Probability Distributions

Exploration: The Memoryless Property of the geometric Distribution

Problem Context

The geometric distribution is uniquelly memoryless. Mathematically , if \(Y\) represents the trial of the first success with probability \(p\) , then for positive integers \(a\) and \(b\) :

\[P(Y > a + b \mid Y > a) = P(Y > b)\]

This means that if we for example, flip a coint \(a\) times, without a success, the probability that we need to flip it more that \(b\) additional times is exactly the same as if we were flipping it \(b\) times from the very beginning.

The Objective

We will simulate a geometric process in R to empirically demonstrate this property.We will compare a baseline geometric distribution against a conditional distribution that has survived \(a\) times.

# Set up variables
prob_success <- 0.2
cutoff <- 5
total_simulations <- 50000

# 1. Baseline: Standard geometric simulation
baseline <- rgeom(total_simulations, prob = prob_success) + 1

# 2. Conditional: Generate a large pool, then keep only the ones greater than our cutoff
large_pool <- rgeom(total_simulations * 5, prob = prob_success) + 1
survived <- large_pool[large_pool > cutoff]

# Take the first 50,000 that survived, and subtract the cutoff
extra_hidden_trials <- survived[1:total_simulations] - cutoff

# Split the plotting screen
par(mfrow =c(1,2))

# Plot 1: The Baseline 
hist(baseline,
     breaks = 30,
     main = "Standard Distribution",
     xlab = "Trials until success",
     col = "green")
     
# Plot 2: The Extra Trials
hist(extra_hidden_trials,
     breaks = 30,
     main = "After Already Failing 5 Times",
     xlab = "Extra trials until success",
     col = "red")

Conclusion

From the generated plots, we can visually observe that the distributions of both the baseline (non-conditional) trials and the shifted conditional trials are virtually identical.

Why This Happens in Code

When we filtered our large_pool to only include games that lasted longer than our cutoff (\(a = 5\)), we isolated the simulations where a player experienced an initial streak of bad luck. By subtracting the cutoff from these survivors (survived - cutoff), we effectively reset their timeline back to “Time Zero.”

If the process had a “memory,” the players who already failed 5 times should show a different probability pattern moving forward. Instead, R’s random generation reveals that their remaining path looks exactly like a brand-new game.

Connection to Exercise 3.71

This simulation provides empirical weight to the algebraic proof required in Exercise 3.71: