Exam 2 Cheat Sheet v1

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San Diego State University**We aren't endorsed by this school

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STAT 550

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Statistics

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Dec 16, 2024

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DistributionParametersProbability Mass FunctionExpectationVarianceUniform𝑛𝑃(𝑋 = 𝑠௡) =ଵ௡⎯ for 𝑖 = 1, … , 𝑛𝐸[𝑋] = ෍ 𝑥௜𝑃(𝑋 = 𝑥௜) = ෍ 𝑥௜ቆ1𝑛⎯⎯ቇ =𝑥ଵ+ ⋯ + 𝑥௡𝑛⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯௡௜ୀଵ௡௜ୀଵ𝑉[𝑋] = 𝐸[𝑋ଶ] − (𝐸[𝑋])ଶ=(𝑛 + 1)(2𝑛 + 1)6⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯− ቆ𝑛 + 12⎯⎯⎯⎯⎯ቇଶ=𝑛ଶ− 112⎯⎯⎯⎯⎯⎯Bernoulli𝑝𝑃(𝑋 = 1) = 𝑝and 𝑃(𝑋 = 0) = 1 − 𝑝𝐸[𝑋] = ෍ 𝑘𝑃(𝑋 = 𝑘) =ஶ௞ୀ଴𝑝𝑉[𝑋] = 𝑝(1 − 𝑝)Binomial𝑛, 𝑝𝑃(𝑋 = 𝑘) = ൫௡௞൯𝑝௞(1 − 𝑝)௡ି௞, for 𝑘 = 0, 1, … , 𝑛𝐸[𝑋] = ෍ 𝑘𝑃(𝑋 = 𝑘) =ஶ௞ୀ଴෍ 𝑘𝑛!𝑘! (𝑛 − 𝑘)!⎯⎯⎯⎯⎯⎯⎯⎯⎯𝑝௞(1 − 𝑝)௡ି௞= 𝑝𝑛ஶ௞ୀ଴𝑉[𝑋] = 𝐸[𝑋ଶ] − (𝐸[𝑋])ଶ= 𝑛𝑝(1 − 𝑝) + (𝑛𝑝)ଶ− (𝑛𝑝)ଶ= 𝑛𝑝(1 − 𝑝)Poisson𝜆𝑃(𝑋 = 𝑘) =௘షഊఒೖ௞!⎯⎯⎯⎯⎯, where 𝑘 = 0, 1𝐸[𝑋] = ෍ 𝑘𝑃(𝑋 = 𝑘) =ஶ௞ୀ଴෍ 𝑘𝑒ିఒ𝜆௞𝑘!⎯⎯⎯⎯⎯ = 𝜆ஶ௞ୀ଴𝑉[𝑋] = 𝐸[𝑋ଶ] − (𝐸[𝑋])ଶ= 𝜆(𝜆 + 1) − 𝜆ଶ= 𝜆Geometric𝑝𝑃(𝑋 = 𝑘) = (1 − 𝑝)௞ିଵ𝑝, for 𝑘 =, 1, 2, …𝐸[𝑋] =1𝑝⎯⎯𝑉[𝑋] =1 − 𝑝𝑝ଶ⎯⎯⎯⎯⎯Negative Binomial𝑟, 𝑝𝑃(𝑋 = 𝑘) = 𝐶௥ିଵ௞ିଵ(1 − 𝑝)௞ି௥, 𝑘 = 𝑟 + 1, 𝑟 + 2, …and 𝑟 = 1, 2, …𝐸[𝑋] =𝑟𝑝⎯⎯𝑉[𝑋] =𝑟(1 − 𝑝)𝑝ଶ⎯⎯⎯⎯⎯⎯⎯Hypergeometric*𝑀, 𝑁, 𝑘𝑃(𝑋 = 𝑥) =஼ೣಾ஼ೖషೣಿషಾ஼ೖಿ⎯⎯⎯⎯⎯⎯⎯, 𝑥 = 0, 1, … , min(𝑘, 𝑀)𝐸[𝑋] = 𝑘𝑀𝑁⎯⎯𝑣[𝑋] = 𝐸[𝑋ଶ] − (𝐸[𝑋])ଶ=112⎯⎯⎯(𝑀 + 1)(𝑁 − 1)*𝑁 = 𝑡𝑜𝑡𝑎𝑙 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛, 𝑀 = 𝑡𝑜𝑡𝑎𝑙 𝑠𝑢𝑏𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛, 𝑘 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑜𝑛𝑠 (how many are chosen for study out of total population), 𝑥 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 (the random variable)Midterm 2 Cheat Sheet (page 1)Friday, November 15, 20245:20 PMSTAT 550 Applied Probability Page 1

