Continuous Random Variables Prof. Metin    Outline To use this in your own course/training, please obtain permission from P rof. rof. P from permission obtain please course/training, own your in this use To Moment generating generating Moment Common continuous random variables Continuous random variables anddensity I f you find any inaccuracies, please contact [email protected] for corrections. for [email protected] contact please inaccuracies, any find you f I Çakanyıldırım used used function various various resources to prepare this document for teaching/training. Updated in Fall 2019 Çakanyıldırım .

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metin /~   Acontinuous random variablehas an uncountable sample space Seeking a Density a Seeking Singularly Absolutely continuous rv – – – – Function: E.g Sample variable: Random there exists there Absolutely for » » • • • • ., finitely many disjointintervals Cantor Cantor set: has map are a A probability re re continuous functionorrandom differentiable continuous a correspondinga Continuous functions length function: continuous ∑ 𝛿𝛿 𝑖𝑖 such that that such Uncountable, so it can it so Uncountable, -0 intervals  𝑏𝑏 𝑖𝑖 over over 1 of mass − 𝑭𝑭 . Continuity ⇐ . Cantor set set Cantor 𝑎𝑎 𝒂𝒂 has an absolutely continuous absolutelyan has 𝑖𝑖 almost everywhere. functions space is uncountable but but space is the range of uncountable the = ≤ 𝑓𝑓 � 𝛿𝛿 called density − that increase only over sets whose total length is is zero. length total sets whose only over increase that 𝒂𝒂 implies implies → ∞ length an uncountable uncountable an 𝒇𝒇 [0,1],so it maps 𝒖𝒖 condition above above for 1interval.condition serve as the range of a continuous rv. a continuous of range the as serve ( 𝒅𝒅 𝑎𝑎 -0 intervals ∑ 𝒅𝒅 𝑖𝑖 variable: variable: , 𝑖𝑖 . 𝑏𝑏 𝐹𝐹 𝑖𝑖 such that such every given), forevery ( 𝑏𝑏 𝑖𝑖 rv ) length 0 0 length − range range 𝐹𝐹 cdf ( 𝑎𝑎 of of 𝑖𝑖 )  𝐹𝐹 length length ≤ satisfying length 1 length 𝜖𝜖 𝜖𝜖 rv is is , is a set withzero length. ⇐ singularity . Nothing Nothing Absolutely continuous: Absolutely Nothing Nothing Singularity:   Everything Nothing

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metin /~ 0.0.. Cantor Set Set Cantor 0.0000002.. 0.000002.. 0.00002.. 1/27 0.00? 0.0002.. 1/9 0.002.. For For I 0.0? 0.01.. nitialize 1/6 𝑛𝑛 0.02? – = 1 0.02.. 1/3 𝑅𝑅 Append 0 Append 𝑅𝑅 to to Uncountable legs Uncountable U ∶ = = sing only 0 & 2 Cantor Table ∞ Append 1 Append [ for every remaining remaining every for 𝑅𝑅 0 , 0.? 1/2 0.1.. 1 ∖ ⊂ ] Append Append Middle third of of third Middle [ 𝟎𝟎 0.2 2 0.20? , 𝟏𝟏 0.21.. 0.2? ] 5/6 : Uncountable length- and : 0.22 𝑅𝑅 𝑅𝑅 0.22? 0.222 0.2222 0.22..     the appendixof eventsMeasurability leadsto a contradiction to the Why uncountable: Read the real number generation schemethat What many is real numbers!!! Uncountable left? = length removed, Total Amount removed – – – – – – – – – – – 0.0202 0.0202 3 base .. 0.002 0.02.. 0.22 0.2 3 base 0.02 examples some 1, no include 3 representation base whose numbers realAll General 𝑛𝑛 𝑛𝑛 𝑛𝑛 = = = base 3 base 3 2 1 base 3 base , , , base 3 base base 3 base 9 1 3 1 3 𝑛𝑛 1 3 , , + , 9 2 3 2 = length 3 = = 2 3 1 = 0 , ∪ , length= , = = length= 0 0 ∪ 9 1 1 27 9 7 1 1 2 1 = 9 1 , + 3 ∑ 7 9 8 2 + + 3 countability 2 ∞ 𝑛𝑛 + ∑ , 2 = , 2 3 8 ∞ 𝑛𝑛 3 1 3 length 1 2 0 3 1 3 1 = 0 3 1 . 1 3 3 1 ∪ 3 1 ∑ 3 2 3 1 3 1 𝑛𝑛 3 1 = 27 𝑛𝑛 ∞ 𝑛𝑛 1 = 3 1 = = = + + 9 𝑛𝑛 3 of real numbers in numbers real of 3 1 3 2 0 , + 2 2 = 2 3 1 3 2 3 2 0 3 3 2 1 9 1 9 1 9 1 9 1 2 ∪ 𝑛𝑛 81 0 1 = = = = 3 25 9 8 9 2 9 2 3 = 1 , 3 26 . 81 1 3

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metin /~ 2.Cantor Function 1.Cantor Set Set Cantor Continuousbut Not AbsolutelyContinuous Cantor function is is function Cantor continuous over all over continuous increasing inside Cantor Set constant Cantoroutside Set Exclude from Cantor Set

Exclude from Cantor Set

Exclude from Cantor Set  Exclude from Exclude Cantor Set Set Cantor nrae ifIncreasesno derivative exists if derivative=0 Constant Cantor Function nrae overIncreases length Constant over length over

Exclude from Cantor Set - - 1 0 Exclude from Cantor Set Exclude from Cantor Set Chalice (1991): Any increasing real (1991): Any increasing Chalice  3. Cantor function a as function 3. Cantor a ) 2/3 Cantorrandom thatvariable is singularly continuous 𝐹𝐹 Cantor Cantor 0 = 0 , , b) 𝐹𝐹 Random Variable Random 8/9 𝑥𝑥 / i 3 s Cantor function. Cantor s = cdf 𝐹𝐹 ( yields 1 𝑥𝑥 ) - valued function on that on has [0,1] function valued / 2 , , c) 8/9 𝐹𝐹 1 − 𝑥𝑥 = 26/27 1 − 𝐹𝐹

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metin /~ 1    Finding Many Densities Many Finding Does each Pdf In this – – – – – 𝑓𝑓 functions parametrized as 𝑓𝑓 as functions parametrized Can we generateweCan examplesthat have continuous versions single point? generateweCan interval an examples over thatdiffering have versions point. the last a differsame exampleVersions over and almosteverywhere are of the Ex: Consider of probability sequence construction:Recall ( the taking the 𝐴𝐴 when takingthe of1 onlyvalue is derived fromisderived » » » » course usually,course No No familythe in function Each countably by the at alterations Note that 𝐹𝐹 , because mustversionsover interval. continuousan differ, because , because lead to a unique 𝑓𝑓 a to lead 𝑓𝑓 ∫ 0 − 𝑥𝑥 ∞ 𝑥𝑥 cdf absolutely continuous 𝑓𝑓 𝑎𝑎 𝑓𝑓 = 1 𝐹𝐹 𝑢𝑢 ∫ = 1 𝐴𝐴 𝑑𝑑 𝑓𝑓 for 0 𝑓𝑓 1 2 𝑑𝑑 ? No! 𝑎𝑎 𝑥𝑥 almost everywhere (over all intervals) = 𝑥𝑥 𝑑𝑑 ≤ 𝑥𝑥 𝑑𝑑 { = = 𝑥𝑥 𝑓𝑓 = 𝑎𝑎 is correct. ≤ many points or at the points the singlemanypoint orat 𝐹𝐹 𝑓𝑓 : if and only if 0 ∫ 0 ( 𝐴𝐴 1 𝑥𝑥 𝑥𝑥 ≤ 𝑓𝑓 ) and over same over rangeand of𝑥𝑥 the 2 rv for 0 𝑎𝑎 + 𝑥𝑥 ≤ = 𝑎𝑎 𝑑𝑑 𝐈𝐈 absolutely 1 𝑑𝑑 ≤ 𝑥𝑥 } } densityis for𝐹𝐹 a = for every set set every for 𝑥𝑥 𝑎𝑎 Ω for 0 , ≤ ℑ 1 , P , because integral is not affected integral affected , because isnot ) and then then ) and ≤ continuous ? 𝑎𝑎 ≤ 𝐴𝐴 . 1 . 𝐈𝐈 𝐹𝐹 𝑥𝑥 𝐴𝐴 rv define a family of family a define { 𝑎𝑎 is a binary variable 𝑎𝑎 = } rather than a a than rather : . 𝐈𝐈 = 0 ≤ P 𝑥𝑥 ( ≤ 𝑋𝑋 1 𝑥𝑥 ≤ . 𝑎𝑎 )

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metin /~   Density and MomentsDensity and makes 𝑓𝑓 Ex: For densityProbability functionpdf – – – – E 1 0 𝑓𝑓 𝑋𝑋 » = = 𝑋𝑋 𝑥𝑥 a continuous random variable random continuous a 𝐹𝐹 P Which doesequality fail in a legitimate density so that that it legitimatedensity one? integrates Find to soa = = 𝑋𝑋 1 ∫ 𝑢𝑢 lim 0 𝜔𝜔 1 → =  𝑥𝑥 2 3 ( ∫ = 0 𝐹𝐹 1 𝑋𝑋 𝑢𝑢 Others are those rvs to are Others similarof discrete 𝑐𝑐 𝑥𝑥 3 𝑢𝑢 – – – 𝑥𝑥 + 2 ≠ + − Variance V Variance The value Expected 𝑢𝑢 𝑓𝑓 2 𝑋𝑋 𝑢𝑢 𝐹𝐹 𝑑𝑑 𝑋𝑋 𝑑𝑑 𝑥𝑥 k 𝑑𝑑 th 𝑑𝑑 𝑢𝑢 = = = )/ moment: moment: arbitrary at at arbitrary ( 8 3 3 𝑐𝑐 𝑥𝑥 1 𝑋𝑋 𝑢𝑢 − = 𝑢𝑢 4 : 3 E 𝑢𝑢 P + V + ) 𝑋𝑋 𝑋𝑋 E define almost everywhere almost define Ω let f , let 𝑋𝑋 3 1 𝑋𝑋 2 1 : = 𝑘𝑘 E countably = 𝑢𝑢 𝑢𝑢 3 𝑋𝑋 P 2 E = | | u 𝑢𝑢 𝑢𝑢 𝑢𝑢 𝑢𝑢 ∪ = = 𝑋𝑋 ∫ = = = 0 1 𝑥𝑥 𝑥𝑥 1 0 2 = ∈ 𝑥𝑥 = ∫ 𝐴𝐴 = - 𝑥𝑥 𝑘𝑘 𝐈𝐈 𝑓𝑓 8 3 𝑥𝑥 0 E many points. Let Let points. many 3 𝑐𝑐 𝑋𝑋 𝑋𝑋 𝑓𝑓 ≤ + + 𝑋𝑋 u 𝑋𝑋 𝑥𝑥 𝜔𝜔 3 1 ≤ 2 1 𝑥𝑥 1 = 𝑑𝑑 ( 2 so so = 𝑑𝑑 𝑑𝑑 : 𝑐𝑐 24 17 𝑑𝑑 𝑢𝑢 𝑥𝑥 𝑐𝑐 . 2 E = + = 𝑋𝑋 2 3 . 𝑢𝑢 ∑ . ).What ).What 𝑥𝑥 ∈ 𝐴𝐴 A be the range of range the be P 𝑋𝑋 value of𝑐𝑐 value 𝜔𝜔 = 0 ? rv

