \cdot \mathbb P(X > t) \\ &= G(s) \cdot G(t). \end{aligned}

Therefore, G(x) = a^x for some a > 0. In particular,

a = G(1) = \mathbb P(X > 1) = 1 - \mathbb P(X=1) = 1-p.

Therefore, G(x) = (1-p)^x, so that

\begin{aligned} \mathbb P(X=x) &= \mathbb P(X > x-1) - \mathbb P(X > x) \\ &= (1-p)^{x-1} - (1-p)^x \\ &= (1-p)^{x-1} \cdot (1 - (1-p)) \\ &= (1-p)^{x-1} \cdot p. \end{aligned}

Definition 2. A discrete random variable X : \Omega \to \mathbb N^+ is said to follow a geometric distribution with success probability p, denoted X \sim \mathrm{Geom}(p), if

\displaystyle \mathbb P(X = x) = (1-p)^{x-1} \cdot p,\quad x \in \mathbb N^+.

Suppose X \sim \mathrm{Geom}(p).

Problem 2. Prove that X satisfies the memoryless property.

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Solution. Using a geometric series,

\begin{aligned} \mathbb P(X > x) &= \sum_{k=x+1}^\infty \mathbb P(X = k) \\ &= \sum_{k=x+1}^\infty (1-p)^{k-1} \cdot p \\ &= \frac{p\cdot (1-p)^x}{1 - (1-p)} = (1-p)^x. \end{aligned}

By the definition of conditional probability,

\begin{aligned} \mathbb P(X > s +t \mid X>t) &= \frac{\mathbb P(X > s +t, X>t)}{\mathbb P(X > t)} \\ &= \frac{\mathbb P(X > s +t)}{\mathbb P(X > t)} \\ &= \frac{(1-p)^{s+t}}{(1-p)^t} \\ &= (1-p)^s = \mathbb P(X > s). \end{aligned}

Problem 3. Prove that \displaystyle \mathbb E[X] = \sum_{x = 0}^\infty \mathbb P(X > x). Hence, evaluate \mathbb E[X] and \mathrm{Var}(X).

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Solution. By interchanging sums,

\begin{aligned} \mathbb E[X] &= \sum_{x = 0}^\infty x \cdot \mathbb P(X = x) = \sum_{x=0}^\infty \sum_{y=0}^{x-1} \mathbb P(X = x) \\ &= \sum_{y=0}^\infty \sum_{x=y+1}^\infty \mathbb P(X = x) = \sum_{y=0}^\infty \mathbb P(X > y) = \sum_{x = 0}^\infty \mathbb P(X > x). \end{aligned}

Hence, using the calculations in Problem 2,

\displaystyle \mathbb E[X] = \sum_{x=0}^\infty (1-p)^x = \frac{1}{1-(1-p)} = \frac 1p.

For the variance, we first compute \mathbb E[X^2]. Observe that

\displaystyle \sum_{y=0}^{x-1} (2y + 1) = x^2.

Therefore, by interchanging sums,

\begin{aligned} \mathbb E[X^2] &= \sum_{x = 0}^\infty x^2 \cdot \mathbb P(X = x) \\ &= \sum_{x=0}^\infty \sum_{y=0}^{x-1} (2y + 1) \cdot \mathbb P(X = x) \\ &= \sum_{y=0}^\infty \sum_{x=y+1}^\infty (2y + 1) \cdot \mathbb P(X = x) \\ &= \sum_{y=0}^\infty \left( (2y + 1) \cdot \sum_{x=y+1}^\infty (1-p)^{x-1} \cdot p \right) \\ &= \sum_{y=0}^\infty \left( (2y + 1) \cdot \frac{(1-p)^y}{1 - (1-p)} \cdot p \right) \\ &= \frac 1p \cdot \sum_{y=0}^\infty \left( (2y + 1) \cdot (1-p)^y \cdot p \right) \\ &= \frac 1p \cdot \left( p + (1-p) \cdot \left( 2 \cdot \sum_{y=1}^\infty y\cdot \mathbb P(X = y ) + \sum_{y=1}^\infty \mathbb P(X = y ) \right) \right) \\ &= \frac 1p \cdot \left( p + (1-p) \cdot \left( \frac 2p + 1 \right) \right) \\ &= 1 + \left( \frac 1p - 1 \right) \cdot \left( \frac 2p + 1 \right) \\ &= 1 + \frac 2{p^2} + \frac 1p - \frac 2p - 1 = \frac {2-p}{p^2}. \end{aligned}

Therefore,

\begin{aligned} \mathrm{Var}(X) &= \mathbb E[X^2] -\mathbb E[X]^2 = \frac {2-p}{p^2} - \frac 1{p^2} = \frac{1-p}{p^2}. \end{aligned}

Problem 4. If Y \sim \mathrm{Geom}(q) is independent of X, compute the distribution of \min\{X, Y\}.

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Solution. Denote W := \min \{X, Y\}. Then

\begin{aligned} \mathbb P(W > w) &= \mathbb P(\min\{X,Y\} > w) \\ &= \mathbb P(X > w, Y > w) \\ &= \mathbb P(X > w) \cdot \mathbb P(Y > w) \\ &= (1-p)^w \cdot (1-q)^w = ((1-p)(1-q))^w. \end{aligned}

Therefore, \min\{X, Y\} \sim \mathrm{Geom}(1-(1-p)(1-q)).

Problem 5. Fix \lambda > 0. Suppose X_n \sim \mathrm{Geom}(\lambda /n). For any x > 0, evaluate

\displaystyle \lim_{n \to \infty} \mathbb P \left( \frac {X_n}n > x \right) .

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Solution. Using the tail-probability,

\begin{aligned} \lim_{n \to \infty} \mathbb P \left( \frac {X_n}n > x \right) &= \lim_{n \to \infty} \mathbb P(X_n > nx) \\ &= \lim_{n \to \infty} \left( 1 - \frac{\lambda}n \right)^{nx} \\ &= \left( \lim_{n \to \infty} \left( 1 - \frac{\lambda}n \right)^{n} \right)^x \\ &= (e^{-\lambda})^x = e^{-\lambda x}. \end{aligned}

—Joel Kindiak, 3 Aug 25, 0004H

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