Expectation, aka Expected Value (Ch 4.1)𝐸[𝑋] = ෍ 𝑥𝑃(𝑋 = 𝑥)௫∈ௌ, 𝑎𝑛𝑑 𝐸[𝑋 + 𝑌] = 𝐸[𝑋] + 𝐸[𝑌]The sum is taken over all values of 𝑋.Expectation of a Function of Random Variable (Ch 4.2)Method 1: Let X be a random variable, and let 𝑌 = 𝑔(𝑋). If the probability distribution of 𝑌is known, then the expectation of 𝑌is𝐸[𝑌] = ෍ 𝑦𝑃(𝑌 = 𝑦)௬∈ௌ೤Note: we need to find the PMF 𝑃(𝑌 = 𝑦)before finding 𝐸[𝑌]when using this method.Method 2: Let 𝑋be a random variable with PMF 𝑃(𝑋 = 𝑥),𝑥 ∈ 𝑆. Let 𝑔(𝑋)be a funtion of 𝑋. Then the expectation of 𝑦 = 𝑔(𝑋)is 𝐸[𝑌] = 𝐸[𝑔(𝑥)] = ෍𝑔(𝑥)𝑃(𝑋 = 𝑥)௫∈ௌVariance, a measure of variability or spread (Ch 4.6)Let 𝑋be a random variable with mean 𝐸[𝑋] = 𝜇 < ∞. Then the variance of 𝑋is𝑉[𝑋] = 𝐸[(𝑋 − 𝜇)ଶ] = ෍ (𝑥 − 𝜇)ଶ𝑃(𝑋 = 𝑥)௫∈ௌೣ= 𝐸[𝑋ଶ] − (𝐸[𝑋])ଶStandard Deviation, measure of spread (Ch 4.6)Let 𝑋be a random variable with variance 𝑉[𝑋]. Then the standard deviation of 𝑋is𝑆𝐷[𝑋] = ඥ𝑉[𝑋]⎯⎯⎯⎯Properties of Variance and Standard Deviation (Ch 4.6)Let 𝑋be a random variable where 𝐸[𝑋]and 𝑉[𝑋]exist, then for constants 𝑎and 𝑏,𝐸[𝑎𝑋 + 𝑏] = 𝑎𝐸[𝑋] + 𝑏,𝑉[𝑎𝑋 + 𝑏] = 𝑎ଶ𝑉[𝑋],𝑆𝐷[𝑎𝑋 + 𝑏] = |𝑎| × 𝑆𝐷[𝑋]General Formula for Variance of a Sum (Ch 4.7)For RVs 𝑋and 𝑌with finite variance,𝑉[𝑋 + 𝑌] = 𝑉[𝑋] + 𝑉[𝑌] + 2𝐶𝑜𝑣(𝑋, 𝑌)and,𝑉[𝑋 − 𝑌] = 𝑉[𝑋] + 𝑉[𝑌] − 2𝐶𝑜𝑣(𝑋, 𝑌)If 𝑋and 𝑌are uncorrelated (aka independent),𝑉[𝑋 + 𝑌] = 𝑉[𝑋] + 𝑉[𝑌]Expectation and Variance of a Sum of INDEPENDENT Random Variables (Ch 4.7)If 𝑋and 𝑌are uncorrelated, i.e. 