Ch. 7.2 Notes : Market Failure. I.5 variables that cause a market system to fail…..
7.2 Day 1: Mean & Variance of Random Variables
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Transcript of 7.2 Day 1: Mean & Variance of Random Variables
7.2 Day 1: Mean & Variance of
Random Variables
Law of Large Numbers
The Mean of a Random Variable
The mean x of a set of observations is their ordinary average, but how do you find the mean of a discrete random variable whose outcomes are not equally likely?
The Mean of a Random
Variable is known as its
expected value.
Ex 1: The Tri-State Pick 3
In the Tri-State Pick 3 game that New Hampshire shares with Maine and Vermont, you choose a 3-digit number and the state chooses a 3-digit winning number at random and pays you $500 if your number is chosen.
Since there are 1000 possible 3 digit numbers, your
probability of winning is 1/1000.
The probability distribution of X (the amount your ticket pays you)
Payoff X: $0 $500
Probability: 0.999 0.001
The ordinary average of the two possible outcomes is $250, but that makes no sense as the average because $0 is far more likely than $500.
In the long run, you would only receive $500 once in every
1,000 tickets and $0 in the remaining 999 of
the tickets
So what is the mean?
The long-run average payoff or mean for this random variable X is fifty cents.
This is also known as the Expected Value.
1 999$500 $0 $0.50
1000 1000
We will say that
μx = $0.50.
Mean of a Discrete Random Variable
Suppose that X is a discrete random variable whose distribution is
Value of X: x1 x2 x3 … xk
Probability: p1 p2 p3 … pk
To find the mean of X, multiply each possible vlaue by its probability, then add all the products
μx = x1p1 + x2p2 + … + xkpk
= Σxipi
We will use μx to signify that this is
the mean of a random variable and not of a data
set.
Ex 2: Benford’s LawCalculating the expected first digit
What is the expected value of the first digit if each digit is equally likely?
μx = 1(1/9) + 2(1/9) + 3(1/9) + 4(1/9) + 5(1/9) + 6(1/9) + 7(1/9) + 8(1/9) + 9(1/9)
= 5
First Digit X 1 2 3 4 5 6 7 8 9
Probability 1/9 1/9 1/9 1/9 1/9 1/9 1/9 1/9 1/9
The expected value is μx = 5.
What is the expected value if the data obeys Benford’s Law?
μx = 1(.301) + 2(.176) + 3(.125) + 4(.097) + 5(.079) + 6(.067) + 7(.058) + 8(.051) + 9(.046)
= 3.441
First Digit X 1 2 3 4 5 6 7 8 9
Probability .301 .176 .125 .097 .079 .067 .058 .051 .046The expected
value is μx = 3.441.
Probability Histogram for equally likely outcomes 1 to 9
In this uniform distribution, the
mean 5 is located at the
center.
Probability Histogram for Benford’s Law
The mean is 3.441 in this right skewed distribution.
Recall…
Computing a measure of spread is an important part of describing a distribution (SOCS)
The variance and the standard deviation are the measures of spread that accompany the choice of the mean to measure center.
Variance of a Discrete Random Variable
Suppose that X is a discrete random variable whose distribution is
Value of X: x1 x2 x3 … xk
Probability: p1 p2 p3 … pk
And that the mean μ is the mean of X. The variance of X is
σx2 = (x1 – μx)2p1 + (x2 – μx)2p2 + … + (xk – μx)2pk
The standard deviation σx of X is the square root of the variance.
We will use σx2 to
signify the variance and σx for the
standard deviation.
Ex 3: Linda Sells Cars
Linda is a sales associate at a large auto dealership. She motivates herself by using probability estimates of her sales. For a sunny Saturday in April, she estimates her car sales as follows:
Cars Sold: 0 1 2 3
Probability: 0.3 0.4 0.2 0.1
Find the mean and variance.
μx = 1.1 σx2 = 0.890
xi pi xipi (xi – μx)2pi
0 0.3 0.0 (0 – 1.1)2(0.3) = 0.363
1 0.4 0.4 (1 – 1.1)2(0.4) = 0.004
2 0.2 0.4 (2 – 1.1)2(0.2) = 0.162
3 0.1 0.3 (3 – 1.1)2(0.1) = 0.361
The standard
deviation is σx = 0.943
The Law of Large Numbers
Draw independent observations at random from any population with finite mean μ.
Decide how accurately you would like to estimate μ.
As the number of observations drawn increases, the mean x of the observed values eventually approaches the mean μ of the population as closely as you specified and then stays that close.
Ex 4: Heights of Young Women(Law of Large Numbers)
The average height of young women is 64.5
in.
The Law of Small Numbers
The law of small numbers does not exist, although psychologists have found that most people believe in the law of small numbers.
Most people believe that in the short run, general rules of probability with be consistent.
This is a misconception because the general rules of probability only exist over the long run.
In the short run, events can only be characterized as random.
How large is a large number?
The law of large numbers does not state how many trials are necessary to obtain a mean outcome that is close to μ.
The number of trials depends on the variability of the random outcomes.
The more variable the outcomes, the more trials that are needed to ensure that the mean outcome x is close the distribution mean μ.