Ce Central Li Limit The heorem number of measurements needed - - PowerPoint PPT Presentation
Ce Central Li Limit The heorem number of measurements needed to estimate the mean convergence of the est estim imated ed m mean ean a better approximation than the Chebyshev Inequality (LLN). Xingru Chen
§ number
measurements needed to estimate the mean § convergence
the est estim imated ed m mean ean § a better approximation than the Chebyshev Inequality (LLN).
Xingru Chen xingru.chen.gr@dartmouth.edu
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𝑌! 𝑌" 𝑌# 𝑌$ 𝑌% ⋯ 𝑌& 𝑌' no normal density funct ction
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𝑌! 𝑌" 𝑌# 𝑌$ 𝑌% ⋯ 𝑌& 𝑌' no normal de density function
If 𝑇' is the sum
𝑜 mutually independent random variables, then the distribution function
𝑇' is well-approximated by a normal density function.
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binomial distribution
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A Bernou
trials proce
a sequence
𝑜 chance experiments such that § Each experiment has two pos
ble
comes, which we may call success and failure. § The probability 𝑞 of success
each experiment is the same for each experiment, and this probability is not affected by any knowledge
previous
probability 𝑟 of failure is given by 𝑟 = 1 − 𝑞. 50% 50% Toss a Coin
head & tail
60% 60% Weather Forecast
not rain & rain
12 12.5% Wheel of Fortune
50 points & others
Bernoulli Trials
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§ Let 𝑜 be a positive integer and let 𝑞 be a real number between and 1. § Let 𝐶 be the random variable which counts the number
successes in a Ber Bernoulli trials proce
parameters 𝑜 and 𝑞. § Then the distribution 𝑐(𝑜, 𝑞, 𝑙) of 𝐶 is called the binomial distribution. Bin Binomia ial D Dist istrib ibutio ion 𝑐 𝑜, 𝑞, 𝑙 = 𝑜 𝑙 𝑞(𝑟')(
1 2 3 4 5 ✗ ✓ ✗ ✗ ✓
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§ Consider a Bernoulli trials process with probability 𝑞 for success
each trial. § Let 𝑌* = 1 or 0 according as the 𝑗th outcome is a success
failure. § Let 𝑇' = 𝑌& + 𝑌% + ⋯ + 𝑌'. Then 𝑇' is the number
successes in 𝑜 trials. § We know that 𝑇' has as its distribution the binomial probabilities 𝑐(𝑜, 𝑞, 𝑙).
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𝑜 increases drifts
to the right flatten
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§ Consider a Bernoulli trials process with probability 𝑞 for success
each trial. § Let 𝑌* = 1 or 0 according as the 𝑗th outcome is a success
failure. § Let 𝑇' = 𝑌& + 𝑌% + ⋯ + 𝑌'. Then 𝑇' is the number
successes in 𝑜 trials. § We know that 𝑇' has as its distribution the binomial probabilities 𝑐(𝑜, 𝑞, 𝑙). § The standardized sum
𝑇' is given by 𝑇'
∗ = ,!)'- '-. .
𝜈 = ⋯ 𝜏% = ⋯
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§ Consider a Bernoulli trials process with probability 𝑞 for success
each trial. § Let 𝑌* = 1 or 0 according as the 𝑗th outcome is a success
failure. § Let 𝑇' = 𝑌& + 𝑌% + ⋯ + 𝑌'. Then 𝑇' is the number
successes in 𝑜 trials. § We know that 𝑇' has as its distribution the binomial probabilities 𝑐(𝑜, 𝑞, 𝑙). § The standardized sum
𝑇' is given by 𝑇'
∗ = ,!)'- '-. .
§ 𝑇'
∗ always
has expected value 0 and variance 1.