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metin /~      Independence Ex: For two independent FortwoEx: variables Ex:A random variable is not independent of itself so E independent FortwoEx: variables This definition is inherited fromthe independence of events Random variables – E where 𝑋𝑋 1 𝑋𝑋 2 𝐴𝐴 = = = E ∫ 𝑥𝑥 1 𝑋𝑋 ( , 𝑥𝑥 1 − P 𝑋𝑋 2 E 𝑥𝑥 ∞ P 𝑋𝑋 and 1 𝑋𝑋 𝑥𝑥 , ∈ 𝑋𝑋 2 𝑎𝑎 2 ] A 𝑓𝑓 ≤ 𝑌𝑌 ( , and 𝑌𝑌 𝑥𝑥 𝑎𝑎 r needn f ifindependent are 1 , ∈ ) 𝑌𝑌 𝑓𝑓 𝐵𝐵 𝐵𝐵 ≤ 𝑥𝑥 = 2 𝑏𝑏 = 𝑋𝑋 𝑋𝑋 𝑑𝑑 1 1 − = P 𝑥𝑥 , , 1 ∞ 𝑋𝑋 𝑋𝑋 𝑋𝑋 P 𝑑𝑑 2 2 , 𝑥𝑥 ∈ 𝑋𝑋 , we, have V we, have E 𝑏𝑏 2 𝐴𝐴 ≤ = . P 𝑎𝑎 ∫ 𝑥𝑥 1 𝑌𝑌 P 𝑥𝑥 ∈ 𝑌𝑌 1 𝐵𝐵 𝑓𝑓 ≤ ( 𝑋𝑋 𝑋𝑋 𝑥𝑥 𝑏𝑏 1 1 1 X ) 𝑋𝑋 𝑑𝑑 + 2 2 𝑥𝑥 1 𝑋𝑋 = = ∫ 2 𝑥𝑥 2 E E = 𝑥𝑥 𝑋𝑋𝑋𝑋 2 𝑋𝑋 V 𝑓𝑓 1 ( 𝑋𝑋 𝑥𝑥 E 1 2 ≠ ) 𝑋𝑋 𝑑𝑑 E + 2 𝑥𝑥 2 𝑋𝑋 V 𝑋𝑋 E 2

𝑋𝑋

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metin /~      UniformRandom Variable CommonContinuous Random Variables Find Find Ex Uniform randomA variable 𝑋𝑋 uniform in any subinterval of – – 𝑥𝑥 : What is the pdf of a random pdfofa : [ variable over uniformisthe What 2 [ We We Since 𝑥𝑥 P E P 𝑢𝑢 ( ( , 𝑋𝑋 𝑈𝑈 need need 𝑥𝑥 𝑎𝑎 1 random [ variable over ( 𝑜𝑜 𝑥𝑥 P 𝑥𝑥 ] 1 ≤ = . 𝑢𝑢 𝑎𝑎 ≥ 1 , 𝑋𝑋 𝑈𝑈 ≤ 𝑥𝑥 = 𝑥𝑥 𝑜𝑜 2 ≤ E 𝑓𝑓 𝑋𝑋 0 ) 𝐹𝐹 𝑥𝑥 , ) 1 𝑎𝑎 ≤ ( 𝑈𝑈 𝑢𝑢 1 . P 𝑥𝑥 + 𝑜𝑜 𝑎𝑎 𝑋𝑋 𝑥𝑥 [ = ≥ ) 𝑥𝑥 𝑢𝑢 1 + , that is 1 , that 𝛿𝛿 , 𝑢𝑢 𝐈𝐈 ≥ 𝑋𝑋 𝑥𝑥 𝑥𝑥 𝛿𝛿 , 𝑜𝑜 2 𝑢𝑢 𝑥𝑥 = 𝑋𝑋 ≤ 𝑜𝑜 = 𝑢𝑢 = 2 ] ≤ P = 𝑥𝑥 a givenlengthof a P 𝑈𝑈 = takes values between 𝑜𝑜 𝑏𝑏 ∫ = ( 𝑥𝑥 𝑏𝑏 𝑥𝑥 𝑥𝑥 � 𝑥𝑥 𝑜𝑜 0 𝑢𝑢 ≤ 𝑥𝑥 𝑢𝑢 𝑜𝑜 − ∫ ≤ 1 , 1 𝑥𝑥 1 𝑥𝑥 𝑢𝑢 , 𝑥𝑥 ≥ 𝑢𝑢 𝑋𝑋 𝑢𝑢 𝑥𝑥 𝑜𝑜 ) 𝑋𝑋 𝑥𝑥 𝑥𝑥 𝑜𝑜 ) 2 𝑐𝑐 𝑜𝑜 ≤ ≤ . 𝑐𝑐𝑐𝑐 ] − 1 1 𝑥𝑥 is denoted is denoted Note each 𝑢𝑢 𝑏𝑏 𝑏𝑏 = 𝑑𝑑 1 + + 𝑑𝑑 c 𝑑𝑑 𝛿𝛿 𝛿𝛿 = 𝑥𝑥 x 1 , the density density the , o 𝛿𝛿 𝑥𝑥 𝑑𝑑 − 𝑜𝑜 for 𝑥𝑥 𝑥𝑥 is the the is − 1 2 𝑥𝑥 x by 𝑈𝑈 X 𝑢𝑢 u = 𝑥𝑥 i 𝑢𝑢 2 𝑢𝑢 . Hence, Hence, . = 𝑢𝑢 2 � | 0 same: and 𝑥𝑥 and ( ≤ 𝑢𝑢 𝑢𝑢 1 𝑥𝑥 𝑈𝑈 𝑥𝑥 = = � 𝑢𝑢 𝑢𝑢 𝑓𝑓 𝑥𝑥 𝑥𝑥 0 𝑎𝑎 , 𝑢𝑢 𝑜𝑜 𝑥𝑥 0 , , 𝑥𝑥 1 𝑥𝑥 𝑢𝑢 , 𝑏𝑏 = 𝑑𝑑 1 𝑜𝑜 𝑜𝑜 𝑜𝑜 and the probability of probability the , and 𝑥𝑥 ] ≤ ) 𝑥𝑥 ? = 2 . 𝑢𝑢 is independent and identical. and independent is 𝑑𝑑 2 + 𝑎𝑎 𝑐𝑐 𝑥𝑥 𝑥𝑥 1 must must 𝑜𝑜 + . = 𝛿𝛿 � , 0 be constant over over constant be 𝑏𝑏 1 𝑥𝑥 + 1 𝑑𝑑 𝛿𝛿 𝑥𝑥 1 ≤ = 𝑥𝑥 𝑥𝑥 𝑜𝑜 2 1 2 . � 𝑥𝑥 𝑋𝑋 𝑥𝑥 1 1 = = being 0 1

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metin /~      Exponential Random Variable CommonContinuous Random Variables Exponential distribution has memorylessproperty: Ex: Find moments of Exponential distribution The Start with the equalityabove, and first variable random uniform A subinterval of of subinterval – – – – – – – – cdf after its installation. its after If a machine is functioning 𝑎𝑎 at time functioning is machine a If for hypothesis induction the establishes line last The E for holds also It Hypothesisholds for that suppose hypothesis As induction an E E time time Here P 𝑋𝑋 𝑋𝑋 𝑋𝑋 and pdf are respectively and respectively pdf are 2 𝑘𝑘 𝑎𝑎 𝑎𝑎 + = 1 ≤ = + 𝜆𝜆 ∫ 0 𝑋𝑋 ∫ 𝑏𝑏 = = is the parameter of distribution the of parameter the is ∞ 0 [ ∞ (will (will ≤ 𝑎𝑎 0 ∫ 𝑥𝑥 0 𝑥𝑥 , ∞ + 𝜆𝜆휆휆휆 𝑎𝑎 𝑎𝑎 2 𝑥𝑥 + 𝑘𝑘 𝜆𝜆휆휆휆 + k over over fail not 𝑘𝑘 𝜆𝜆 + 𝛿𝛿 + = 1 𝛿𝛿 1 ∫ − ] 𝑘𝑘 0 2 𝜆𝜆휆휆휆 = ∞ 𝜆𝜆 of a given length length given a of , which we consider for getting intuition. The step of of step The intuition. getting for consider we which , − = 𝑥𝑥 exp 𝜆𝜆 𝑘𝑘 1 𝜆𝜆 𝑑𝑑 𝑋𝑋 𝜆𝜆휆휆휆 . 𝑑𝑑 − 𝑑𝑑 takes values between between values takes = − 𝜆𝜆 [ 𝜆𝜆 𝑎𝑎 𝜆𝜆 𝐹𝐹 = − , − 𝑋𝑋 𝑎𝑎 𝑑𝑑 𝑎𝑎 𝑥𝑥 − 𝜆𝜆 𝑥� + − 𝑥𝑥 𝜆𝜆 𝑥𝑥 = 2 𝑏𝑏 𝑏𝑏 𝑑𝑑 ( 𝛿𝛿 exp ] = has not failed over [ over failed not has − − ) E → 𝑥𝑥 = is is P 𝜆𝜆 𝐈𝐈 𝛿𝛿 𝑘𝑘 𝑋𝑋 𝜆𝜆 𝑥𝑥 . − + the the ∞ 𝑘𝑘 𝑏𝑏 ≥ 𝑘𝑘 exponentially decreases as as decreases exponentially 1 + 𝜆𝜆 | 𝜆𝜆 0 ≤ exp 0 ∞ 1 𝜆𝜆 and then 𝑎𝑎 then and [ = same same E 1 + 𝑋𝑋 | ( 0 ∞ 𝑘𝑘 𝑋𝑋 − ∫ ≤ − ! 0 𝑘𝑘 + ∞ / 𝜆𝜆 0 ) exp 𝜆𝜆 𝑏𝑏 probability probability exp ∫ 𝑘𝑘 = + 0 and and 𝑘𝑘 . ∞ P | 0 ∞ 𝛿𝛿 + 𝜆𝜆 2𝑥𝑥 𝑘𝑘 − 𝑘𝑘 − + 𝑋𝑋 + = + 0 1 ∞ 1 𝜆𝜆 𝜆𝜆 1 exp for for , . ∫ > 𝜆𝜆 𝑥𝑥 𝑎𝑎 ! 0 , and the probability of of , probability the and 0 ∞ ] : ) 𝑎𝑎 0 𝑑𝑑 ] , the probability that it will be functioning at functioning be it will that probability the , − that it functioned for the first first the for functioned it that 𝑘𝑘 P ≤ + and and 𝜆𝜆 + = 𝜆𝜆 𝑋𝑋 𝑎𝑎 𝑏𝑏 1 0 ≤ 𝑑𝑑 ≥ − 𝑥𝑥 𝑓𝑓 = 𝑏𝑏 𝑋𝑋 𝑘𝑘 𝑏𝑏 = exp 𝑎𝑎 exp and and P 𝑥𝑥 increases: 0 ( = k 𝜆𝜆 − 𝑋𝑋 + 𝛿𝛿 = 𝜆𝜆 = − exp > 𝜆𝜆 2 ≥ 𝜆𝜆 2 𝐈𝐈 ∫ | 𝜆𝜆 𝑥𝑥 0 0 ∞ 𝑎𝑎 0 can be dropped. dropped. be can ∞ ( ≥ . ) 𝑑𝑑 − 0 = 𝑥𝑥 P 𝜆𝜆휆휆휆 𝜆𝜆 𝜆𝜆 ( 𝑋𝑋 𝜆𝜆 1 exp 𝑋𝑋 being in anyin being ) > ( − − 𝑏𝑏 𝜆𝜆 𝑏𝑏 ) 𝜆𝜆 𝜆𝜆 . time units units time 𝑑𝑑 ) = 𝜆𝜆 2 2

