MATH 20: PROBABILITY
Sums
- f
Independent Random Variables Xingru Chen xingru.chen.gr@dartmouth.edu
XC 2020
MATH 20: PROBABILITY Sums of Independent Random Variables - - PowerPoint PPT Presentation
MATH 20: PROBABILITY Sums of Independent Random Variables Xingru Chen xingru.chen.gr@dartmouth.edu XC 2020 Syllabus Su Sums of Random Va Variables La Law of o La Large N Numbers Ce Central Limit Theorem Ge Generating
Sums
Independent Random Variables Xingru Chen xingru.chen.gr@dartmouth.edu
XC 2020
Su Sums
Random Va Variables La Law
La Large N Numbers Ce Central Limit Theorem Ge Generating Functions Marko kov Chains
Mon 17 Tue 18 Wed 19 Thu 20 Fri 21 Sat 22 Sun 23 Mon 24 Tue 25 Wed 26 β¦ Sun 30 Quiz Homework (due Fri 28) Quiz Final Quiz
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Two real numbers π and π are chosen at random and uniformly from [0, 1]. Let π = π + π. Please derive expressions for the cumulative distribution and the density function
π.
π ππ¨ πΊ! π¨ = π π¨ = β―
πΊ! π¨ = π π β€ π¨ = β― range
π is β¦
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(π β π) π¨ = 6
"# $#
π π¨ β π§ π π§ ππ§ π% π = ;
&
π'(π)π((π β π)
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distribution function π'(π¦) ra random vari riable le X distribution function π((π§) ra random vari riable le Y
ra random vari riable le π = π + π distribu bution
ction
ππ(π) in independent
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π π β π the probability that π takes
the value π¨ π¨ = 1 + (π¨ β 1) π¨ = 2 + (π¨ β 2) π¨ = 3 + (π¨ β 3) π π β π π π β π
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π π β π the probability that π takes
the value π¨ π¨ = 1 + (π¨ β 1) π¨ = 2 + (π¨ β 2) π¨ = 3 + (π¨ β 3) π π β π π π β π π π = π¨ = ;
&*"# $#
π π = π π(π = π¨ β π)
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Β§ Let π and π be two in independent integer-valued random variables, with distribution functions π!(π¦) and π"(π¦) respectively. Β§ Then the co convo volut ution of π!(π¦) and π"(π¦) is the distribution function π# = π! β π" given by π# π = β$ π!(π)π"(π β π), for π = β― , β2, β1, 0, 1, 2, β―. Β§ The function π# π¦ is the distribution function
the random variable π = π + π.
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Β§ A die is rolled
π' and π( be the
and let π( = π' + π( be the sum
these
Β§ Then π' and π( have the common distribution function: π =
' ( % + ,
" ! " ! " ! " ! " ! "
. Β§ The distribution function
π( is then the convolution
this distribution with itself.
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π π β π π π( = 2 = π 1 π(1) π π( = 3 = π 1 π 2 + π 2 π(1) π π( = 4 = β― π π β π π π β π
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π( = π' + π(
π! π(π!) 2 and 12 1 36 3 and 11 2 36 4 and 10 3 36 5 and 9 4 36 6 and 8 5 36 7 6 36
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in independent random varia iables
π' + π( + β― + π. π»π ππ π»π + ππ π»π + ππ β― π»π"π + ππ
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π π" = 3 = π π! = 2 π(π" = 1) π π" = 4 = π π! = 2 π π" = 2 + π π! = 3 π(π" = 1) π π" = 5 = β―
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π β β π(π.) β β― Ce Central Limit Theorem
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Ra Random vari riable le π Ra Random vari riable le π binomial distribution parameters: π and π Ra Random vari riable le π binomial distribution parameters: π and π 01 01 02 02
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Ra Random vari riable le π binomial distribution parameters: π and π Ra Random vari riable le π binomial distribution parameters: π and π 01 01 02 02
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co common pa paramet eter er π
π' + π( + β― + π& π»π ππ π»π + ππ π»π + ππ β― π»π"π + ππ
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co common pa paramet eter er π
π' + π( + β― + π& π»π ππ π»π + ππ π»π + ππ β― π»π"π + ππ π3: the number
trials up to and including the the first succe ccess π4: β―
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co common pa paramet eter er π
π' + π( + β― + π& π»π ππ π»π + ππ π»π + ππ β― π»π"π + ππ π3: the number
trials up to and including the the first succe ccess π4: the number
trails up to and include the πth successes
negative binomial distribution parameters: π and π
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density function π(π¦) ra random vari riable le X density function π(π§) ra random vari riable le Y
ra random vari riable le π = π + π density funct ction
π(π) in independent
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the probability that π takes
the value π¨ π¨ = 1 + (π¨ β 1) π¨ = 2 + (π¨ β 2) π¨ = 3 + (π¨ β 3)
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the probability that π takes
the value π¨ π¨ = π + (π¨ β π) π¨ = π + (π¨ β π) π¨ = π + (π¨ β π)
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Β§ Let π and π be two continuous random variables with density functions π(π¦) and π(π§), respectively. Β§ Assume that both π(π¦) and π(π§) are defined for all real numbers. Β§ Then the con convol
π and π is the function given by (π β π) π¨ = β«
"# $#π π¨ β π§ π π§ ππ§.