𝜈 = 0 𝜏% = 1
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𝑇'
∗ = 𝑇' − 𝑜𝑞
𝑜𝑞𝑟
𝑐(𝑜, 𝑞, 1) 𝑐(𝑜, 𝑞, 2) 𝑐(𝑜, 𝑞, 3) 𝑐(𝑜, 𝑞, 4) 𝑐(𝑜, 𝑞, 5) 𝑐(𝑜, 𝑞, 0)
𝑐(𝑜, 𝑞, 𝑜)
1 2 3 4 5 … 𝑜
0 − 𝑜𝑞 𝑜𝑞𝑟 1 − 𝑜𝑞 𝑜𝑞𝑟 2 − 𝑜𝑞 𝑜𝑞𝑟 3 − 𝑜𝑞 𝑜𝑞𝑟 4 − 𝑜𝑞 𝑜𝑞𝑟 5 − 𝑜𝑞 𝑜𝑞𝑟 … 𝑜 − 𝑜𝑞 𝑜𝑞𝑟
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𝑇'
∗ = 𝑇' − 𝑜𝑞
𝑜𝑞𝑟 𝜈 = 0 𝜏% = 1
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standardized sum standard normal distribution shapes: same heights: different why?
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standardized sum standard normal distribution shapes: same heights: different why?
standardized sum: ∑(/0
'
ℎ* = 1 standard normal distribution : ∫
)1 21𝑔 𝑦 𝑒𝑦 = 1
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standardized sum: ∑(/0
'
ℎ* = 1 standard normal distribution : ∫
)1 21𝑔 𝑦 𝑒𝑦 = 1 ≈ ∑(/0 '
𝑔 𝑦* 𝑒𝑦 C
(/0 '
ℎ* = 1 C
(/0 '
𝑔 𝑦* 𝑒𝑦 ≈ 1
ℎ( = 𝑔 𝑦( 𝑒𝑦
𝑔 𝑦( = ℎ( 𝑒𝑦
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ℎ) = 𝑐(𝑜, 𝑞, 𝑙)
𝑇'
∗ = 𝑇' − 𝑜𝑞
𝑜𝑞𝑟
𝑒𝑦 = 𝑦)*& − 𝑦) = 1 𝑜𝑞𝑟 𝑐(𝑜, 𝑞, 1) 𝑐(𝑜, 𝑞, 2) 𝑐(𝑜, 𝑞, 3) 𝑐(𝑜, 𝑞, 4) 𝑐(𝑜, 𝑞, 5) 𝑐(𝑜, 𝑞, 0)
𝑐(𝑜, 𝑞, 𝑜) 1 2 3 4 5 … 𝑜
0 − 𝑜𝑞 𝑜𝑞𝑟 1 − 𝑜𝑞 𝑜𝑞𝑟 2 − 𝑜𝑞 𝑜𝑞𝑟 3 − 𝑜𝑞 𝑜𝑞𝑟 4 − 𝑜𝑞 𝑜𝑞𝑟 5 − 𝑜𝑞 𝑜𝑞𝑟 … 𝑜 − 𝑜𝑞 𝑜𝑞𝑟
𝑙 = 𝑜𝑞𝑟𝑦) + 𝑜𝑞
𝑏 : the integer nearest to 𝑏
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ℎ( = 𝑐(𝑜, 𝑞, 𝑙) 𝑒𝑦 = 𝑦(2& − 𝑦( = 1 𝑜𝑞𝑟 𝑙 = 𝑜𝑞𝑟𝑦( + 𝑜𝑞
𝑔 𝑦( = 3"
45 =
𝑜𝑞𝑟 𝑐(𝑜, 𝑞, 𝑜𝑞𝑟𝑦( + 𝑜𝑞 )
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§ For the binomial distribution 𝑐(𝑜, 𝑞, 𝑙) we have lim
'→1
𝑜𝑞𝑟𝑐 𝑜, 𝑞, 𝑜𝑞 + 𝑦 𝑜𝑞𝑟 = 𝜚(𝑦), where 𝜚(𝑦) is the standard normal density. § The proof
this theorem can be carried
using Stirling’s approximation. St Stirling’s ’s Formula The sequence 𝑜! is asymptotically equal to 𝑜'𝑓)' 2𝜌𝑜.