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metin /~   ParetoRandom Variable CommonContinuous Random Variables Heavy tail in practice in tail Heavy E Find Ex: – – – – – – – – Number Number Number Phone W Heavy tail causes diverging moments. Since For For This is known as Pareto rule in Quality Management. Management. Qualityin rule Pareto as known is This ealth Two pdfs Two Exponential decay 𝛼𝛼 𝛼𝛼 ∫ = > conversation conversation 0 1 𝑋𝑋 distribution distribution of failures caused by a root aroot by caused failuresof sent tweets of 1 1 𝑥𝑥 , , Slower decay Slower E E and moments moments and 𝑥𝑥 𝑋𝑋 𝑋𝑋 𝑑𝑑 = = is finite and 𝑥𝑥 and finite is in a a in ∫ ∫ duration 𝑥𝑥 𝑥𝑥 ∞ ∞ 𝑢𝑢 𝑢𝑢 𝑥𝑥 𝑥𝑥 by an individual an by population: 𝑥𝑥 𝑥𝑥 𝛼𝛼 𝑥𝑥 1 𝛼𝛼 + 𝑢𝑢 + 𝑢𝑢 𝛼𝛼 1 1 : 𝑑𝑑 𝑑𝑑 there are plenty of phone conversations with a very long duration. a with duration. long very conversations of plenty are phone there 𝑘𝑘      = = ≥ 𝑥𝑥 − there are plenty of individuals with very large amount of of very wealth. withamount large of plenty are individuals there cause 𝑢𝑢 𝑥𝑥 exp 𝑓𝑓 The Since P with variable a random Seek tail distributionheavy yields adecay Slower P tail the in decay slower For 𝑥𝑥 𝛼𝛼 ln 𝑋𝑋 : 𝑢𝑢 for for − 𝛼𝛼 1 there are plenty of individuals sending a large number of a of number sending largetweets. of plenty are there individuals and and 𝑥𝑥 𝑥𝑥 𝑥𝑥 are plenty of root causes leading to a large number of failures. failures. of number large a to leading causes root of plenty : are there cdf 𝑥𝑥 𝑥𝑥 − 𝛼𝛼 | 𝑢𝑢 𝛼𝛼 − 𝑥𝑥 ∞ P ≥ = 𝜆𝜆 1 𝑢𝑢 𝛼𝛼 | and pdf are respectively and respectively pdf are 1 𝑋𝑋 → 𝑥𝑥 ∞ 𝐈𝐈 ≥ 𝑢𝑢 , , diverging mean implies diverging moments 𝑥𝑥 ≥ ≥ ≤ ∞ = 𝑥𝑥 0 . For For . 𝑢𝑢 . 𝑥𝑥 𝛼𝛼 𝑥𝑥 − 𝛼𝛼 𝑢𝑢 − 𝑥𝑥 𝛼𝛼 1 𝛼𝛼 𝑎𝑎 𝑥𝑥 + = 𝑢𝑢 𝛼𝛼 𝛼𝛼 𝑢𝑢 o ufcety large sufficientlyfor 1 < 1 , 1 set set , E P 𝑋𝑋 𝑋𝑋 ( = 𝑋𝑋 ( ≥ 𝑋𝑋 𝛼𝛼 ≥ 𝐹𝐹 ∫ 𝑥𝑥 ≥ 𝑋𝑋 𝑥𝑥 ∞ 𝑥𝑥 𝑢𝑢 𝑥𝑥 𝑥𝑥 𝑥𝑥 ) = , note note , ) 𝑥𝑥 𝑥𝑥 𝛼𝛼 = proportional to 𝑥𝑥 ( 𝑢𝑢 𝛼𝛼 + 𝑥𝑥 1 . 𝑢𝑢 𝐈𝐈 𝑑𝑑 𝑥𝑥 / exp ≥ 𝑥𝑥 𝑥𝑥 > ) 𝑢𝑢 𝛼𝛼 𝛼𝛼 [ 𝜆𝜆 1 for for ∫ 𝑥𝑥 ∞ − 𝑢𝑢 𝑥𝑥 𝑥𝑥 ( ≥ 𝑥𝑥 𝑥𝑥 𝑥𝑥 ≥ 1 𝑥𝑥 + 𝑢𝑢 𝑢𝑢 𝑥𝑥 1 − / 𝑥𝑥 𝑎𝑎 𝑑𝑑 𝛼𝛼 𝑥𝑥 𝑢𝑢 or or ) . for for 𝛼𝛼 → ]

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metin /~   the limit witha series diverging to . .   . .       Gamma and Beta Functions Beta and Gamma Normalization Functions: 𝐵𝐵 function Beta Ex: Ex: ∫ Ex: Ex: Γ Then function Gamma integer nonnegative For Γ 0 ∞ – 𝑛𝑛 2 7 𝑘𝑘 𝑥𝑥 + Γ Γ Γ Γ , − This factorial definition does not apply to real numbers or negative integers. Let us extend the definition. the extend us Let integers. negative or numbers real to apply not does definition factorial This 𝑚𝑚 = 1 1 / 0 1 2 5 2 7 2 1 2 Γ = = = = exp = = = = = 2 5 ∫ 2 𝑛𝑛 0 Γ 0 Γ ∞ ∞ ∫ ∫ 15 ∫ − − 0 𝑘𝑘 0 = 0 ∞ ∞ 𝑘𝑘 ∞ 𝑥𝑥 𝐵𝐵 𝜋𝜋 , by lower bounding bounding bylower , 8 ∫ 𝑥𝑥 + Γ 2 5 𝑛𝑛 0 exp exp , because , 𝑥𝑥 𝜋𝜋 ∞ ( 𝑚𝑚 2 3 𝑑𝑑 𝑚𝑚 exp 𝑚𝑚 𝑛𝑛 for any real number number real any for because 𝑛𝑛 Γ − ) , 𝑥𝑥 1 = 𝑘𝑘 − bandi h oe sn hne fvrals variables. of change a using notes the in obtained 2 3 𝑛𝑛 − exp − − 𝑢𝑢 ∫ 𝑥𝑥 1 2 0 𝑥𝑥 = = ∞ 𝑛𝑛 2 5 𝑑𝑑 𝑑𝑑 𝑑𝑑 − exp − ∫ , 𝑑𝑑 2 3 0 𝑥𝑥 exp 1 𝑛𝑛 2 1 = = ! 𝑥𝑥 𝑑𝑑 Γ − = 𝑑𝑑 𝑚𝑚 ∞ 1 − 𝑢𝑢 2 1 𝜋𝜋 = − 2 𝑛𝑛 𝑥𝑥 . 1 𝑛𝑛 = 2𝑢𝑢 𝑢𝑢 𝑛𝑛 Γ 1 15 𝑑𝑑 Γ 𝑑𝑑 − − 𝑛𝑛 8 𝑛𝑛 𝑛𝑛 𝜋𝜋 , 1 𝑥𝑥 = 𝑘𝑘 𝑛𝑛 − � 1 − 0 𝑑𝑑 ∞ 𝑑𝑑 2 𝑥𝑥 𝑛𝑛 = … − ∫ 1 0 1 exp 2 𝑥𝑥 𝑘𝑘 1 − and and − 1 𝑥𝑥 1 0 − 𝑑𝑑 ! 𝑑𝑑 𝑥𝑥 = 𝑚𝑚 1 . − 1 𝑑𝑑 𝑑𝑑 = 𝐵𝐵 ( 𝑘𝑘 , 𝑚𝑚 ) .

