CMU 15-251

Probability II

Teachers:

Victor Adamchik

Ariel Procaccia (this time)

Deep questions

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Great expectations

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Random variable

• 𝑆

• 𝑋: 𝑆 → ℝ

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o 𝑋 =

𝑋 3,4 = 3, 𝑋 1,5 = 1

o 𝑋 =

𝑋 3,4 = 7, 𝑋 1,5 = 6

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Two coins tossed

• 𝑋: 𝑇𝑇, 𝐻𝑇, 𝑇𝐻,𝐻𝐻 → *0,1,2+

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𝑇𝑇 1

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𝐻𝑇 1

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𝑇𝐻 1

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𝐻𝐻 1

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𝑆

0 1

4

1 1

2

2 1

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ℝ

𝑋

From RVs to events

• 𝑋 𝑎 ∈ ℝ 𝐸𝑋 = 𝑎

Pr 𝐸 = Pr 𝑋 = 𝑎 = Pr, 𝑡 ∈ 𝑆 𝑋 𝑡 = 𝑎+-

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𝑇𝑇 1

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𝐻𝑇 1

4

𝑇𝐻 1

4

𝐻𝐻 1

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𝑆

0 1

4

1 1

2

2 1

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ℝ

𝑋

Pr 𝑋 = 1 = Pr 𝐻𝑇, 𝑇𝐻 = 1/2

From events to RVs

• 𝐸𝐸

𝕀𝐸 𝑡 = 1 𝑡 ∈ 𝐸0 𝑡 ∉ 𝐸

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𝑇𝑇 1

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𝐻𝑇 1

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𝑇𝐻 1

4

𝐻𝐻 1

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𝑆

0 1

2

1 1

2

ℝ

𝕀𝐸

𝐸

𝐸 =

Independent RVs

•

𝑎, 𝑏, 𝑋 = 𝑎 𝑌 = 𝑏

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Expectation

• 𝑋

𝔼 𝑋 = Pr 𝑡 × 𝑋 𝑡 = Pr 𝑋 = 𝑘 × 𝑘

𝑘𝑡∈𝑆

• 𝑋 𝔼 𝑋 = ?

1. 1

2. 4/3

3. 3/2

4. 2

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Pr,𝑋 = 𝔼 𝑋 -

Expectation

• 𝕀𝐸 𝐸𝔼 𝕀𝐸 = Pr 𝕀𝐸 = 1 × 1 = Pr,𝐸-

• 𝑋 𝑌𝑆 𝑍 = 𝑋 + 𝑌

𝑍 𝑡 = 𝑋 𝑡 + 𝑌 𝑡

• 𝑋 𝑌𝑍

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Linearity of expectation

• 𝑍 = 𝑋 + 𝑌𝔼 𝑍 = 𝔼 𝑋 + 𝔼,

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𝑋 𝑌

Linearity of expectation

𝔼 𝑍 = Pr 𝑡 𝑍 𝑡

𝑡∈𝑆

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= Pr 𝑡 (𝑋 𝑡 + 𝑌 𝑡 )

𝑡∈𝑆

= Pr 𝑡 𝑋 𝑡 + Pr 𝑡 𝑌(𝑡)

𝑡∈𝑆𝑡∈𝑆

= 𝔼 𝑋 + 𝔼 𝑌 ∎

using linearity of expectation

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Using linearity of expectation

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𝑘 × Pr 𝑘 correct letters = 𝑘 × ,..aargh!!..-

𝑘𝑘

Using linearity of expectation

• 𝐴𝑖 𝑖

• 𝕀𝐴𝑖𝐴𝑖

o

• 𝔼 𝕀𝐴𝑖= Pr 𝐴𝑖 = 1 = 1/100

• 𝑍 = 𝕀𝐴𝑖

100𝑖=1

• 𝔼 𝑍 = 100 ×1

100= 1

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Using linearity of expectation

• 𝑛 𝑝

1. 1

2. 𝑝

3. 𝑛

4. 𝑛𝑝

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Balls and bins

• 𝑛 𝑛

•

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Balls and bins

• 𝑋𝑖 𝑖

• 𝑛 = 𝔼 𝑋1 + ⋯𝑋𝑛 = 𝔼 𝑋1 + ⋯+ 𝔼 𝑋𝑛

• 𝔼 𝑋𝑖 = 1 𝑖

•

log 𝑛 / log log 𝑛

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Balls and bins

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• 𝑛 𝑛

•

1. ½

2. 1

3. 2

4. log 𝑛/ log log 𝑛

• ∼ log log 𝑛

Conditional expectation

• 𝔼 𝑋 𝐸- = Pr 𝑋 = 𝑘 𝐸- × 𝑘𝑘

•

𝔼 𝑋 = 𝔼 𝑋 𝐸- × Pr 𝐸 + 𝔼 𝑋 𝐸 - × Pr,𝐸 -

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geometric RVs

• 𝑝

• 𝑋 =

• 𝔼 𝑋 = 𝔼 𝑋 𝐻- Pr,𝐻- + 𝔼 𝑋 𝑇- Pr,𝑇-= 1 ⋅ 𝑝 + 𝔼 𝑋 + 1 1 − 𝑝= 1 + 𝔼 𝑋 1 − 𝑝

• 𝔼 𝑋 = 1/𝑝

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CMUland

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What we have learned

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