𝐶𝑜𝑣(𝑋, 𝑌) = 0, then𝑉[𝑋 ± 𝑌] = 𝑉[𝑋] + 𝑉[𝑌]Let 𝑋ଵ,… , 𝑋௡be 𝑛independent random variables, then𝐸[𝑋ଵ+ ⋯ + 𝑋௡] = 𝐸[𝑋ଵ] + ⋯ + 𝐸[𝑋௡]and 𝑉[𝑋ଵ+ ⋯ + 𝑋௡] = 𝑉[𝑋ଵ] + ⋯ + 𝑉[𝑋௡]General Formula for Variance of a Linear Combination, aka Sum (Ch 4.7)For RVs 𝑋and 𝑌with finite variance,𝑉[𝑎𝑋 ± 𝑏𝑌] = 𝑎ଶ𝑉[𝑋] + 𝑏ଶ𝑉[𝑌] + 2𝑎𝑏𝐶𝑜𝑣(𝑋, 𝑌)In general, let 𝑋ଵ, … , 𝑋௡be random variables, then𝑉[𝑋ଵ+ ⋯ + 𝑋௡] = ෍ ෍𝐶𝑜𝑣൫𝑋௜, 𝑌௝൯ = ෍ 𝑉[𝑋௜] + ෍ 𝐶𝑜𝑣(𝑋௜, 𝑌௝)௜ஷ௝௡௜ୀଵ௡௝௡௜In particular,𝑉[𝑋ଵ+ 𝑋ଶ+ 𝑋ଷ] = 𝑉[𝑥ଵ] + 𝑉[𝑥ଶ] + 𝑉[𝑥ଷ] + 2𝐶𝑜𝑣(𝑋ଵ,𝑋ଶ) + 2𝐶𝑜𝑣(𝑋ଵ, 𝑋ଷ) + 2𝐶𝑜𝑣(𝑋ଶ, 𝑋ଷ)Covariance (Ch 4.7)For random variables 𝑋and 𝑌, with repsective means 𝐸[𝑋] = 𝜇௑and 𝐸[𝑌] = 𝜇௒, the covariance between 𝑋and 𝑌is𝐶𝑜𝑣(𝑋, 𝑌) = 𝐸[(𝑋 − 𝜇௑)(𝑌 − 𝜇௒)] = 𝐸[𝑋𝑌] − 𝜇௑𝜇௒= 𝐸[𝑋𝑌] − 𝐸[𝑋]𝐸[𝑌]Correlation (Ch 4.7)The correlation between 𝑋and 𝑌is𝐶𝑜𝑟𝑟(𝑋, 𝑌) =𝐶𝑜𝑣(𝑋, 𝑌)𝑆𝐷[𝑋]𝑆𝐷[𝑌]⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ =𝐸[(𝑋 − 𝜇௑)(𝑌 − 𝜇௒)]ඥ𝑉[𝑋]𝑉[𝑌]⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯Joint Probability Mass Function, aka Joint Distribution (Ch 4.3)𝑝௑,௒(𝑥, 𝑦) = 𝑃(𝑋 = 𝑥, 𝑌 = 𝑦), 𝑥 ∈ 𝑆௫, 𝑦 ∈ 𝑆௬Marginal Distributions (Ch 4.3)If 𝑋takes values in a set 𝑆௫and 𝑌takes values in a set 𝑆௬, then the marginal distribution of 𝑿is𝑝௑(𝑥) = 𝑃(𝑋 = 𝑥) = ෍ 𝑃(𝑋 = 𝑥, 𝑌 = 𝑦)௬∈ௌ೤and the marginal distribution of 𝒀is𝑝௒(𝑦) = 𝑃(𝑌 = 𝑦) = ෍ 𝑃(𝑋 = 𝑥, 𝑌 = 𝑦)௫∈ௌೣTable of Joint PMF and Marginal PMFConditional Probability Mass Function (Ch 4.8)If 𝑋and 𝑌are two discrete random variables, then the conditional probability mass function of 𝑌given 𝑋 = 𝑥is𝑃(𝑌 = 𝑦|𝑋 = 𝑥) =𝑃(𝑌 = 𝑦, 𝑋 = 𝑥)𝑃(𝑋 = 𝑥)⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯=𝑝௑,௒(𝑥, 𝑦)𝑝௑(𝑥)⎯⎯⎯⎯⎯⎯⎯⎯when 𝑃(𝑋 = 𝑥) > 0.Conditional Probability Mass Function for Independent Random Variables(Ch 4.8)When 𝑋and 𝑌are independent, then the conditional probability of 𝑌given 𝑋 = 𝑥is𝑃(𝑌 = 𝑦|𝑋 = 𝑥) =𝑃(𝑌 = 𝑦, 𝑋 = 𝑥)𝑃(𝑋 = 𝑥)⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 𝑃(𝑌 = 𝑦)Conditional Expectation and Variance of 