0% found this document useful (0 votes)
4 views6 pagesLecture 5
The document discusses the concepts of expectation and variance in the context of random variables, using the Jones family's height data as an example. It explains how to calculate the mean and variance through integration and provides properties of expectation. Additionally, it highlights the importance of numerical outcomes for random variables in calculating expectations.
Uploaded by
joshuaboryerCopyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
0% found this document useful (0 votes)
4 views6 pagesLecture 5
The document discusses the concepts of expectation and variance in the context of random variables, using the Jones family's height data as an example. It explains how to calculate the mean and variance through integration and provides properties of expectation. Additionally, it highlights the importance of numerical outcomes for random variables in calculating expectations.
Uploaded by
joshuaboryerCopyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
/ 6
DATA 208 Lecture 5 A
DATA 203 Lecture 5: Expectation
The Jones’ family height data is
hy=18, m=16, Ay=18, hy hy = 17
If we randomly pick a Jones’ 18,
family member, their height is a oa oa
random variable H with p.d.. fu =
The mean height jin is 1G 7B |
1 1 2 1
y= 1GX E417 X SH18x2419x=1.76 (1)
Effectively, this is,
"Sum over all h values of {value x probability that value occurs}”
which can be written in general terms as Ha
mean ie fe. fin(te) dh
where the integration picks up each >
probability mass, multiplies it by
the h value at which it occurs,
and sums them up. This yields (1)
The variance of the Joneses’ heights is of = (Iki — un)” :
which, in words is
"Sum over all A values of {(value - mean)® % probability that value occurs}
which we can write in integral form as
variancx i. (hi)? fir) dh DATA 208 Lacture 5 B
Expectation
For a random variable X with a pdf. fx() we can define the
expectation of a function g(X) as
Bg) = Btg1X)) = fafa). fala) ae
In particular g(X) = X gives
a)= [ x. f(z) de = mean
and g(X) = (X — 2)? gives
B((x-y)) -{ (e—)?.. fx(a) de = variance
# Example: Let X be a
uniformly distributed random
variable (RLV.) with a p.d.f
as shown,
To calculate the mean:
For the variance
B((x-29)= f° (@-2¥ fx) de DATA 208 Loeture 5 ©
-f (x2)? .0de+
(@-9"
6
(2)? 0dr
=04
+0= 3 = variance
lem 3
Basic properties of expectation
(2) E(k) =k for any constant k € R
(b) B(kg(X)) = kE(g(X)) for any constant & € Rand function
aX)
B(g(X) + W(X) = E(g(X)) + E(A(X)) for any functions
and h.
«© For (a) we have
kfe(a) de = af” fla) do =k
EX)
since [%, fxx() dx must equal 1 as
outcome occurs.
«For (b) we have
Beg) = [hate fuley de =k fafa) de = KEL)
© For (c)
the probability that any
Bg(x)+H(3)) = J (ola) +X) fal) dr
-[ 9 (2) Fx) + Rw) fx) dx
= [Oabertale) der J nla) fue) de = Bta(X))+B(H(0) DATA 208 Lecture 5 D
‘* Example: X is a continuous random variable with p.d-f
Pr 2/2 O