Geometric Distribution Cheatsheet

Cheatsheet Content

Geometric Distribution Models the number of trials until the first success in a sequence of independent Bernoulli trials. Each trial has a probability of success $p$ and probability of failure $1-p$. There are two common definitions for the random variable. Sample Space for Geometric Distribution The sample space $S$ for this experiment is countably infinite: $E_1$: $S$ (success on first trial) $E_2$: $FS$ (failure on first, success on second) $E_3$: $FFS$ (first success on the third trial) $E_k$: $\underbrace{FFFF...F}_{k-1}S$ (first success on the $k^{th}$ trial) Scenario 1: Total Trials ($X$) Random Variable $X$: Number of total trials until the first success. Support: $X \in \{1, 2, 3, \dots\}$ Example: Interviewing people until the first one passes (including the one who passed). Probability Mass Function (PMF) $$ P(X=k) = (1-p)^{k-1}p $$ $k-1$ failures followed by 1 success. Example: If $p=0.2$, $P(X=1) = (0.8)^0(0.2) = 0.2$. Example: If $p=0.2$, $P(X=2) = (0.8)^1(0.2) = 0.16$. Example: If $p=0.2$, $P(X=4) = (0.8)^3(0.2) = 0.1024$. Expected Value $$ E[X] = \frac{1}{p} $$ Example: If $p=0.2$, $E[X] = 1/0.2 = 5$. Variance $$ Var(X) = \frac{1-p}{p^2} $$ Example: If $p=0.2$, $Var(X) = (1-0.2)/(0.2^2) = 0.8/0.04 = 20$. Scenario 2: Failures Before Success ($Y$) Random Variable $Y$: Number of failures before the first success. Support: $Y \in \{0, 1, 2, \dots\}$ Example: Interviewing people until the first one passes (counting only those who failed before). Probability Mass Function (PMF) $$ P(Y=k) = (1-p)^{k}p $$ $k$ failures followed by 1 success. Example: If $p=0.2$, $P(Y=0) = (0.8)^0(0.2) = 0.2$. Example: If $p=0.2$, $P(Y=1) = (0.8)^1(0.2) = 0.16$. Example: If $p=0.2$, $P(Y=3) = (0.8)^3(0.2) = 0.1024$. Expected Value $$ E[Y] = \frac{1-p}{p} $$ Example: If $p=0.2$, $E[Y] = (1-0.2)/0.2 = 0.8/0.2 = 4$. Variance $$ Var(Y) = \frac{1-p}{p^2} $$ Note: The variance is the same for both scenarios. Example: If $p=0.2$, $Var(Y) = (1-0.2)/(0.2^2) = 0.8/0.04 = 20$. Relationship Between $X$ and $Y$ $X = Y + 1$ $E[X] = E[Y] + 1$ $Var(X) = Var(Y)$ Key Properties Memoryless Property: The geometric distribution is the only discrete distribution with the memoryless property. $$ P(X > m+n | X > m) = P(X > n) $$ This means that if you've already failed $m$ trials, the probability of needing $n$ more trials for success is the same as if you were starting fresh. CDF (Cumulative Distribution Function) for $X$: $$ F(k) = P(X \le k) = 1 - (1-p)^k $$ This is the probability of success occurring on or before the $k$-th trial. Example: Engine Malfunction Probability Problem Suppose that the probability of engine malfunction during any one-hour period is $p = 0.02$. Find the probability that a given engine will survive two hours. Solution using $Y$ (failures before success) Let $Y$ denote the number of one-hour intervals until the first malfunction. A "success" here is a malfunction. For the engine to survive two hours, it must NOT malfunction in the first hour AND NOT malfunction in the second hour. This means the first malfunction must occur in the 3rd hour or later. Therefore, we are looking for $P(Y \ge 2)$ (i.e., 2 or more successful 1-hour periods before malfunction). If $Y=0$, malfunction in 1st hour. If $Y=1$, malfunction in 2nd hour. If $Y=2$, malfunction in 3rd hour. $P(\text{survive two hours}) = P(Y \ge 2)$ (i.e., 2 or more successful 1-hour periods before malfunction). Using the complement rule: $P(Y \ge 2) = 1 - P(Y $P(Y=0) = (1-p)^0 p = p = 0.02$. $P(Y=1) = (1-p)^1 p = (0.98)(0.02) = 0.0196$. $P(\text{survive two hours}) = 1 - (0.02 + 0.0196) = 1 - 0.0396 = 0.9604$. Solution using $X$ (total trials until success) Let $X$ denote the total number of one-hour intervals (trials) until the first malfunction. For the engine to survive two hours, the first malfunction must occur on the 3rd hour or later. So, we want to find $P(X \ge 3)$. Using the complement rule: $P(X \ge 3) = 1 - P(X $P(X=1) = (1-p)^{1-1}p = p = 0.02$. $P(X=2) = (1-p)^{2-1}p = (1-p)p = (0.98)(0.02) = 0.0196$. $P(\text{survive two hours}) = 1 - (0.02 + 0.0196) = 1 - 0.0396 = 0.9604$.