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lim
'→1
𝑜𝑞𝑟𝑐 𝑜, 𝑞, 𝑜𝑞 + 𝑦 𝑜𝑞𝑟 = 𝜚(𝑦)
𝑐 𝑜, 𝑞, 𝑙 ≈ ⋯ normal binomial binomial normal
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lim
'→,
𝑜𝑞𝑟𝑐 𝑜, 𝑞, 𝑜𝑞 + 𝑦 𝑜𝑞𝑟 = 𝜚(𝑦)
𝑐 𝑜, 𝑞, 𝑙 ≈ ⋯ 𝑙 = 𝑜𝑞 + 𝑦 𝑜𝑞𝑟 𝑦 = 𝑙 − 𝑜𝑞 𝑜𝑞𝑟 𝑐 𝑜, 𝑞, 𝑙 ≈ 𝜚(𝑦) 𝑜𝑞𝑟 𝑐(𝑜, 𝑞, 𝑙) ≈ 1 𝑜𝑞𝑟 𝜚(𝑙 − 𝑜𝑞 𝑜𝑞𝑟 )
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Tos
a coi coin Find the probability
exactly 55 heads in 100 tosses
a coin.
lim
'→,
𝑜𝑞𝑟𝑐 𝑜, 𝑞, 𝑜𝑞 + 𝑦 𝑜𝑞𝑟 = 𝜚(𝑦)
𝑐(𝑜, 𝑞, 𝑙) ≈ 1 𝑜𝑞𝑟 𝜚(𝑙 − 𝑜𝑞 𝑜𝑞𝑟 )
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Tos
a coi coin Find the probability
exactly 55 heads in 100 tosses
a coin. 𝑐(𝑜, 𝑞, 𝑙) ≈ 1 𝑜𝑞𝑟 𝜚(𝑙 − 𝑜𝑞 𝑜𝑞𝑟 ) 𝑜 = 100 𝑞 = 1 2 𝑜𝑞 = 50 𝑜𝑞𝑟 = 5 𝑙 = 55 𝑦 = 𝑙 − 𝑜𝑞 𝑜𝑞𝑟 = 1 𝑐(100, 1 2 , 55) ≈ 1 5 𝜚(1)
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Tos
a coi coin Find the probability
exactly 55 heads in 100 tosses
a coin. 𝑐(𝑜, 𝑞, 𝑙) ≈ 1 𝑜𝑞𝑟 𝜚(𝑙 − 𝑜𝑞 𝑜𝑞𝑟 ) 𝑐 100, 1 2 , 55 ≈ 1 5 𝜚 1 = 1 5 ( 1 2𝜌 𝑓)&/%) St Standard normal distribu bution 𝒂 𝜈 = 0 and 𝜏 = 1 𝑔
8 𝑦 = & %9: 𝑓) 5); #/%:# = & %9 𝑓)5#/%.
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𝑐(𝑜, 𝑞, 1) 𝑐(𝑜, 𝑞, 2) 𝑐(𝑜, 𝑞, 3) 𝑐(𝑜, 𝑞, 4) 𝑐(𝑜, 𝑞, 5) 𝑐(𝑜, 𝑞, 0)
𝑐(𝑜, 𝑞, 𝑜) 1 2 3 4 5 … 𝑜
0 − 𝑜𝑞 𝑜𝑞𝑟 1 − 𝑜𝑞 𝑜𝑞𝑟 2 − 𝑜𝑞 𝑜𝑞𝑟 3 − 𝑜𝑞 𝑜𝑞𝑟 4 − 𝑜𝑞 𝑜𝑞𝑟 5 − 𝑜𝑞 𝑜𝑞𝑟 … 𝑜 − 𝑜𝑞 𝑜𝑞𝑟
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𝑐(𝑜, 𝑞, 1) 𝑐(𝑜, 𝑞, 2) 𝑐(𝑜, 𝑞, 3) 𝑐(𝑜, 𝑞, 4) 𝑐(𝑜, 𝑞, 5) 𝑐(𝑜, 𝑞, 0)
𝑐(𝑜, 𝑞, 𝑜) 1 2 3 4 5 … 𝑜
0 − 𝑜𝑞 𝑜𝑞𝑟 1 − 𝑜𝑞 𝑜𝑞𝑟 2 − 𝑜𝑞 𝑜𝑞𝑟 3 − 𝑜𝑞 𝑜𝑞𝑟 4 − 𝑜𝑞 𝑜𝑞𝑟 5 − 𝑜𝑞 𝑜𝑞𝑟 … 𝑜 − 𝑜𝑞 𝑜𝑞𝑟
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§ Let 𝑇' be the number
successes in n Bernoulli trials with probability 𝑞 for success, and let 𝑏 and 𝑐 be two fixed real numbers. § Then lim
'→21𝑄 𝑏 ≤ ,!)'- '-. ≤ 𝑐 = ∫ < = 𝜚 𝑦 𝑒𝑦 = NA(𝑏, 𝑐).
