5.5.3 Means and Variances for Linear Combinations
What is called a linear combination of random variables?
aX+b, aX+bY+c, etc. 2X+3 5X+9Y+10
If a and b are constants, then E(aX+b)=aE(X)+b.__________________________
β’ Corollary 1. E(b)=b. β’ Corollary 2. E(aX)=aE(X).
Let π= π0 + π1π+ π2π+ π3π+ β―
Means of Linear Combinations
πΈαΊπα»= π0 + π1πΈπ+ π2πΈπ+ π3πΈπ+ β―
Example X, Y, Z = values of three dice You win π= 5+ 2π+ 3π+ 4π dollars πΈαΊπα»= πΈαΊπα»= πΈαΊπα»= 3.5 Your total expected winnings are πΈαΊπα»= 5+ 2αΊ3.5α»+ 3αΊ3.5α»+ 4αΊ3.5α»= 36.50
Expected value (constant) = constant πΈαΊ5α»= 5
πΈαΊβα»= βαΊππ₯ππππ‘ππ π£πππ’ππ α» πΈαΊ5+ 3π+ 4π+ 5πα»= πΈαΊ5α»+ πΈαΊ3πα»+ πΈαΊ4πα»+ πΈαΊ5πα» πΈαΊππππ π‘πππ‘βπα»= ππππ π‘πππ‘ βπΈπ πΈαΊ3πα»= 3πΈαΊπα» πΈαΊ4πα»= 4πΈαΊπα» EαΊ5πα»= 5πΈαΊπα» πΈαΊπ0 + π1π+ π2π+ π3πα» = πΈαΊπ0α»+ πΈαΊπ1πα»+ πΈαΊπ2πα»+ πΈαΊπ3πα» = π0 + π1πΈπ+ π2πΈπ+ π3πΈπ
Variances of Linear Combinations
πππαΊπ0 + π1π+ π2π+ π3πα» = π12πππαΊπα»+ π22πππαΊπα»+ π32πππαΊπα» when X, Y and Z are independent.
πππαΊπ0α»= 0 πππαΊππππ π‘πππ‘α»= 0
πππαΊπ0 + πα»= πππαΊπα» Adding a constant doesnβt change variance
πππαΊπ1πα»= π12ππππ Variances are in units2.
ππ‘.π·ππ£αΊπ1πα»= Θ&π1Θ& ππ‘.π·ππ£αΊπα»
If X, Y, Z are independent or uncorrelated. πππαΊπ+ π+ πα»= πππαΊπα»+ πππαΊπα»+ πππαΊπα» Putting these facts together πππαΊπ0 + π1π+ π2π+ π3πα» = αΊif independentα» πππαΊπ0α»+ πππαΊπ1πα»+ πππαΊπ2πα»+ πππαΊπ3πα» = π12πππαΊπα»+ π22πππαΊπα»+ π32πππαΊπα» For independent X, Y
ππ+π = ΰΆ₯πππαΊπα»+ πππαΊπα»
ππ+π2 = ππ2 + ππ2
ππ+π = ΰΆ§ππ2 + ππ2
Like Pythagorean Theorem π2 = Θ2 + b2, π= ΞΎπ2 + π2
Most useful facts:
If X1, X2, X3, β¦ , Xn are independent measurements each with πΈ(π1) = π and πππαΊπ₯πα»= π2 then
πΈαΊπΰ΄€α»= π πππαΊπΰ΄€α»= π2π ππ‘.π·ππ£αΊπΰ΄€α»= πΞΎπ
To see this:
πΈΰ΅¬1π β ππΰ΅°= 1π πΈαΊβ ππα»= 1π β πΈαΊππα»= 1π ππ
π=1 = 1παΊππα»= π
The sample mean πΰ΄€ is an unbiased estimator of the population mean π.
πππ࡬1π β ππΰ΅°= ࡬1πΰ΅°2 πππαΊβ ππα»= 1π2 πππαΊππα»= 1π2 π2π
π=1 = 1π2αΊππ2α»= π2π
5.5.5 The Central Limit Effect
β’ If X1, X2,β¦, Xn are independent random variables with mean m and variance s2, then for large n, the variable is approximately normally distributed.
If a population has the N(m,s) distribution, then the sample mean x of n independent observations has the distribution.
For ANY population with mean m and standard deviation s the sample mean of a random sample of size n is approximately
when n is LARGE.( , )Nn
sm
( , )Nn
sm
x
Central Limit Theorem
If the population is normal or sample size is large, mean follows a normal distribution
and
follows a standard normal distribution.
x( , )N
ns
m
n
xxzx
x
sm
sm
β’ The closer xβs distribution is to a normal distribution, the smaller n can be and have the sample mean nearly normal.
β’ If x is normal, then the sample mean is normal for any n, even n=1.
β’ Usually n=30 is big enough so that the sample mean is approximately normal unless the distribution of x is very asymmetrical.
β’ If x is not normal, there are often better procedures than using a normal approximation, but we wonβt study those options.
β’ Even if the underlying distribution is not normal, probabilities involving means can be approximated with the normal table.
β’ HOWEVER, if transformation, e.g. , makes the distribution more normal or if another distribution, e.g. Weibull, fits better, then relying on the Central Limit Theorem, a normal approximation is less precise. We will do a better job if we use the more precise method.
If X1, X2, β¦ , Xn are normal, then πΰ΄€ is exactly normal.
Example: Bags of potatoes weigh
π= 5.0 pounds
π= 0.1 pounds
If we buy 4 bags, what is PαΊπΰ΄€< 4.9α»?
πΈαΊπΰ΄€α»= π= 5 ΰΆ₯πππαΊπΰ΄€α»= πΞΎπ = 0.1ΞΎ4 = 0.05
PαΊπΰ΄€< 4.9α»= Pαπΰ΄€β πΈπΰ΄€ΰΆ₯VΘrαΊXΰ΄₯α»< 4.9β 5.00.05 α= PαΊπ§< β2α»= 0.0228
Exampleβ’ X=ball bearing diameterβ’ X is normal distributed with m=1.0cm and s=0.02cmβ’ =mean diameter of n=25β’ Find out what is the probability that will be off by
less than 0.01 from the true population mean.
%76.98)5.25.2(
)004.0
0.101.1004.0
0.199.0()01.199.0(
004.02502.0
0.1
zP
zPxP
nx
x
ss
m
xx
Exercise
The mean of a random sample of size n=100 is going to be used to estimate the mean daily milk production of a very large herd of dairy cows. Given that the standard deviation of the population to be sampled is s=3.6 quarts, what can we assert about he probabilities that the error of this estimate will be more then 0.72 quart?