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metin /~      Erlang CommonContinuous Random Variables The random variable: random integer For Ex: UTD's Sum of 𝛼𝛼 independent two up Add Ex: An the second floor had 4 routers and running the 5th now. now. 5th the running and 4routers had hand, floor other On the second the now. 8th the running and routers 7 failed had floor first the that told are We immediately. replaced – – – – – – – Erlang = = pdf F 𝑓𝑓 P variables: E(Gamma( Since P(Gamma(8, SOM building is is building If the SOM Suppose all routers 𝑋𝑋 or or − � 𝑋𝑋 0 𝑥𝑥 exp the the 𝑥𝑥 of ≤ ( SOM has routers at every at routers has SOM many independent many independent 𝛼𝛼 1 𝛼𝛼 = rv 𝑥𝑥 Gamma , 2 − is integer, we can applycan we integer, is − find find nd 𝑑𝑑 𝑑𝑑 is the sum of an integer number of exponential random variables. variables. random exponential of number integer an of sum the is and Gammaand Random Variables 𝜆𝜆 exp = floor, P(Gamma(5, floor, 𝜆𝜆 P E(Expo 𝜆𝜆 P E(Gamma ) 𝑋𝑋 + ≥ − 𝑋𝑋 ≤ 10) 1 1 𝜆𝜆 𝛼𝛼 ( + − 𝑥𝑥 ( 𝑥𝑥 , are are 𝜆𝜆 𝜆𝜆 represents the probability the represents 𝑋𝑋 𝜆𝜆 𝛼𝛼 − ))=1/ = 𝜆𝜆 2 , 8 years old, old, 8 years : identical with Expo with identical ( 𝜆𝜆 휆휆휆 𝑢𝑢 𝛼𝛼 ≤ 𝜆𝜆휆휆휆 ))= ) , 𝑓𝑓 𝜆𝜆 𝜆𝜆 𝑥𝑥 ) 𝐸𝐸 𝑋𝑋 ( 𝑋𝑋 )) 𝛼𝛼 𝜆𝜆휆휆휆 and and − 𝐸𝐸𝐸𝐸𝐸𝐸 1 / 𝜆𝜆 = 𝑥𝑥 and and 𝜆𝜆 𝜆𝜆 , − ) floor. floor. 𝑋𝑋 𝜆𝜆 ≥ the formulas for the expected value and variance of sum of independent random random independent of sum of variance and value expected the for formulas the ∫ and and E(Expo 𝜆𝜆 ) 0 2 = 10) and P(Gamma(5, and 10) ( 𝑥𝑥 V(Gamma − 𝜆𝜆 then then P ∼ 𝜆𝜆 𝜆𝜆 ) V(Gamma − 𝑢𝑢 𝜆� 𝑋𝑋 Since the the Since 𝐸𝐸 a itiuinprmtrzd by parameterized has a distribution 𝜆𝜆휆휆휆 2 revise probability probability revise 𝐸𝐸𝐸𝐸𝐸𝐸 𝑑𝑑 𝜆𝜆 ≤ − ( Γ 𝜆𝜆 2 𝑥𝑥 𝜆𝜆 = ( )=2 ( ( ) − 𝛼𝛼 𝛼𝛼 − 𝜆𝜆 ieie with lifetime parameter that the � ( ) , 𝜆𝜆 ) 0 𝜆𝜆 𝜆𝜆 𝛼𝛼 service is provided is without anservice 𝑢𝑢 / 𝜆𝜆 𝑥𝑥 𝜆𝜆 )). We We )). , what is thewhat, , 𝜆𝜆 2 𝜆𝜆 𝜆𝜆휆휆휆 exp , 𝛼𝛼 + ))= − which imply imply which 1 𝜆𝜆 routers routers 𝜆𝜆 𝛼𝛼 2 − ) can start with the first two moments of Exponential two of moments withfirst the start can exp / ≥ − for for 𝜆𝜆 𝜆𝜆 as P(Gamma(8, as 10| 𝜆𝜆 𝜆𝜆 2 𝑢𝑢 . 𝑥𝑥 − 𝑑𝑑 on the first first the on Gamma(5, 𝑑𝑑 𝜆𝜆 ≥ cdf − V(Expo 0 � = of of 0 and and 𝑥𝑥 𝜆𝜆휆휆휆 𝜆𝜆휆휆휆 𝜆𝜆 𝑋𝑋 measured in 1/year. 1/year. in measured 𝜆𝜆 floor last last floor ( 𝜆𝜆 𝛼𝛼 = ) 𝜆𝜆 ) ≥ ))=1 ≥ > − − 8). 8). 𝑋𝑋 10|Gamma(8 𝜆𝜆 1 𝜆𝜆 interruption, interruption, 0 / 𝑥𝑥 𝛼𝛼 + 𝜆𝜆 2 , 𝑑𝑑 𝜆𝜆 . for 10 years. 10 years. for 𝜆𝜆 𝑋𝑋 𝜆𝜆 2 . ? . , 𝜆𝜆 a failed router is is router a failed 𝑥𝑥 ) ≥ P 𝑥𝑥 8). 8). 2 0 𝜆𝜆휆휆휆 𝑋𝑋 2 ≤ 𝑥𝑥 ≤ 1 𝑢𝑢 + 𝑥𝑥 − ≤ − 𝜆𝜆 𝑥𝑥 𝑥𝑥 2 𝑢𝑢 ≤

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metin /~     BetaRandom Variable CommonContinuous Random Variables In general, Beta random variable variable random Beta general, In 𝑋𝑋 Suppose Consider – – – – – – – is the posterior 𝑓𝑓 𝑋𝑋 𝛼𝛼 Prior: Observation: To simplify further suppose 𝑓𝑓 For P Replace P 𝑝𝑝 − 𝑃𝑃 where where = 𝑋𝑋 1 𝑛𝑛 constant but but trialsBernoulli with constant = Uniform distribution over [0,1] over distribution Uniform ≡ ∫ successes P trials yield 0 1 𝑝𝑝 𝑓𝑓 𝑓𝑓 Bin 𝑃𝑃 𝑃𝑃 𝑃𝑃 𝑓𝑓 Bin ( 𝑃𝑃 Bin 𝑃𝑃 𝑢𝑢 𝑝𝑝 𝑚𝑚 is the pdf of 𝑃𝑃 of pdf the is n probability of success success of probability = P P , 𝑛𝑛 successes out of 𝑃𝑃 𝑛𝑛 & , Bin Bin 𝑢𝑢 , 𝑃𝑃 𝑃𝑃 𝜆𝜆 = , 𝑚𝑚 Bin − 𝑓𝑓 𝑋𝑋 m = 𝑛𝑛 𝑛𝑛 = successes. Given Given successes. 1 , , 𝑝𝑝 𝑢𝑢 𝑝𝑝 𝑚𝑚 ( 𝑚𝑚 failures = 𝑛𝑛 = = , = 0 with a continuous support and support continuous a with 𝑃𝑃 𝑃𝑃 < � = ∫ 𝑚𝑚 . 𝑚𝑚 𝑢𝑢 ) 0 is uniform over over uniform is Youcan skipthe middle step by conditioning. 1 < 𝑓𝑓 = P 𝑛𝑛 1 𝑢𝑢 𝑝𝑝 𝑑𝑑 𝑋𝑋 P 𝑚𝑚 𝑚𝑚 𝑃𝑃 P 𝑚𝑚 𝑥𝑥 𝑃𝑃 = = Bin 1 1 𝑝𝑝 ) = , − − Bin ∫ by by = given 𝑢𝑢 0 1 n , 𝑝𝑝 𝑢𝑢 Bin 𝑓𝑓 𝑥𝑥 𝑓𝑓 , 𝑃𝑃 𝑓𝑓 unknown success success unknown 𝑃𝑃 𝑚𝑚 𝑃𝑃 𝑛𝑛 𝛼𝛼 𝑛𝑛 𝑛𝑛 𝑃𝑃 , − − − = 𝑃𝑃 𝑢𝑢 ( 𝑝𝑝 𝑛𝑛 1 B 𝑚𝑚 𝑚𝑚 the the update successes, m [ 𝑢𝑢 , = 0 ( 𝑃𝑃 𝐶𝐶 1 𝑑𝑑 𝐶𝐶 ) 𝛼𝛼 , 𝑚𝑚 − 𝑚𝑚 1 𝑚𝑚 P 𝑛𝑛 𝑛𝑛 , 𝜆𝜆 = ] 𝑥𝑥 ( 𝑢𝑢 𝑝𝑝 ) , uninformative prior, prior, uninformative , = Bin = 𝑚𝑚 𝑚𝑚 𝜆𝜆 𝑚𝑚 − 𝐵𝐵 1 ∑ 1 1 ( ≈ ( 0 𝑚𝑚 − − 𝑝𝑝 𝑛𝑛 for for < � 𝑓𝑓 𝑚𝑚 0 , 𝑢𝑢 P 𝑃𝑃 + 𝑢𝑢 𝑝𝑝 1 𝑢𝑢 < 𝑓𝑓 prior 𝑃𝑃 𝑃𝑃 0 1 1 ) 1 𝑛𝑛 𝑛𝑛 = P probability , − − 𝑢𝑢 − = ≤ 𝑛𝑛 𝑝𝑝 𝑚𝑚 𝑚𝑚 𝑃𝑃 , P Bin 𝑝𝑝 − = 𝑑𝑑 𝑚𝑚 𝑥𝑥 Bin 𝑢𝑢 𝑚𝑚 𝑛𝑛 ) ≤ , − Bin = 𝑛𝑛 to to leads which distribution of 𝑚𝑚 + , 𝑛𝑛 𝑃𝑃 1 ∫ , 0 1 𝑢𝑢 n = 1 and and ) , 𝑓𝑓 𝑚𝑚 𝑓𝑓 𝑃𝑃 = 𝑃𝑃 𝑃𝑃 𝑃𝑃 = 𝑚𝑚 𝑢𝑢 is. is. 𝑝𝑝 𝛼𝛼 m , 𝑑𝑑 𝑢𝑢 𝑝𝑝 𝜆𝜆 𝑑𝑑 𝑚𝑚 𝑚𝑚 if if > 𝑃𝑃 1 1 0 − − 𝑃𝑃 has a discrete support adiscrete has , 𝑝𝑝 𝑢𝑢 we are seeking seeking are we 𝑛𝑛 𝑛𝑛 − − 𝑚𝑚 𝑚𝑚 𝑑𝑑