§ We denote this area by NA(𝑏, 𝑐). 𝑏 ≤ 𝑇' − 𝑜𝑞 𝑜𝑞𝑟 = 𝑇'
∗ ≤ 𝑐
𝑏 𝑜𝑞𝑟 + 𝑜𝑞 ≤ 𝑇' ≤ 𝑐 𝑜𝑞𝑟 + 𝑜𝑞
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lim
'→21𝑄 𝑏 ≤ ,!)'- '-. ≤ 𝑐 = NA(𝑏, 𝑐).
𝑏 ≤ 𝑇' − 𝑜𝑞 𝑜𝑞𝑟 = 𝑇'
∗ ≤ 𝑐
𝑏 𝑜𝑞𝑟 + 𝑜𝑞 ≤ 𝑇' ≤ 𝑐 𝑜𝑞𝑟 + 𝑜𝑞 𝑗 ≤ 𝑇' ≤ 𝑘 𝑏 = 𝑗 − 𝑜𝑞 𝑜𝑞𝑟 ≤ 𝑇'
∗ ≤ 𝑘 − 𝑜𝑞
𝑜𝑞𝑟 = 𝑐 lim
'→21𝑄 𝑗 ≤ 𝑇' ≤ 𝑘 = lim '→21𝑄 *)'- '-. ≤ 𝑇' ∗ ≤ >)'- '-. = NA(*)'- '-. , >)'- '-.).
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𝑗 ≤ 𝑇' ≤ 𝑘 𝑏 = 𝑗 − 𝑜𝑞 𝑜𝑞𝑟 ≤ 𝑇'
∗ ≤ 𝑘 − 𝑜𝑞
𝑜𝑞𝑟 = 𝑐 lim
'→21𝑄 𝑗 ≤ 𝑇' ≤ 𝑘 = lim '→21𝑄 *)'- '-. ≤ 𝑇' ∗ ≤ >)'- '-. = NA(*)'- '-. , >)'- '-.).
@→BC𝑄 𝑗 ≤ 𝑇@ ≤ 𝑘 = NA( DE!
"E@F
@FG , HB!
"E@F
@FG ).
𝑗 𝑘
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Tos
a coi coin A coin is tossed 100
the probability that the number
heads lies between 40 and 60 (including the end points).
lim
'→21𝑄 𝑗 ≤ 𝑇' ≤ 𝑘 = NA(
𝑗 − 1 2 − 𝑜𝑞 𝑜𝑞𝑟 , 𝑘 + 1 2 − 𝑜𝑞 𝑜𝑞𝑟 )
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Tos
a coi coin A coin is tossed 100
the probability that the number
heads lies between 40 and 60 (including the end points). 𝑜 = 100 𝑞 = 1 2 𝑜𝑞 = 50 𝑜𝑞𝑟 = 5 𝑗 = 40 𝑗 − 1 2 − 𝑜𝑞 𝑜𝑞𝑟 = −2.1 𝑘 = 60 𝑘 + 1 2 − 𝑜𝑞 𝑜𝑞𝑟 = 2.1
the
will not deviate by more than two standard deviations from the expected value
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Tos
a coi coin A coin is tossed 100
the probability that the number
heads lies between 40 and 60 (including the end points).
