Discrete Random Variables: PMF, CDF, Binomial & Poisson
“Expectation E[X], Variance Var(X), Bernoulli trials, Binomial PMF, and Poisson limit theorem.”
1.Discrete Random Variables, Moments & Distributions
§ 2.1.A discrete random variable X maps outcomes from sample space Omega to countable values in R, characterized by a Probability Mass Function (PMF) p(x) = P(X = x) satisfying p(x) >= 0 and Sum p(x) = 1.
Binomial Distribution:
Models the number of successes k in n independent Bernoulli trials with constant success probability p.
P(X = k) = (n choose k) p^k (1 - p)^{n - k}.
Expectation: E[X] = np. Variance: Var(X) = np(1 - p).
Poisson Distribution:
Models the count of rare events occurring in a fixed interval of time or space with constant average arrival rate lambda.
P(X = k) = (e^{-lambda} lambda^k) / k!.
Crucial Property: E[X] = Var(X) = lambda.
2. Worked Practice Problem: Poisson Queueing Arrivals
Remember that P(X >= 2) = 1 - P(X = 0) - P(X = 1).
Problem: Network requests arrive at an API server following a Poisson process with an average rate of lambda = 3 requests per minute. Find the exact probability that at least 2 requests arrive in a given 1-minute interval.
Step 1: State the Poisson PMF with parameter lambda = 3:
P(X = k) = (e^{-3} * 3^k) / k!
Step 2: Express 'at least 2' via the complement:
P(X >= 2) = 1 - P(X = 0) - P(X = 1)
Step 3: Evaluate P(X = 0):
P(X = 0) = (e^{-3} * 3^0) / 0! = e^{-3} / 1 ≈ 0.04979
Step 4: Evaluate P(X = 1):
P(X = 1) = (e^{-3} 3^1) / 1! = 3 e^{-3} ≈ 0.14936
Step 5: Compute P(X >= 2):
P(X >= 2) = 1 - (e^{-3} + 3 e^{-3}) = 1 - 4 e^{-3}
P(X >= 2) = 1 - 4 * 0.04979 = 1 - 0.19915 = 0.80085 (approx 80.09%).
Conclusion: The probability that at least 2 requests arrive in one minute is 1 - 4e^{-3} ≈ 0.8009.
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