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metin /~     Expected Value of Beta Random Variable Variable Random Beta ofValue Expected Question: Question: Thecorresponding expectedvalueprobability of success of is with variable random Beta the of pdf The Ex: – E Answer: Prior distribution plays a role. Consider two special cases: special two Consider role. a plays distribution Prior Answer: 𝑋𝑋 » » Why = 0.5 Infinitely many trials. trials. many Infinitely 𝛼𝛼 i.e., a trial,before success of probability of value expected the is What No trials: = ∫     0 does the expected value expected the does 1 E 𝑛𝑛 to converge to this constant: constant: this to converge to You can say that the effect of prior prior of effect the that can say You parameter true the to converges probability success of value expected the Hence lim the into this Inserting T 𝑥𝑥 ( → he success probability is a unknown constant constant unknown a is probability success he 𝑈𝑈 𝑥𝑥 ∞ 𝛼𝛼 Number − 0 , 1 𝐵𝐵 1 Number ( 1 𝛼𝛼 ) − 𝛼𝛼 , 𝛼𝛼 𝜆𝜆 from the prior, and this is what you get from the formula the from get you what is this and prior, the from 𝑥𝑥 ) + 𝛼𝛼 + 𝛼𝛼 Number 𝜆𝜆 𝜆𝜆 − of 𝜆𝜆 1 = What is the expected value of probability of success success of probability of value expected the is What of = successes 𝑑𝑑 𝑑𝑑 Number trials Number = Number Number formula, of differ 𝑓𝑓 𝐵𝐵 𝑋𝑋 𝛼𝛼 + successes 𝛼𝛼 1 − 𝑥𝑥 2 + , 𝜆𝜆 of of 𝑛𝑛 from the ratio of of the the ratiofrom 1 lim 1 = → of 𝐵𝐵 successes of successes successes ∞ = � trials 0 trials 𝛼𝛼 P 𝑛𝑛 1 lim 𝑈𝑈 → ≈ 𝑥𝑥 + and Interpretation and ∞ Bin 𝑥𝑥 𝑝𝑝 0 1 + 𝑝𝑝 𝛼𝛼 + 𝑢𝑢 , Number 1 , − 𝑛𝑛 Number & 2 + 𝑛𝑛 𝜆𝜆 𝑛𝑛 1 + 2 𝐵𝐵 , 𝑝𝑝 aihsatrifnt trials. infinite after vanishes 𝜆𝜆 1 ( 1 1 𝑢𝑢 = 𝛼𝛼 as − = − ≠ − , 𝑝𝑝 𝜆𝜆 Γ 1 𝑛𝑛 0 𝑥𝑥 0 𝑢𝑢 Γ Number ) 𝑝𝑝 numberof successes of , the ratio of of ratio the , 𝛼𝛼 failures failures + → + of 𝑢𝑢 𝛼𝛼 𝜆𝜆 Number − trials + Γ 2 1 ∞ 1 trials ≤ ( 𝜆𝜆 𝜆𝜆 𝑑𝑑 = ) 𝜖𝜖 Γ 2 1 Γ is is + of = 𝛼𝛼 + 𝛼𝛼 + 2 of successes + 1 1 1 𝜆𝜆 number of successes of number trials for for = + Γ 1 𝜆𝜆 𝑛𝑛 lim 𝜖𝜖 → after after > ∞ = to the to 𝑝𝑝 0 𝛼𝛼 𝑛𝑛 𝑢𝑢 . So So . 𝛼𝛼 + 𝛼𝛼 𝑛𝑛 + 𝜆𝜆 + → , ratio of successes of ratio , = 2 number of trialsof number 1 ∞ 𝜆𝜆 = 𝑝𝑝 , = 𝑢𝑢 𝜆𝜆 𝑝𝑝 1 to to → fe niie trials. infinite after 𝑢𝑢 ? number of trialsof number ∞ You expect expect You ? ?

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metin /~      Normal Random Variable CommonContinuous Random Variables Ex: Ex: The variable random normal the for range The Normal densityis symmetric around its mean (mode) andhas light tails.    The density integrates to 1. integrates to Thedensity cdf Finally, Finally, Also note First E . to polarcoordinates transformation from 𝑋𝑋 can be computed only numerically. numerically. only can be computed ∫ The last equality is is equality last The − ∞ ∞ = ∫ 0 exp 2𝜋𝜋 ∫ − ∫ ∞ ∞ ∫ − = ∞ 0 ∞ ∞ − 2𝜋𝜋 𝜇𝜇 𝑟𝑟𝑟� 1 𝑥𝑥 𝑧𝑧 � 2 1 2 𝜎𝜎 − ∞ exp ∞ 2𝜋𝜋 𝑑𝑑 1 𝑧𝑧 − 𝜎𝜎 1 2𝜋𝜋 1 𝑟𝑟 ∫ 2 − exp 2 − ∞ ∞ exp 𝑥𝑥 𝑑𝑑 𝑓𝑓 2 − 𝑋𝑋 exp 𝜎𝜎 𝜇𝜇 𝑑𝑑𝑑𝑑 =1 2 − 𝑥𝑥 2 𝜃𝜃 𝑟𝑟 − = − = 𝑥𝑥 𝑑𝑑 2 𝑧𝑧 𝑧𝑧 − 2 𝜎𝜎 𝑧𝑧 ∫ 1 2 2 𝜇𝜇 0 2 2 2 2𝜋𝜋 = 2𝜋𝜋 𝑧𝑧 2 1 2 𝑑𝑑 𝑑𝑑 𝑑𝑑𝑑𝑑 𝑧𝑧 𝑧𝑧 𝜎𝜎 2𝜋𝜋 1 𝑁𝑁 𝑑𝑑 2 + exp 𝑑𝑑 ∫ = ∫ 𝜃𝜃 0 𝑟𝑟 𝜎𝜎 ∞ 𝑁𝑁 − ∞ = 2 𝑁𝑁𝑁𝑁 ∞ ∫ = � 𝑟𝑟𝑟� − ∞ − = exp ∫ − ∞ ∞ arccos ∞ − 𝑁𝑁 ∞ ∫ 𝑧𝑧 ( ∞ − 𝑥𝑥 1 ∞ 𝜇𝜇 2 ∞ 2 2𝜋𝜋 ( − − 𝑧𝑧 − , + 𝜎𝜎 𝜇𝜇 𝜎𝜎 exp 𝑟𝑟 𝑧𝑧 2 𝜇𝜇 2 2 2 2 2 𝑧𝑧 exp + ) 𝑧𝑧 2 2 2 𝑟𝑟 𝑑𝑑 1 𝑑𝑑𝑑𝑑 is all real numbersandthedensity is 𝑧𝑧𝑧𝑧 − = = 𝑧𝑧 − 1 2 ) with 0 integrating + 2 2𝜋𝜋 𝑧𝑧 2 𝑧𝑧 2 2𝜋𝜋 2 2 1 ∫ 0 𝑑𝑑 𝑑𝑑 𝑧𝑧 ∞ exp 𝑧𝑧 𝑧𝑧 exp = 2 = 𝑑𝑑 𝑥𝑥 𝑧𝑧 𝜇𝜇 − 1 𝜎𝜎 − − 𝑔𝑔 𝜇𝜇 = 𝑢𝑢 𝑧𝑧 𝑧𝑧 2 ∫ 2 𝑑𝑑 0 2𝜋𝜋 = 𝑑𝑑 ∫ = 𝑧𝑧 − 0 ∞ 𝑔𝑔 2 𝑟𝑟𝑟� 𝜋𝜋 − . 𝑧𝑧 over over − 𝑟𝑟 2 2 ( 𝑑𝑑 − 𝑎𝑎 𝑑𝑑𝑑𝑑 , 𝑎𝑎 )

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metin /~   Normal Random Variable CommonContinuous Random Variables Ex: Ex:   What What is the distribution of We obtain the density for for the density We obtain P Partial expected expected for𝑋𝑋 value Partial 𝑑𝑑 . E 𝑎𝑎 𝑑𝑑 𝑢𝑢 𝑋𝑋 Theequalitylast is from differentiating an integral,akaLeibniz rule. P + I 𝑋𝑋 𝑏𝑏 ≤ 𝑋𝑋 𝑎𝑎 ≤ ≤ = = = = 𝑢𝑢 𝑢𝑢 𝜇𝜇 𝜇𝜇 𝜇𝜇 � 𝑎𝑎 − − = 𝐹𝐹 𝐹𝐹 𝐹𝐹 𝑏𝑏 𝑎𝑎 ∞ 𝑁𝑁 𝑁𝑁 𝑁𝑁 P 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑥𝑥 = 𝑋𝑋 2𝜋𝜋 𝑑𝑑 𝑑𝑑 1 𝑁𝑁 𝑁𝑁 𝑁𝑁 ≤ 𝑢𝑢 0 0 0 𝑁𝑁 ∫ 𝜎𝜎 , , , 𝑢𝑢 1 1 1 − 𝑎𝑎 − 𝑢𝑢 exp ∞ 𝑁𝑁 𝑎𝑎 − 𝑏𝑏 𝑎𝑎 𝑁𝑁𝑁𝑁 𝑏𝑏 𝑎𝑎 𝑎𝑎 𝑎𝑎 and differentiate with respect to to respect with differentiate and ∼ − − − 2𝜋𝜋 𝑁𝑁 + − 1 𝑁𝑁 𝜎𝜎 𝜇𝜇 𝜎𝜎 𝜇𝜇 𝜎𝜎 𝜇𝜇 𝑎𝑎 𝑏𝑏 𝜎𝜎 𝑁𝑁 𝑥𝑥 exp − − + if if 𝑁𝑁 2 + − 𝑁𝑁𝑁𝑁 𝜎𝜎 𝜎𝜎 𝜎𝜎 𝜎𝜎 𝑋𝑋 𝑏𝑏 𝜇𝜇 2 𝑓𝑓 , ∫ 𝑁𝑁 𝑎𝑎 ∞ is is 𝑁𝑁 2𝜋𝜋 1 2 − 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑎𝑎 2 𝜇𝜇 𝜎𝜎 − 𝑁𝑁 𝜎𝜎 exp 𝜇𝜇 𝑑𝑑 , 𝑥𝑥 2 𝑁𝑁 2 𝑑𝑑 𝜎𝜎 − 2 𝑁𝑁 𝜎𝜎 𝑁𝑁 . / 𝜇𝜇 2 = 0 2 𝑁𝑁𝑁𝑁 2 − , 1 2 � up to 𝑎𝑎 up to 2𝜋𝜋 𝑢𝑢 1 − 𝑁𝑁 𝑎𝑎 𝑑𝑑 ∞ 𝑎𝑎 | 𝜎𝜎 − exp ∞ 𝜎𝜎 − 𝜇𝜇 𝑑𝑑 𝑎𝑎 𝜇𝜇 𝜇𝜇 , 𝜎𝜎 − ( = 𝜎𝜎 𝜇𝜇 𝜇𝜇 − . 2 2 + / 𝑢𝑢 2𝜋𝜋 2 . 𝜎𝜎 𝑢𝑢 1 𝑧𝑧 𝑑𝑑 We can considerWe can 𝑎𝑎 ) 𝑢𝑢 𝑎𝑎 exp with 2𝜋𝜋 1 exp 𝑢𝑢 − = 𝑢𝑢 𝑧𝑧 − 2 2 − 2 𝑧𝑧 𝑎𝑎 𝑏𝑏 2 2 2 − 𝜎𝜎 𝑎𝑎 2 𝑑𝑑 𝑎𝑎 𝑧𝑧 2