𝑗 = 40 𝑗 − 1 2 − 𝑜𝑞 𝑜𝑞𝑟 = −2.1 𝑘 = 60 𝑘 + 1 2 − 𝑜𝑞 𝑜𝑞𝑟 = 2.1
lim
$→&'𝑄 𝑗 ≤ 𝑇$ ≤ 𝑘 = NA(
𝑗 − 1 2 − 𝑜𝑞 𝑜𝑞𝑟 , 𝑘 + 1 2 − 𝑜𝑞 𝑜𝑞𝑟 )
lim
$→&'𝑄 40 ≤ 𝑇$ ≤ 60 = NA −2.1,2.1
= 2NA 0, 2.1
2𝜏 2.1𝜏
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𝑻𝒐
∗ = 𝒚
lim
$→'
𝑜𝑞𝑟𝑐 𝑜, 𝑞, 𝑜𝑞 + 𝑦 𝑜𝑞𝑟 = 𝜚(𝑦)
𝑻𝒐 = 𝒍
𝑐(𝑜, 𝑞, 𝑙) ≈ 1 𝑜𝑞𝑟 𝜚(𝑙 − 𝑜𝑞 𝑜𝑞𝑟 )
𝒃 ≤ 𝑻𝒐
∗ ≤ 𝒄
lim
'→*, 𝑄 𝑏 ≤
'/0 ≤ 𝑐 = NA(𝑏, 𝑐).
𝒋 ≤ 𝑻𝒐 ≤ 𝒌
lim
$→&'𝑄 𝑗 ≤ 𝑇$ ≤ 𝑘 = NA( ()!
")$*
$*+ , ,&!
")$*
$*+ ).
CLT
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any independent trials process such that the individual trials have finite variance
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§ Consider an independent trials process with common distribution function 𝑛 𝑦 defined
the integers, with expected value 𝜈 and variance 𝜏%. § Let 𝑇' = 𝑌& + 𝑌% + ⋯ + 𝑌' be the sum
𝑜 independent discrete random variables
the process. § The standardized sum
𝑇' is given by 𝑇'
∗ = ,!)'; ':# .
§ 𝑇'
∗ always
has expected value 0 and variance 1. 𝐹(𝑇') = ⋯ 𝑊(𝑇') = ⋯
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independent trials process 𝑄(𝑇' = 𝑙)
𝑦)= 𝑙 − 𝑜𝜈 𝑜𝜏%
𝑇'
∗ = 𝑇' − 𝑜𝜈
𝑜𝜏% normal
𝑜𝜏-𝑄(𝑇$ = 𝑙)
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§ Let 𝑌&, 𝑌%, ⋯, 𝑌' be an independent trials process and let 𝑇' = 𝑌& + 𝑌% + ⋯ + 𝑌'. § Assume that the greatest common divisor
the differences
all the values that the 𝑌( can take
is 1. § Let 𝐹 𝑌( = 𝜈 and 𝑊 𝑌( = 𝜏%. § Then for 𝑜 large, 𝑄(𝑇' = 𝑙) ≈
& ':# 𝜚(()'; ':#).
§ Here 𝜚(𝑦) is the standard normal density.
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§ Let 𝑇' = 𝑌& + 𝑌% + ⋯ + 𝑌' be the sum
n discrete independent random variables with common distribution having expected value 𝜈 and variance 𝜏%. § Then, for 𝑏 < 𝑐, lim
'→21𝑄 𝑏 ≤ ,!)'; ':# ≤ 𝑐 = & %9 ∫ < = 𝑓)5#/%𝑒𝑦 = NA(𝑏, 𝑐).
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𝑄(𝑇' = 𝑙) ≈ 1 𝑜𝜏% 𝜚(𝑙 − 𝑜𝜈 𝑜𝜏% ) lim
'→*, 𝑄 𝑑 ≤ 𝑇' ≤ 𝑒 =
NA(1.'2
'3" , 4.'2 '3").
CLT
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lim
'→*, 𝑄 𝑗 ≤ 𝑇' ≤ 𝑘 =
NA(
5.#
".'/
'/0 , 6*#
".'/
'/0 ).