𝒀Given 𝑿 = 𝒙For discrete random variables 𝑋and 𝑌, the conditional expectationof 𝑌given 𝑋 = 𝑥is𝐸[𝑌|𝑋 = 𝑥] = ෍ 𝑦𝑃(𝑌 = 𝑦|𝑋 = 𝑥)௬∈ௌ೤and the Conditional Variance of 𝑌given 𝑋 = 𝑥is𝑉[𝑌|𝑋 = 𝑥] = 𝐸[(𝑌 − 𝐸[𝑌|𝑋 = 𝑥])ଶ|𝑋 = 𝑥] = 𝐸[𝑌ଶ|𝑋 = 𝑥] − (𝐸[𝑌|𝑋 = 𝑥])ଶMoment Function (Ch 5.2)Let 𝑋be a random variable. The 𝑘௧௛moment function of 𝑋is defined as𝐸ൣ𝑋௞൧ = ෍ 𝑥௞𝑃(𝑋 = 𝑥)௫∈ௌೣwhere 𝑘 = 1, 2, …When 𝑘 = 1, the expectation 𝐸[𝑋]is the first moment function.Moment Generating Functions MGF (Ch 5.2)Let 𝑋be a random variable. The MGF of 𝑋is the real-valued function𝑚௑(𝑡) = 𝐸[𝑒௧௑] = ෍𝑒௧௫𝑝௑(𝑥) = ෍ 𝑒௧௫𝑃(𝑋 = 𝑥)௫௫Note that MGF is a function of 𝑡and is determined by the PMF 𝑃(𝑋 = 𝑥).MGF for Geometric DistributionLet 𝑋~𝐺𝑒𝑜𝑚(𝑝), then the MGF of 𝑋is𝑚௑(𝑡) = 𝐸[𝑒௧௑] = ෍ 𝑒௧௞(1 − 𝑝)௞ିଵ 𝑝 =𝑝𝑒௧1 − 𝑒௧(1 − 𝑝)⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ஶ௞ୀଵMGF for Binomial DistributionLet 𝑋~𝐵𝑖𝑛𝑜𝑚(𝑛, 𝑝), then the MGF of 𝑋is𝑚௑(𝑡) = 𝐸[𝑒௧௑] ෍ 𝑒௧௞൬𝑛𝑘൰ 𝑝௞(1 − 𝑝)௡ି௞= (𝑒௧𝑝 + 1 − 𝑝)௡௡௞ୀ଴MGF for Poisson DistributionLet 𝑋~𝑃𝑜𝑖𝑠(𝜆), then the MGF of 𝑋is𝑚௑(𝑡) = 𝐸[𝑒௧௑] = ෍ 𝑒௧௞𝑒ିఒ𝜆௞𝑘!⎯⎯ = 𝑒ఒ൫௘೟ିଵ൯ஶ௞ୀ଴Properties of Moment Generating Functions (Ch 5.2)Let 𝑚௑(𝑡)be the MGF of 𝑋 where the 𝑘௧௛derivative exists, then𝐸ൣ𝑋௞൧ = 𝑚௑(௞)(0), for 𝑘 = 1, 2, …1.If 𝑋and 𝑌are independent random variables, then the MGF of their sum is the product of their MGFs,𝑚௑ା௒(𝑡) = 𝐸ൣ𝑒௧(௑ା௒)൧ = 𝐸[𝑒௧௑]𝐸[𝑒௧௒] = 𝑚௑(𝑡)𝑚௒(𝑡)2.Let 𝑋be a random variable with MGF 𝑚௑(𝑡)and constants 𝑎 ≠ 0and 𝑏, then𝑚௔௑ା௕(𝑡) = 𝐸ൣ𝑒௧(௔௑ା௕)൧ = 𝑒௧௕𝐸[𝑒௔௧௑] = 𝑒௧௕𝑚௑(𝑎𝑡)3.MGFs uniquely determine the underlying probability distribution. That is, if two random variables have the same MGF, then they have the same probability distribution.4.Midterm 2 Cheat Sheet (page 2)Friday, November 15, 20246:17 PMSTAT 550 Applied Probability Page 1