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metin /~ Log CommonContinuous Random Variables In particular, In particular, Finally,     Ex: Find the moment moment the Find Ex: etc. demand, of price, pdf the Find modelling Ex: for advantage an is This reals. nonnegative is lognormal for range The 𝑋𝑋 E is lognormal random variable if if variable random is lognormal n 𝑋𝑋 E ormal RandomVariable 𝑘𝑘 𝑋𝑋 = = 𝑘𝑘 = E V exp exp E = 𝑋𝑋 𝑋𝑋 exp exp = 𝑘𝑘 𝑘𝑘 = 𝑘𝑘 𝜇𝜇 P exp exp 𝑘𝑘 𝑘𝑘 exp 𝑋𝑋 ∫ 𝑋𝑋 𝑘𝑘 𝑘𝑘 − 𝑓𝑓 ∞ ( E ≤ ∞ 𝑋𝑋 𝜇𝜇 + 2𝜇𝜇 ( 𝑋𝑋 exp = + 𝑥𝑥 𝑥𝑥 𝑘𝑘 2 1 2 𝑘𝑘 + 2 𝜎𝜎 𝜎𝜎 ) ∫ = 𝑘𝑘 = − . 2 ∞ 2 2 𝑘𝑘 𝑘𝑘 ∞ / 𝜎𝜎 𝑑𝑑 P 𝑢𝑢 2 ∫ 𝑑𝑑 exp 𝑑𝑑 2 2 − ) ∞ 𝑌𝑌 ∞ P and and 𝑋𝑋 − 2𝜋𝜋 ≤ 1 𝑋𝑋 𝑘𝑘 2𝜋𝜋 = 𝜎𝜎 1 𝑦𝑦 ln E ≤ exp exp 𝜎𝜎 exp 𝑥𝑥 𝑋𝑋 exp 𝑥𝑥 2𝜋𝜋 2 1 ( = = 𝜇𝜇 𝑌𝑌 − 𝜎𝜎 = ) � + exp − 2 𝑥𝑥 1 − 𝑢𝑢 =1 for for 𝜎𝜎 exp ln ∞ 2 𝜎𝜎 ( 2 2 𝑢𝑢 𝑥𝑥 2 2𝜋𝜋 − 1 𝑌𝑌 2 𝑑𝑑 − 𝑘𝑘 𝜎𝜎 2 𝑢𝑢 𝜎𝜎 2𝜋𝜋 ∼ 2 𝜎𝜎 𝜇𝜇 2 1 2 𝑦𝑦 2 ) exp − 𝑁𝑁 = 2 + 𝜎𝜎 with 𝜎𝜎 𝜇𝜇 2 2 𝑑𝑑 exp exp 𝑁𝑁 2 𝜎𝜎 𝑑𝑑 𝑁𝑁𝑁𝑁 − 𝑢𝑢 2 𝑑𝑑 = 𝑑𝑑 2𝜇𝜇 𝑁𝑁 = − ln ( exp & 𝜇𝜇 𝑦𝑦 + , 𝑥𝑥 2 𝑦𝑦 𝜎𝜎 − 𝜎𝜎 − 2 𝜎𝜎 2 𝑘𝑘 − 2 𝜎𝜎 𝜇𝜇 ) 2 𝑘𝑘 𝜇𝜇 2 𝜇𝜇 exp 2 exp 2 𝑑𝑑 𝜎𝜎 𝑘𝑘 𝑑𝑑 2 2 2 𝜎𝜎 2 − 1

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metin /~ Multiplicative Evolution Forward p ..... (t price p p p p p -1, ( ( ( ( p p p ( T T t t T ( ( ( + − T T t t − − − , , ) 1 1 , T T 1 , 1 2 , T , T T , ) ) , T ) T Lognormal Prices in Every Period in Every Prices Lognormal T Application: Price Evolution with ) ) t ) -1 ) ) = = = = ε ε ε ε T t − 1 − 1 2 t , T = : ε T − ( t − Forward 1 ) p price (t , T ) from now from Looking Looking at future future at ln

p 𝑝𝑝 t ( T 𝑇𝑇 , , T 𝑇𝑇

) = = = = = ε ln .... ε ε 1 1 1 ε ε p 𝑝𝑝 2 2 ( .... 𝑡𝑡 T p , ( 𝑇𝑇 − ε T T 1 + − − , t T 𝑇𝑇 � 2 𝑖𝑖 p = − ) , ( 1 T 𝑡𝑡 t ln , ) T 𝜖𝜖 ) 𝑖𝑖 Price Price Exponential Each Each At Price Price Let Let 𝑡𝑡 𝑝𝑝 , i ( n discrete time discrete n known known ln 𝑇𝑇 𝑃𝑃 , { 𝑇𝑇 𝜖𝜖 𝑡𝑡 𝑃𝑃 Brownian Motion 𝑖𝑖 ) T 1 = is Normal is , is lognormal is 𝑃𝑃 𝑝𝑝 ln 2 ( , p 𝑇𝑇 price Spot … 𝑝𝑝 (T , 𝑇𝑇 } 𝑡𝑡 , , ) T is 𝑇𝑇 )

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metin /~      

p p ( Sep (Aug,Dec for November for prices Forward 4 normal distributions fit to 4 logarithms of epsilons. Good fit ⇒ Good epsilons. of logarithms 4 to distributionsfit normal 4 E Future December for prices Forward Forward prices for from Decembermonths Aug  , 𝑆𝑆

Dec 𝑡𝑡 Aug 𝑋𝑋 ε =   𝑡𝑡 (forward) price model 0 from price(forward) ) 4 )=$100 = 𝑠𝑠 = = Forward Price EvolutionExample 0 ln ε ln 95 E ln ln 𝜖𝜖 4 1 ε ε p 𝜖𝜖 𝑋𝑋 𝑋𝑋 1 𝑋𝑋 / 1 2 ( 100 𝑡𝑡 𝑡𝑡 = 𝑡𝑡 = Aug … is normal with parameters parameters with normal is = ln( ln( = 𝜖𝜖 ∑ 𝑡𝑡 112 p , 105 𝑠𝑠 Dec 𝑖𝑖 𝑡𝑡 and and is lognormal each epsilon where (Sep,Dec 0 = exp 1 / / ln 104 ) 101 Sep 𝜖𝜖

𝑡𝑡 p 𝑖𝑖 ); ); , sum of ( )=$95 from monthsAug

from monthsJul

Oct ln ln + ε ε 𝑡𝑡 2 , 2

𝜎𝜎 2 Dec 2 = = to to ln( ln( 𝑡𝑡 ε & ) 3 identical normal 𝑡𝑡 104 101 = = : p 𝑉𝑉 𝑝𝑝 ε 92 (Oct,Dec 𝑡𝑡 3 / / 𝑡𝑡 𝑆𝑆 p 90 92 / 𝑡𝑡 ( -Nov and and 95 Sep = -Dec: : -Dec Oct ); ); =

ln ln 𝑡𝑡 𝐿𝐿 , 𝑠𝑠 , )=$92 𝜎𝜎 ε Dec ε 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 0 2 using similar processdata not shown explicitly 3 3 2 V = = if each if ) 𝑋𝑋 rvs

p ln( ln( 𝑡𝑡 ( so is their multiplication T 90 92 is also normal also is = = /

/ ln p

𝑠𝑠 Dec 93 95 0 𝑟𝑟 p ( 2 Nov 𝑟𝑟 𝜖𝜖 (Nov,Dec ( Lognormal prices Lognormal ); ); 𝑖𝑖 exp , 𝑝𝑝

is is ln T ln (

, 0 ε 𝑁𝑁 ε = Dec Nov 𝑡𝑡 2𝑡𝑡 ) 4 4 Dec = = 𝑁𝑁 ε or 𝑆𝑆 or ε 𝑁𝑁𝑁𝑁 2 ) )=$104 1 ln( ln( + = = = ) 93 𝑁𝑁 95 104 ε = 112 𝑡𝑡 2𝑡𝑡 2 = 𝜇𝜇 ε 𝜎𝜎 p / / 1 , 100 98 ( 𝜎𝜎 2 / p 𝑋𝑋 / Oct 104 92 . 𝑋𝑋 ( 2 𝑡𝑡 ). T 𝑠𝑠 − 𝑡𝑡 ). 0 − , p exp Dec 1 )=$112(Dec,Dec = ) Nov 𝑡𝑡 2𝑡𝑡 Dec , + T 𝑡𝑡 = : 𝜎𝜎

. Dec 2

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) metin /~ )      Mixtures of Random variables Random of Mixtures Mixing allows us to create new create to us allows Mixing have we Then Given Ex: Consider the mixture of a mass of Random Each Each – – If Product rv rvs you want only the new customer demand for products, you should consider consider you should products, for demand customer new the only want you = via max/min max/min via others from derived variables 𝑋𝑋 𝜆𝜆 , 𝐷𝐷 returns due to unmet expectations to a retailer can be treated as negative demand, then the demand 𝐷𝐷 demand the then demand, negativeas be treated can aretailer to expectations unmet to due returns 𝑌𝑌 [A discrete 𝐹𝐹 , the mixture the, 𝑍𝑍 𝑧𝑧 = 𝑝𝑝 𝐹𝐹 𝐹𝐹 rv 𝑋𝑋 𝑍𝑍 ] 𝑧𝑧 𝑧𝑧 𝑍𝑍 + can be obtained as can be obtained + 𝑍𝑍 𝜆𝜆 rvs = 𝑆𝑆𝑆𝑆 = Random variable decomposition: variable Random 1 and obtain richer examples. richer obtain and 𝑞𝑞 [A [A − 𝑌𝑌 + or for 𝑞𝑞 singularly continuous 𝑝𝑝 𝑋𝑋 with at zero with with zero at 1 𝜆𝜆 𝐹𝐹 w 𝐷𝐷 − 𝑌𝑌 ith ( + probability 𝑧𝑧 𝑞𝑞 ) 𝜆𝜆 probability . 𝑆𝑆𝑆𝑆 1 𝑞𝑞 − + prtr a aemse tteedo hi ags ranges. their of end the at masses have can operators if 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 exp 𝜆𝜆 𝐴𝐴 𝑧𝑧 = 1 = − 0 − 𝜆𝜆 𝑝𝑝 𝜆𝜆 1 rv 𝑝𝑝 . random variable whose whose variable random ] + if 𝜆𝜆 𝑧𝑧 𝐴𝐴 > [An absolutelycontinuous 0 max . { 0 , 𝐷𝐷 } which has a mass at 0. 0. at mass a has which cdf is rv ]

