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What does the Method of Moments (M.O.M.) estimation give for estimators of the parameters of geometric, Poisson, and gamma distributions? For the geometric distribution, we use the fact that the first moment is E[X] = 1/p and make replacements with estimators to get X̄ = 1/p̂. Then algebraically solve for p̂ to get p̂ = 1/X̄. For the Poisson distribution, it's almost too easy. The first moment is E[X] = k = λ, so we get X̄ = k̂ and we are done. The sample mean is the M.O.M. estimator. For the gamma distribution, there are two parameters α and β, so we need two equations for their estimators. The key equations relating these parameters to the moments are E[X] = αβ and E[X^2] - (E[X])^2 = αβ^2. The resulting system of algebraic equations is nonlinear, but relatively easy to solve by division and substitution. Links and resources =============================== 🔴 Subscribe to Bill Kinney Math: https://www.youtube.com/user/billkinn... 🔴 Subscribe to my Math Blog, Infinity is Really Big: https://infinityisreallybig.com/ 🔴 Follow me on Twitter: / billkinneymath 🔴 Follow me on Instagram: / billkinneymath 🔴 You can support me by buying "Infinite Powers, How Calculus Reveals the Secrets of the Universe", by Steven Strogatz, or anything else you want to buy, starting from this link: https://amzn.to/3eXEmuA. 🔴 Check out my artist son Tyler Kinney's website: https://www.tylertkinney.co/ 🔴 Desiring God website: https://www.desiringgod.org/ AMAZON ASSOCIATE As an Amazon Associate I earn from qualifying purchases.