𝒋 ≤ 𝑻𝒐 ≤ 𝒌
𝑄(𝑇' = 𝑙) ≈ 1 𝑜𝜏% 𝜚(𝑙 − 𝑜𝜈 𝑜𝜏% )
𝑻𝒐 = 𝒍
𝑐(𝑜, 𝑞, 𝑙) ≈ 1 𝑜𝑞𝑟 𝜚(𝑙 − 𝑜𝑞 𝑜𝑞𝑟 )
𝑻𝒐 = 𝒍
lim
'→*, 𝑄 𝑑 ≤ 𝑇' ≤ 𝑒 =
NA(
1.'2 '3" , 4.'2 '3").
𝒅 ≤ 𝑻𝒐 ≤ 𝒆 Bernoulli
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A surveying instrument makes an error
0, 1,
2 feet with equal probabilities when measuring the height
a 200-foot tower.
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Find the expected value and the variance for the height
using this instrument
height = error + 200
er error
1 2 probability 1 5 1 5 1 5 1 5 1 5
𝐹 error = 1 5 −2 − 1 + 0 + 1 + 2 = 0 𝑊 error = 1 5 (−2)%+(−1)%+0% + 1% + 2% = 2 𝐹 height = 0 + 200 = 200 𝑊 height = 2
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Estimate the probability that in 18 independent measurements
this tower, the average
the measurements is between 199 and 201, inclusive.
𝜈 = 200 𝜏% = 2
lim
'→21𝑄 𝑑 ≤ 𝑇' ≤ 𝑒 = NA(𝑑 − 𝑜𝜈
𝑜𝜏% , 𝑒 − 𝑜𝜈 𝑜𝜏% ). 199 ≤ 𝑇' 𝑜 = 𝑇' 18 ≤ 201 𝑜 = 18
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𝜈 = 200 𝜏% = 2
lim
'→*, 𝑄 𝑑 ≤ 𝑇' ≤ 𝑒 = NA(𝑑 − 𝑜𝜈
𝑜𝜏% , 𝑒 − 𝑜𝜈 𝑜𝜏% ).
199 ≤ 𝑇' 𝑜 = 𝑇' 18 ≤ 201 𝑜 = 18 𝑄 199 ≤ 𝑇' 18 ≤ 201 = 𝑄 3582 ≤ 𝑇' ≤ 3618 ≈ NA 3582 − 3600 36 , 3618 − 3600 36 = NA(−3, 3).
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§ independent
§ independent § identical
§ independent § identical
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§ Let 𝑌&, 𝑌%, ⋯, 𝑌' be a sequence
independent discrete random
exist a constant 𝐵, such that 𝑌* ≤ 𝐵 for all 𝑗. § Let 𝑇' = 𝑌& + 𝑌% + ⋯ + 𝑌' be the
that 𝑇' → ∞. § For each 𝑗, denote the expected value and variance
𝑌* by 𝜈* and 𝜏*
%,
respectively. § Define the expected value and variance
𝑇' to be 𝑛' and 𝑡'
%,
respectively. § For 𝑏 < 𝑐, lim
'→21𝑄 𝑏 ≤ ,!)@! A!
≤ 𝑐 =
& %9 ∫ < = 𝑓)5#/%𝑒𝑦 = NA(𝑏, 𝑐).
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continuous random variables with a common density function
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§ Let 𝑇' = 𝑌& + 𝑌% + ⋯ + 𝑌' be the sum
𝑜 independent continuous random variables with common density function 𝑞 having expected value 𝜈 and variance 𝜏%. § Let 𝑇'
∗ = ,!)'; ':# .
§ Then we have, for all 𝑏 < 𝑐, lim
'→21𝑄 𝑏 ≤ 𝑇' ∗ ≤ 𝑐 = & %9 ∫ < = 𝑓)5#/%𝑒𝑦 = NA(𝑏, 𝑐).
XC 2020
§ Suppose a surveyor wants to measure a known distance, say of 1 mile, using a transit and some method of triangulation. § He knows that because of possible motion of the transit, atmospheric distortions, and human error, any one measurement is apt to slightly in error. § He plans to make several measurements and take an average. § He assumes that his measurements are independent random variables with common distribution of mean 𝜈 = 1 and standard deviation 𝜏 = 0.0002.
XC 2020
§ He can say that if 𝑜 is large, the average ,!