∈

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metin /~ .      Moment Generating FunctionsMoment Recovering moments from a given given a from moments Recovering Since function generating is moment Afunction such A function that yieldsall of the momentsuniquely identifies the associatedrandom variable 𝑋𝑋 variable Random exp 𝑑𝑑 𝑑𝑑 𝑡𝑡 𝑘𝑘 𝑘𝑘 𝑡𝑡 𝑚𝑚 𝑋𝑋 = 𝑚𝑚 𝑡𝑡 1 𝑚𝑚 𝑋𝑋 = + can be described uniquely by all of its moments: moments: its of all by uniquely described be can 𝑋𝑋 𝑡𝑡 ( � 𝑖𝑖 𝑡𝑡 𝑡𝑡 = ∞ ) = 0 + = E 𝑥𝑥 � = 𝑡𝑡 ∑ ∞ 𝑋𝑋 2 − 𝑡𝑡 − ∞ ! 𝑖𝑖 ∞ 2 ∞ 𝑖𝑖 + 1 ! exp 1 𝑑𝑑 𝑑𝑑 = mgf 𝑡𝑡 + 𝑡𝑡 ⇒ 3 𝑡𝑡 𝑘𝑘 𝑘𝑘 ! 𝑖𝑖 𝑡𝑡 𝑡𝑡 = 3 � 𝑡𝑡 ∞ E − + 𝑖𝑖 𝑚𝑚 ∞ + = = 𝑝𝑝 𝑋𝑋 𝑋𝑋 ⋯ 𝑋𝑋 𝑡𝑡 𝑖𝑖 ( 𝑘𝑘 E E ! 𝑖𝑖 ( 𝑡𝑡 𝑡𝑡 + 2 E 𝑥𝑥 ) 𝑋𝑋 ! 𝑋𝑋 : ) = 𝑚𝑚 𝑋𝑋 2 𝑡𝑡 𝑘𝑘 𝑘𝑘 𝑖𝑖 𝑡𝑡 and and 𝑑𝑑 ! + i 𝑑𝑑 𝑋𝑋 𝑖𝑖 𝑡𝑡 + + 𝑘𝑘 + 𝑘𝑘 𝑡𝑡 𝑡𝑡 𝑖𝑖 𝑚𝑚 𝑖𝑖 3 𝑚𝑚 = ⋯ = � � = ! 𝑘𝑘 ∞ 𝑘𝑘 ∞ 𝑋𝑋 𝑋𝑋 3 + + ( E 1 1 𝑡𝑡 + 𝑡𝑡 ) E E exp ⋯ = � 𝑋𝑋 𝑋𝑋 𝑡𝑡 + = ∫ 𝑖𝑖 𝑖𝑖 − 0 ∞ 𝑡𝑡 ∞ 𝑖𝑖 ( 𝑡𝑡 𝑖𝑖 𝑖𝑖 exp 𝑡𝑡 𝑖𝑖 ! − E 𝑖𝑖 − − 𝑖𝑖 𝑘𝑘 𝑘𝑘 + 𝑋𝑋 1 ) 𝑡𝑡 ! ⋯ , … E 𝑓𝑓 𝑖𝑖 ( 𝑋𝑋 ! 𝑋𝑋 𝑖𝑖 𝑝𝑝 2 − 𝑥𝑥 𝑋𝑋 , 𝑘𝑘 𝑑𝑑 𝑥𝑥 … 𝑑𝑑 + , E 1 ) 𝑋𝑋 𝑡𝑡 𝑘𝑘 𝑖𝑖 − 𝑘𝑘 , …

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metin /~    Examples of MGFs of Examples Ex: For variable. random of the Binomial function to obtain generating function the this Use moment variable. random Bernoulli the for function generating moment the Find Ex: Ex: If function function – – – where – Then Binomial For We 𝑋𝑋 𝑚𝑚 independence of of independence an exponential random variable variable random exponential an a Bernoulli random variable random Bernoullia rce iety as directly proceed and and 𝑚𝑚 𝑋𝑋 𝑋𝑋 𝑡𝑡 𝑚𝑚 + 𝑌𝑌 E 𝑋𝑋 = 𝑌𝑌 random variable random 𝑖𝑖 𝑚𝑚 are two independent random variables with with variables random are two independent 𝑋𝑋 of E 𝑡𝑡 𝑋𝑋 𝑘𝑘 + 𝑒𝑒 the sum of these variables? sum of the variables? these 𝑌𝑌 𝑡𝑡 = = 𝑡𝑡 𝑡𝑡 𝑚𝑚 E 𝑑𝑑 exp 𝑑𝑑 = 𝑡𝑡 𝑘𝑘 ∑ 𝑘𝑘 exp = � 𝑖𝑖 𝑛𝑛 𝑚𝑚 ( = 0 𝑡𝑡 ∞ 1 E( 𝑡𝑡 𝑋𝑋 𝑒𝑒 𝑋𝑋 𝐵𝐵 ) 𝑡𝑡 𝑡𝑡 𝑖𝑖 exp( 𝑡𝑡 ( 𝑡𝑡 and and 𝑋𝑋 𝑛𝑛 𝑡𝑡 𝜆𝜆 � 𝑖𝑖 𝑡𝑡 , 𝑒𝑒 = 𝑝𝑝 𝑋𝑋 − exp 0 = ) 𝑡𝑡 𝜆𝜆 𝑖𝑖 = ( = is a sum of with success probability probability success with 𝑋𝑋 � 𝑑𝑑 𝑖𝑖 ( = exp 𝑛𝑛 𝜆𝜆 + 𝑡𝑡 1 𝑡𝑡 𝑑𝑑 𝑌𝑌 𝑑𝑑 = ) 𝑋𝑋 𝑚𝑚 𝑡𝑡 𝑘𝑘 )) is used in the middle equality. middle the in used is 𝑘𝑘 𝜆𝜆 0𝑡𝑡 𝑋𝑋 find find ) � 𝑖𝑖 𝜆𝜆 0 = ∞ − 𝑛𝑛 𝑡𝑡 𝑒𝑒 the the E(exp( 1 Bernoulli random variables variables random Bernoulli 𝑡𝑡 = − − − 𝜆𝜆 1 mgf − 𝑝𝑝 � 𝑡𝑡 𝑡𝑡 = 1 𝑥𝑥 0 𝑡𝑡 + 𝑑𝑑 − 𝑚𝑚 = )) exp 𝑝𝑝 𝑋𝑋 𝑝𝑝 = 𝜆𝜆 𝑚𝑚 ( , E(exp( 𝑡𝑡 𝜆𝜆 𝑘𝑘 + 𝑋𝑋 ) ! 1𝑡𝑡 & and and − 𝑝𝑝 𝜆𝜆 𝜆𝜆 exp the moments the 𝑝𝑝 − 𝑡𝑡 − 1 𝑚𝑚 = 𝑡𝑡 𝑡𝑡 )) 𝑌𝑌 𝑡𝑡 − 𝑒𝑒 , 𝑘𝑘 − 1 = what is the moment generating generating is moment the what so − 𝑛𝑛 𝜆𝜆 1 − − 𝑚𝑚 � 𝑡𝑡 𝑡𝑡 = 𝑝𝑝 𝑋𝑋 𝑥𝑥 0 � 0 ∞ = 𝑡𝑡 E + ( 𝜆𝜆 𝑚𝑚 𝑘𝑘 𝑝𝑝 𝑋𝑋 𝑘𝑘 = ! 𝑌𝑌 𝑘𝑘 exp ( 𝜆𝜆 ) 𝑡𝑡 . − 𝜆𝜆 ) ( 𝑡𝑡 𝑡𝑡 ) converges for for Integral 𝑡𝑡 <

𝜆𝜆

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metin /~    Examples of MGFs of Examples Ex: Find the the Find Ex: Ex Ex – 𝑚𝑚 : Find the the Find : the Find : Normal summing mean by and parameters. obtained variance are parameters whose variablerandom normal another is variables random normal independent of Sum 𝑚𝑚 Normal 𝜇𝜇 1 mgf mgf mgf , 𝜎𝜎 1 2 + 𝜇𝜇 for for for 𝑚𝑚 Normal , 𝜎𝜎 Normal 2 𝑚𝑚 𝑋𝑋 Gamma 𝐺𝐺 𝑡𝑡 = 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺 𝜇𝜇 = Normal = 2 = 𝜇𝜇 , 𝜎𝜎 � 1 𝑒𝑒 2 2 𝑒𝑒 ( , − 𝜎𝜎 𝜇𝜇 𝑛𝑛 ∞ 𝑛𝑛 𝜇𝜇 ∞ 1 2 𝜇𝜇 , 𝜆𝜆 𝜇𝜇 , 𝑡𝑡 + + 𝜆𝜆 + 𝑒𝑒 0 ( 0 ) = Normal 𝑡𝑡 ( . = 𝑡𝑡 . 5 𝑡𝑡 𝜇𝜇 5 = for integerfor ) 𝜎𝜎 𝜎𝜎 , 𝑚𝑚 2 𝑚𝑚 = 𝜎𝜎 2 𝑒𝑒 𝑡𝑡 2𝜋𝜋 Normal 𝑡𝑡 1 2 2 ( Normal 2 � 𝜇𝜇 ) 𝑖𝑖 � = 𝜎𝜎 𝑛𝑛 1 𝜇𝜇 − 1 + ∞ 2 ∞ 𝑒𝑒 𝜇𝜇 , 𝜎𝜎 𝑚𝑚 − 2 2 2 0 ) 𝜇𝜇 𝐸𝐸 𝑡𝑡 1 𝜇𝜇 2𝜋𝜋 𝑛𝑛 . ( 1 5 + 𝐸𝐸𝐸𝐸𝐸𝐸 1 𝑡𝑡 + 0 , ≥ 𝑥𝑥 ) 𝜎𝜎 𝜇𝜇 . 𝑒𝑒 5 − 1 2 2 ( − 𝜆𝜆 𝜇𝜇 1 , 𝜎𝜎 𝜎𝜎 0 . 1 1 2 2 2 𝑡𝑡 . 5 + + 𝑡𝑡 / 𝜎𝜎 𝜎𝜎 𝑚𝑚 𝜎𝜎 𝑧𝑧 2 2 2 2 − 2 = Normal ) 𝑑𝑑 𝜎𝜎 𝑡𝑡 𝑑𝑑 2 𝜎𝜎 𝑡𝑡 2 𝜆𝜆 = 𝑑𝑑 − 𝜆𝜆 𝑑𝑑 𝑒𝑒 𝜇𝜇 𝜇𝜇 𝑡𝑡 2 𝜇𝜇 ⇐ , 𝜎𝜎 � 𝑛𝑛 2 − 2 − ∞ ∞ 0 𝑡𝑡 for . 5 1 𝑧𝑧 2𝜋𝜋 = 2 𝑡𝑡 + 𝑒𝑒 𝑒𝑒 < 𝜎𝜎𝜎𝜎𝜎𝜎 𝜇𝜇 𝜎𝜎 1 𝜎𝜎𝜎𝜎 𝜆𝜆 𝑡𝑡 + 𝑒𝑒 = 0 − − . 5 0 𝜎𝜎 0 . 5 1 . 2 𝑧𝑧 5 𝑡𝑡 2 2 𝑧𝑧 𝑑𝑑 𝑒𝑒 − 𝑧𝑧 𝜇𝜇 2 𝜎𝜎 𝑡𝑡 ⇐ + 2 0 𝑥𝑥 . + 5 = 𝜎𝜎 0 2 2 𝜇𝜇 . 𝑡𝑡 5

