Bandit Algorithms

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2.6 Conditional expectation 32

Proposition2.8.IfX≥ 0 is a nonnegative random variable, then

E[X] =

∫∞


0

P(X > x)dx.

The integrand in Proposition 2.8 is called thetail probability function
x7→P(X > x)ofX. This is also known as the complementary cumulative
distribution function ofX. Thecumulative distribution function(CDF) of
Xis defined asx7→P(X≤x)and is usually denoted byFX. These functions
are defined for all random variables, not just nonnegative ones. One can check
thatFX:R→[0,1] is increasing, right-continuous andlimx→−∞FX(x) = 0
andlimx→∞FX(x) = 1. The CDF of a random variable captures every aspect
of the probability measurePXinduced byX, while still being just a function
on the real line, a property that makes it a little more human-friendly thanPX.
One can also generalize CDFs to random vectors: IfXis anRk-valued random
vector then its CDF is defined as theFX:Rk→[0,1] function that satisfies
FX(x) =P(X≤x), where, in line with our conventions,X≤xmeans that all
components ofXare less-than-equal to the respective component ofx.

2.6 Conditional expectation


Conditional expectationallows us to talk about the expectation of a random
variable given the value of another random variable, or more generally, given
someσ-algebra.
Example2.9.Let (Ω,F,P) model the outcomes of an unloaded dice: Ω = [6],
F = 2Ω and P(A) = |A|/6. Define two random variables X and Y by
Y(ω) =I{ω > 3 }andX(ω) =ω. Suppose we are interested in the expectation
ofXgiven a specific value ofY. Arguing intuitively, we might notice thatY= 1
means that the unobservedXmust be either 4, 5 or 6, and that each of these
outcomes is equally likely and so the expectation ofXgivenY = 1 should
be (4 + 5 + 6)/3 = 5. Similarly, the expectation ofXgivenY = 0 should be
(1 + 2 + 3)/3 = 2. If we want a concise summary, we can just write that ‘the
expectation ofXgivenY’ is 5Y+ 2(1−Y). Notice how this is a random variable
itself.
The notation for this conditional expectation isE[X|Y]. Using this notation,
in Example 2.9 we can concisely writeE[X|Y]= 5Y+ 2(1−Y). A little more
generally, ifX: Ω→XandY: Ω→YwithX,Y ⊂Rand|X|,|Y|<∞. Then
E[X|Y] : Ω→Ris the random variable given byE[X|Y](ω) =E[X|Y=Y(ω)]
where

E[X|Y=y] =


x∈X

xP(X=x|Y=y) =


x∈X

xP(X=x,Y=y)
P(Y=y)

. (2.8)


This is undefined whenP(Y=y) = 0 so thatE[X|Y](ω) is undefined on the
measure zero set{ω:P(Y=Y(ω)) = 0}.
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