π¨ = π¦ + π§
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Β§ Let π and π be two in independent random variables with density functions π
5(π¦) and
π
6 π§
defined for all π¦. Β§ The the sum π = π + π is a random variable with density function π
!(π¨),
where π
! is
the convolution
π
5 and
π
6.
π
! π¨ = (π 5 β π 6) π¨ = β« "# $#π 5 π¨ β π§ π 6 π§ ππ§.
π¨ = π¦ + π§
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Β§ Suppose we choose independently two numbers at random from the interval [0, 1] with uniform probability density. Β§ What is the density
their sum? Un Unifor
distribu bution
π
5 π¦ = π 6 π¦ = Y1,
0 β€ π¦ β€ 1 0.
π
! π¨ = 6 "# $#
π
5 π¨ β π§ π 6 π§ ππ§
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Un Unifor
distribu bution
π
5 π¦ = π 6 π¦ = Y1,
0 β€ π¦ β€ 1 0.
π
! π¨ = 6 "# $#
π
5 π¨ β π§ π 6 π§ ππ§
π
! π¨ = 6 7 '
π
5 π¨ β π§ ππ§
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Un Unifor
distribu bution
π
5 π¦ = π 6 π¦ = Y1,
0 β€ π¦ β€ 1 0.
π
! π¨ = 6 "# $#
π
5 π¨ β π§ π 6 π§ ππ§
π
! π¨ = 6 7 '
π
5 π¨ β π§ ππ§
0 β€ π¨ β π§ β€ 1, π¨ β 1 β€ π§ β€ π¨ π β€ π β€ π π
! π¨ = 6 7 8
ππ§ = π¨ π β€ π β€ π π
! π¨ = 6 8"' '
ππ§ = 2 β π¨
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Un Unifor
distribu bution
π
5 π¦ = π 6 π¦ = Y1,
0 β€ π¦ β€ 1 0.
π
! π¨ = 6 "# $#
π
5 π¨ β π§ π 6 π§ ππ§
π
! π¨ = 6 7 '
π
5 π¨ β π§ ππ§
π
! π¨ = d
π¨, 0 β€ π¨ β€ 1 2 β π¨, 1 β€ π¨ β€ 2 0.
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Β§ Suppose we choose two numbers at random from the interval 0, β with an exponential density with parameter π. Β§ What is the density
their sum? Ex Exponen ential al di distribution π
5 π¦ = π 6 π¦ = Yππ"9:,
π¦ β₯ 0 0.
π
! π¨ = 6 "# $#
π
5 π¨ β π§ π 6 π§ ππ§
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Ex Exponen ential al di distribution π
5 π¦ = π 6 π¦ = Yππ"9:,
π¦ β₯ 0 0.
π
! π¨ = 6 "# $#
π
5 π¨ β π§ π 6 π§ ππ§
π
! π¨ = 6 7 8
π
5 π¨ β π§ π 6 π§ ππ§
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Ex Exponen ential al di distribution π
5 π¦ = π 6 π¦ = Yππ"9:,
π¦ β₯ 0 0.
π
! π¨ = 6 "# $#
π
5 π¨ β π§ π 6 π§ ππ§
π
! π¨ = 6 7 8
ππ"9(8"<)ππ"9<ππ§ = 6
7 8
π(π"98ππ§ = π(π¨π"98 π
! π¨ = Yπ(π¨π"98,
π¨ β₯ 0 0.
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Β§ Suppose π and π are two independent random variables, each with the standard normal density. Β§ What is the density
their sum? Nor Norma mal distribution
π
5 π¦ = π 6 π¦ =
1 2π π":#/( π
! π¨ = 6 "# $#
π
5 π¨ β π§ π 6 π§ ππ§
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Nor Norma mal distribution
π
5 π¦ = π 6 π¦ =
1 2π π":#/( π
! π¨ = 6 "# $#
π
5 π¨ β π§ π 6 π§ ππ§
= 1 2π 6
"# $#
π"(8"<)#/(π"<#/(ππ§ = 1 4π π"8#/+
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convolution Β§ Means: π# and π! Β§ Variances: π#
! and
π!
!
Normal + Normal = Normal Β§ Means: π# + π! Β§ Variances: π#
! + π! !
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