2 +

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𝜎𝜎

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metin 2 /~    Some functions notMGFare The following satisfies 𝑚𝑚 Question: Iseachfunction with that We know – – If there is arv is there If counterexample. by No, Answer: » » » » This is a contradiction is This We Let E 𝑋𝑋 𝑌𝑌 have have 𝑚𝑚 = = 𝑋𝑋 1 𝑋𝑋 X, we must have through differentiating differentiating through have must we X, V , 3 𝑡𝑡 E 𝑌𝑌 a new random variable. variable. random a new = 𝑋𝑋 = 2 0 E = 𝑋𝑋 = 𝑌𝑌 2 0 2 1 . and in general in and = − 𝑚𝑚 E 1 𝑚𝑚 𝑋𝑋 𝑌𝑌 but is not an mgf an not is but 0 𝑡𝑡 2 = = = E 1 1 E 𝑋𝑋 an an + 6 𝑋𝑋 � 𝑖𝑖 mgf 𝑘𝑘 = ∞ − 1 E = 𝑖𝑖 for a random variable. arandom for 𝑋𝑋 𝑘𝑘 − 𝑚𝑚 𝑡𝑡 3 𝑖𝑖 , 1 2 = ! 6 − 3 2 = − 3 < 0

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metin /~ Geometric Summary    Bernoulli Space of Space of RVs Discrete Moment generating function generating Moment Common continuous random variables Continuous random variables andexistence/uniqueness density of success u t Wai ntil Sum Hypergeometric Binomial Poisson Limit Limit Condition Uniform Beta Posterior Prior/ Decaying Space of Absolutely Continuous RVs Tail Exponential Lognormal Normal Pareto Any RV Any Tail Heavier exp Tail Lighter Sum Erlang Gamma

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metin Space of Singularly Continuous RVs /~ AbsoluteContinuity UniformContinuity vs.      Uniformcontinuity imply doesAbsolutenot continuity Absolute continuity ⇒ – Absolutely givenevery for Uniformly Continuous givenfor every Continuous for every givenevery for Counterexample: Cantor function Counterexample: Cantor » » Cantor functionis not absolutely continuous becauseit is not differentiable Cantor functionis uniformlycontinuous  Continuous el suiomy continuous.is uniformlyreals Heine 𝐹𝐹 : At every fixed: 𝑎𝑎 Atevery 𝜖𝜖 𝜖𝜖 𝜖𝜖 ∑ -Cantor , there exists there , exists there , , there exists there , 𝑖𝑖 𝑏𝑏 | 𝑏𝑏 𝑏𝑏 0 0 𝑖𝑖 − − − 𝐹𝐹 𝐹𝐹 Uniformcontinuity, definitionsabove 𝑎𝑎 𝑎𝑎 Theorem:Acontinuous functionfrom a compactset to 𝑎𝑎 : : For 0 0 𝑖𝑖 For singleinterval a | ≤ ≤ ≤ 𝛿𝛿 𝛿𝛿 𝛿𝛿 finitely many disjointintervals 𝛿𝛿 𝛿𝛿 𝛿𝛿 0 such that that such such that that such , 𝑎𝑎 implies implies implies implies implies implies 0 such that that such ∑ 𝑖𝑖 𝐹𝐹 𝐹𝐹 | 𝐹𝐹 𝑏𝑏 𝑏𝑏 𝑎𝑎 0 0 𝑏𝑏 0 𝑖𝑖 , − − 𝑏𝑏 − 0 𝐹𝐹 𝐹𝐹 𝐹𝐹 , 𝑎𝑎 𝑎𝑎 ( 0 0 𝑎𝑎 𝑖𝑖 ) | ≤ ≤ ( ≤ 𝑎𝑎 𝜖𝜖 𝜖𝜖 𝑖𝑖 𝜖𝜖 , 𝑏𝑏 𝑖𝑖 ),

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metin /~  Normal Random Variable - Ex: Partialexpected value � − 𝑎𝑎 ∞ 𝜎𝜎 − 𝜇𝜇 1 1 1 1 𝑧𝑧 𝑎𝑎 𝑎𝑎 𝑎𝑎 𝑎𝑎 > > exp ≤ > 𝜇𝜇 𝜇𝜇 𝜇𝜇 𝜇𝜇 � � � � − − − 𝑎𝑎 − 𝑎𝑎 𝑎𝑎 ∞ ∞ ∞ = 𝑎𝑎 − 𝜎𝜎 − 𝑎𝑎 𝜎𝜎 − 𝜎𝜎 − 𝜎𝜎 − 𝜎𝜎 − 𝜇𝜇 𝜇𝜇 𝜇𝜇 𝑧𝑧 𝜇𝜇 1 2 𝜇𝜇 𝑧𝑧 𝑧𝑧 𝑧𝑧 2 𝑎𝑎 𝑒𝑒 𝑧𝑧 𝑒𝑒 𝑒𝑒 > 𝑒𝑒 − − − 𝜇𝜇 𝑑𝑑𝑑𝑑 𝑧𝑧 − 𝑧𝑧 𝑧𝑧 � 2 � 𝑧𝑧 2 2 − / 0 E 2 / 𝑎𝑎 / ∞ = 2 𝑎𝑎 2 2 / 𝜎𝜎 − 𝑑𝑑𝑑𝑑 2 𝑑𝑑𝑑𝑑 𝜎𝜎 − 𝑑𝑑𝑑𝑑 𝜇𝜇 𝑋𝑋 𝑑𝑑𝑑𝑑 1 𝜇𝜇 𝑧𝑧 ( 𝑎𝑎 I = = 𝑋𝑋 = − ≤ exp = 𝜇𝜇 ≤ 𝑧𝑧 1 1 1 𝑎𝑎 � ) 1 𝑎𝑎 𝑎𝑎 𝑎𝑎 − 𝑒𝑒 > 𝑎𝑎 𝑎𝑎 > ≤ ∞ − for 𝑋𝑋 > − 𝜇𝜇 𝜇𝜇 𝜎𝜎 − 𝜇𝜇 𝑧𝑧 𝜇𝜇 𝜇𝜇 � 2 � � 𝑧𝑧 2 − / ∞ � 𝑧𝑧 − 2 2 − − ∞ − ∞ 𝑎𝑎 𝑑𝑑𝑑𝑑 exp 0 𝑎𝑎 𝑎𝑎 𝑎𝑎 𝜎𝜎 − 𝜎𝜎 − ∼ 𝜎𝜎 − 𝜎𝜎 − 𝑑𝑑𝑑𝑑 𝜇𝜇 𝜇𝜇 𝜇𝜇 𝜇𝜇 + 𝑧𝑧 𝑧𝑧 2 𝑁𝑁 𝑧𝑧 = / 1 − 𝑒𝑒 𝑒𝑒 𝑒𝑒 2 𝑁𝑁 − − 𝑎𝑎 − 𝑒𝑒 � 𝑧𝑧 𝑁𝑁 > 𝑧𝑧 𝑧𝑧 2 𝑧𝑧 − ∞ 2 2 2 𝑁𝑁𝑁𝑁 2 𝜇𝜇 𝑢𝑢 / / Manipulation / 𝑎𝑎 2 2 � 𝑑𝑑 2 𝜎𝜎 − 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 0 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑 𝜇𝜇 𝑁𝑁 𝑎𝑎 𝜎𝜎 − 2 𝜇𝜇 with + + 𝜇𝜇 = + / , 2 𝑧𝑧 1 𝜎𝜎 1 1 𝑒𝑒 1 𝑒𝑒 𝑎𝑎 𝑎𝑎 − 𝑎𝑎 𝑎𝑎 2 − > > > 𝑢𝑢 𝑢𝑢 > 𝑧𝑧 𝜇𝜇 𝜇𝜇 𝜇𝜇 . 𝜇𝜇 𝑑𝑑 2 = / � � 𝑑𝑑 � � 2 0 − ∞ − 𝑑𝑑𝑑𝑑 𝑎𝑎 𝑎𝑎 𝑧𝑧 ∞ 𝑎𝑎 𝑎𝑎 𝜎𝜎 𝜎𝜎 − − 𝑎𝑎 2 𝜎𝜎 − 𝜎𝜎 − 𝜇𝜇 / 𝜎𝜎 − 𝜇𝜇 𝜇𝜇 = 𝜇𝜇 𝜇𝜇 2 𝑧𝑧 𝑧𝑧 𝑧𝑧 𝑒𝑒 & 2 0 exp 𝑒𝑒 − / − 2 𝑑𝑑 𝑧𝑧 𝑧𝑧 𝑒𝑒 2 𝑑𝑑 2 − / / 2 − 𝑢𝑢 = 2 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑑𝑑𝑑𝑑 𝑧𝑧 𝑧𝑧𝑧𝑧𝑧𝑧 𝑑𝑑 2 2 𝑑𝑑𝑑𝑑

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metin /~