Transcript The Wide World of Expectation!

The Wide World of Expectation!
We shall examine some useful
properties of the Expected Value
operation, in relation to how it is used
in Game Theory
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Notation
• Discrete random variable r, taking values (r1,r2,...,rn)
with probability (p1,p2,....,pn) respectively.
• E[r] = Expected value of r.
• E[r] := p.r := r1p1 + r2p2 + ... + rnpn
• Note that p is a probability vector so
• 0 ≤ pi ≤ 1 , i=1,2,....,n
• p1 + p2 + ... + pn = 1
• Example
• r = (3, 6, 2)
• p = (0.1, 0.7, 0.2)
• E[r] = 3x0.1 + 6x0.7 + 2x0.2
2
= 0.3 + 4.2 + 0.4 = 4.9
Property 1
• Let C = (c, c, c, …, c), then
• E[C+r] = E[(c+r1, c+r2, …, c+rn)] = c + E[r]
• In words, if we add a constant c to each component
of r, then the expected value of the new variable is
equal to c + the expected value of old variable
• Proof:
• E[C+r]
• = E [(c+r1, c+r2, …, c+rn)] := p [(c+r1, c+r2, …, c+rn)]
• = (c + r1)p1 + (c + r2)p2 + ... + (c+ rn)pn
• = c(p1 + p2 + ... + pn) + r1p1 + r2p2 + ... + rnpn
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• = c + E[r]
Example
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•
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If, as in previous example
r = (3, 6, 2)
p = (0.1, 0.7, 0.2)
But we now also have c = 2.3
E[C + r] = 5.3x0.1 + 8.3x0.7 + 4.3x0.2
= 0.53 + 5.81 + 0.86 = 7.2
• c + E[r] = 2.3 + 4.9 (using previous example)
= 7.2 = E[C + r]
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Property 2
• E[cr] = cE[r] , c = any constant
• In words, if we multiply r by some constant c,
then the expected value of the new random
variable cr is equal to c times the expected
value of the old random variable.
• Proof:
• E[cr] := p(cr) := (cr1)p1 + (cr2)p2 + ... + (crn)pn
•
= c(r1p1 + r2p2 + ... + rnpn)
•
= cE[r]
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Example
• r = (3, 6, 2)
• p = (0.1, 0.7, 0.2)
• E[r] = 3x0.1 + 6x0.7 + 2x0.2
= 0.3 + 4.2 + 0.4 = 4.9
• c=2
• E[cr] = cE[r] = 2x4.9 = 9.8
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Property 3
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E[A + cr] = a + cE[r], where A = (a, a, ..., a)
In words, expectation is a linear operation.
Proof.
Implied by the first two properties
Example
r = (3, 6, 2)
p = (0.1, 0.7, 0.2)
E[r] = 3x0.1 + 6x0.7 + 2x0.2
= 0.3 + 4.2 + 0.4 = 4.9
• a = 2.3 ; c = 2
• E[A + cr] = a + cE[r] = 2.3 + 2x4.9 = 12.1
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Property 4
•
min { ri} ≤ E[r] ≤ max{ ri }
• In words, the expected value of r cannot be smaller
than the smallest value that r can take, neither can
it be larger than the largest value that r can take.
• Proof that E[r] ≤ max { ri} = R, say :
• E[r] := r1p1 + r2p2 + ... + rnpn
• Hence, since pi ≥ 0
• E[r] ≤ Rp1 + Rp2 + ... + Rpn
• E[r] ≤ R(p1 + p2 + ... + pn)
• =Rx1
• = max { ri}
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• Exercise: Proof that min {ri} ≤ E[r]
Example
• r = (3, 6, 2)
• p = (0.1, 0.7, 0.2)
• E[r] = 3x0.1 + 6x0.7 + 2x0.2
= 0.3 + 4.2 + 0.4 = 4.9
• min {ri} = 2; max {ri}= 6
• min { ri} ≤ E[r] ≤ max{ ri } (OK)
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E[r]
ri
1
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i
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Property 5
• Same as Property 4, except that we ignore (drop)
{ri} values associated with pi = 0 (if any).
• This makes Property 4 a bit stronger.
• Proof: {ri} values for which p i= 0 do not contribute
anything to E[r].
• Example r = (3, 6, 2, 1), p = (0.1, 0.7, 0.2, 0),
• min {ri} = 1; max {ri}= 6
• E[r] = 3x0.1 + 6x0.7 + 2x0.2 + 1x0
•
= 0.3 + 4.2 + 0.4 + 0 = 4.9
• min { ri} ≤ E[r] ≤ max{ ri } (Property 4): 1 ≤ E[r] ≤ 6
• 2 ≤ E[r] ≤ 6
(Property 5)
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“Lemma”
• If E[r] < some constant c, then there exists some i
such that ri < c. Also if E[r] > some constant c, then
there exists some i such that ri > c.
• Proof: Property 4 implies that c > E[r] ≥ min {ri}
• This implies that c > min {ri}
• Hence, there exists at least one i such that ri < c.
ri
C
E[r]
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Similarly with inequalities reversed.
Impossible !!
i
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