E[F] ? Albert R Meyer, May 8, 2013 Albert R Meyer, May 8, 2013 - - PowerPoint PPT Presentation

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Mathematics for Computer Science Mean Time to Failure MIT 6.042J/18.062J Flip a coin until a Head comes up Expected Time Pr[Head] = p to Failure F ::= #flips to 1 st Head E[F] ? Albert R Meyer, May 8, 2013 Albert R Meyer, May 8,

SLIDE 1

Albert R Meyer, May 8, 2013

Mathematics for Computer Science MIT 6.042J/18.062J

Expected Time to Failure

ranvarfail.1 Albert R Meyer, May 8, 2013 ranvarfail.3

Flip a coin until a Head comes up Pr[Head] = p

F ::= #flips to 1st Head

E[F] ?

Mean Time to “Failure”

Albert R Meyer, May 8, 2013

Mean Time to “Failure”

Pr[F= 1]

ranvarfail.4

Pr[F= 1] = Pr[H] = p

Albert R Meyer, May 8, 2013

Mean Time to “Failure”

Pr[F= 1] = Pr[H] = p Pr[F= 2] = Pr[TH] = q⋅p Pr[F= 3] = Pr[TTH] = q2⋅p

PDFF(n) = qn-1p Geometric Distribution

ranvarfail.5

SLIDE 2

Albert R Meyer, May 8, 2013

E[F] = ∑n>0 n⋅Pr[F=n]

= ∑n>0 n⋅qn-1p = p ∑n≥0 (n+1)qn

ranvarfail.6

Mean Time to “Failure”

1 (1− q)2 n n q q I qq

Albert R Meyer, May 8, 2013

E[F] = ∑n>0 n⋅Pr[F=n]

= ∑n>0 n⋅qn-1p

ranvarfail.7

Mean Time to “Failure”

= p 1 (1− q)2

Albert R Meyer, May 8, 2013

E[F] = ∑n>0 n⋅Pr[F=n]

= ∑n>0 n⋅qn-1p

ranvarfail.8

Mean Time to “Failure”

= p 1 p2 = 1

p

Albert R Meyer, May 8, 2013

Mean Time to “Failure”

H

p q

H

p q

H

p q

ranvarfail.11

E[F] = ?

SLIDE 3

Albert R Meyer, May 8, 2013

Mean Time to “Failure”

H

p q

B B

now use Total Expectation

ranvarfail.12

E[F] = ?

Albert R Meyer, May 8, 2013

Mean Time to “Failure”

H

p q

B B

E[F |1st is H]⋅p

ranvarfail.13

E[F] =

Albert R Meyer, May 8, 2013

Mean Time to “Failure”

H

p q

B B

+ E[F |1st is T]⋅q

1 E[F+1]

ranvarfail.14

E[F] =

E[F |1st is H]⋅p

Albert R Meyer, May 8, 2013

Mean Time to “Failure”

H

p q

B B

p

+ (E[F] +1) ⋅q

ranvarfail.15

now solve for E[F]

E[F] =

SLIDE 4

Albert R Meyer, May 8, 2013

Mean Time to “Failure”

ranvarfail.16

1 p

E[F] =

Albert R Meyer, May 8, 2013

Mean Time to Failure

application: if space station Mir

has 1/150,000 chance of destruction in any given hour, how may hours expected until destruction? 150,000 hours ≈ 17 years

ranvarfail.17 Albert R Meyer, May 8, 2013

Intuitive argument

E[#fails in 1 try] = p E[#fails in n tries] = np E[#tries between fails]

ranvarfail.18

= # tries # fails = n np = 1 p

SLIDE 5

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6.042J / 18.062J Mathematics for Computer Science

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