' has
a density function that is approximately normal, with mean 1 mile, and standard deviation 0.000%
'
miles. § How many measurements should he make to be reasonably sure that his average lies within 0.0001 of the true value?
XC 2020
Expect cted Value
St Standard De Deviation 𝑇' = 𝑌& + 𝑌% + ⋯ + 𝑌' Su Sum of 𝒐 in independent me measureme ments § He can say that if 𝑜 is large, the average ,!
' has
a density function that is approximately normal, with mean 1 mile, and standard deviation 0.000%
'
miles.
Variance ce 𝐵' = 𝑌& + 𝑌% + ⋯ + 𝑌' 𝑜 𝐹 𝐵' = 𝜈 𝐸 𝐵' = 𝜏 𝑜 𝑊 𝐵' = 𝜏% 𝑜 Av Average of 𝒐 in independent me measureme ments
XC 2020
Chebyshev inequality LLN LLN 𝑇'
∗ = 𝑇' − 𝑜𝜈
𝑜𝜏% CLT § How many measurements should he make to be reasonably sure that his average lies within 0.0001 of the true value?
XC 2020
𝝉𝟑 𝜻𝟑
𝑸(|𝒀 − 𝝂| ≥ 𝜻)
in inequalit ity
𝑸(|𝒀 − 𝝂| ≥ 𝜻) ≤ 𝝉𝟑 𝜻𝟑
This is a sample text. Insert your desired text here. This is a sample text. Insert your desired text here. Let 𝑌 be a discrete random variable with expected value 𝜈 = 𝐹(𝑌) and variance 𝜏% = 𝑊(𝑌), and let 𝜁 > 0 be any positive
no not ne necessarily po posi sitive
XC 2020
Chebyshev inequality LLN LLN § How many measurements should he make to be reasonably sure that his average lies within 0.0001 of the true value? 𝜈 = 1 𝜏% = (0.0002)%
𝑄(|𝑌 − 𝜈| ≥ 𝜁) ≤ 𝜏% 𝜁% 𝑄 𝐵' − 𝜈 ≥ 0.0001 ≤ 𝜏% 𝑜𝜁% = 0.0002 % 𝑜 0.0001 % = 4 𝑜
XC 2020
Chebyshev inequality LLN LLN § How many measurements should he make to be reasonably sure that his average lies within 0.0001 of the true value?
𝑄(|𝑌 − 𝜈| ≥ 𝜁) ≤ 𝜏% 𝜁% 𝑄 𝐵' − 𝜈 ≥ 0.0001 ≤ 𝜏% 𝑜𝜁% = 4 𝑜 = 0.05 𝑜 = 80
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𝑇'
∗ = 𝑇' − 𝑜𝜈
𝑜𝜏% CL CLT
lim
'→21𝑄 𝑏 ≤ 𝑇' ∗ ≤ 𝑐 = NA(𝑏, 𝑐)
𝑄 𝐵' − 𝜈 ≤ 0.0001 = 𝑄 𝑇' − 𝑜𝜈 ≤ 0.5𝑜𝜏 = 𝑄 𝑇' − 𝑜𝜈 𝑜𝜏% ≤ 0.5 𝑜 = 𝑄 −0.5 𝑜 ≤ 𝑇'
∗ ≤ 0.5 𝑜
≈ NA −0.5 𝑜, 0.5 𝑜 𝜈 = 1 𝜏% = (0.0002)%
XC 2020
𝑇'
∗ = 𝑇' − 𝑜𝜈
𝑜𝜏% CL CLT
lim
'→21𝑄 𝑏 ≤ 𝑇' ∗ ≤ 𝑐 = NA(𝑏, 𝑐)
𝑄 𝐵' − 𝜈 ≤ 0.0001 ≈ NA −0.5 𝑜, 0.5 𝑜 = 0.95 0.5 𝑜 = 2 𝑜 = 16
XC 2020
Chebyshev inequality LLN LLN 𝑇'
∗ = 𝑇' − 𝑜𝜈
𝑜𝜏% CLT § How many measurements should he make to be reasonably sure that his average lies within 0.0001 of the true value? 𝑜 = 80 𝑜 = 16
XC 2020