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AMS 5
CHANCE VARIABILITY
The Law of Averages
When tossing a fair coin the chances of tails and heads are the
same: 50% and 50%. So if the coin is tossed a large number of
times, the number of heads and the number of tails should be
approximately equal. This is the law of averages.
The number of heads will be off half the number of tosses by
some amount. That amount is called chance error. The chance
error increases with the number of tosses in absolute terms, but
it decreases in relative terms.
The Law of Averages
Q: A coin is tossed and you win a dollar if there are more than
60% heads. Which is better: 10 tosses or 100?
A: 100 tosses is better. As the number of tosses increase you are
more likely to be close to 50%, according to the law of
averages.
Q: Same as before, but you win one dollar if there are exactly
50% heads.
A: 10 tosses is better, since in absolute terms you are more likely
to be off the expected value when the number of tosses is
large.
Q: 100 tickets are drawn with replacement from one of two
boxes: one contains two tickets with -1 and two with 1. The
other contains one ticket with -1 and one with 1. One hundred
tickets will be drawn at random with replacement from one of
them and the amount on the ticket will be paid to you, which
box do you prefer?
A: In this case the expected payoff is the same for both boxes,
since they both have 50% 1 and 50% -1.
Random Variables
Chance Processes
Variables whose results depend on the
outcome of a random experiment
Random Variables
Example 1: Toss a coin 1000 times and report
the number of heads. If you repeat the
experiment the number of heads will turn out
differently.
Example 2: The amount of money won or lost at
roulette.
Example 3: The percentage of Democrats in a
random sample of voters.
Box Model
• Find an analogy between the process being studied (e.g.
sampling voters) and drawing numbers at random from a box.
• Connect the variability you want to know about (e.g. estimate
for the Democratic vote) with the chance variability in the sum of
the numbers drawn from the box.
The analogy between a chance process and drawing from a box
is called box model.
Consider a box model for a roulette. A roulette wheel has 38
pockets. 1 through 36 are alternatively colored red and black,
plus 0 and 00 which are colored green. So, there are 18 red
pockets and 18 black ones. Suppose you win $1 if red comes out
and loose $1 if either a black number or 0 or 00 come out. Your
chance of winning is 18 to 38 and your chance of loosing is 20 to
38. A box representation is
Box Model
Suppose now that you bet a dollar on a single number and that, if
you win, you get your $1 plus $35, but you loose your $1 if any
other number comes up. A box model of this bet is given by
What is your net gain after 100 plays? This correspond to the sum
of 100 draws made at random with replacement from the above
box. To calculate this amount we need the concept of Expected
value.
Expected Value and Standard Error
A chance process is running. It delivers a number. Then we rerun it and it delivers another number, and so on. The numbers
delivered by the process vary around the expected value, the
amounts off being similar in size to the standard error.
Example: Count the number of heads in 100 tosses of a coin.
The expected value of the number of heads is 50. You actually
toss the coin and the results are:
• 57 heads, you are off by 7
• 46 heads, you are off by -4
• 47 heads, you are off by -3
The amounts off are similar in size to the standard error.
The expected value and the standard error depend on the
random process that generates the numbers.
Expected Value and Standard Error
Consider the box
and draw a ticket at random with
replacement 100 times. What is the expected value of the sum of
the tickets?
The chance of a 1 is 75% and the chance of a 5 is 25%. So we
expect to see 25 x 5 + 75 x 1 = 200.
Notice that this number is equal to 100 x 2 = 200 which is the
number of draws times the average number in the box. As a
general rule we have that the expected value is given by:
Expected Value and Standard Error
Q: Suppose you play Keno, a game where you win $2 and pay $1
to play. You have 1 chance in 4 to win. How much should you
expect to win if you play 100 times?
A: A box representation of the game is given by
so the average in the box is: ($2- 3 x $1)/4=-$0.25.
So you are expected to `win' -$25. Of course, if you keep playing
you are expected to loose more money!
Q: Consider the box
and suppose 25 draws with replacement are made from the box.
What is the expected sum?
A: Each number should appear 1/5 of the time, that is 5 on
average. So the expected value of the sum is
5 x 0 + 5 x 2 + 5 x 3 + 5 x 4 + 5 x 6 = 75 = 25 x 3.
Expected Value and Standard Error
In the last example we will not see each ticket appearing exactly 5
times. The actual sum we observe will be off by a chance error
sum = observed value = expected value + chance error
The standard error gives a measure of how large the chance error
is likely to be. When drawing at random with replacement from a
box we can calculate the standard error for the sum of the draws
as:
where SD of box stands for the standard deviation of the list of
numbers in the box. Notice that the SE increases as the number of
draws increases, but only by a factor equal to the square root of
the number of draws.
Expected Value and Standard Error
Expected Value and Standard Error
Can we make the previous statements more precise?
Consider again the box with the five numbers. We know that in 25
draws the expected value is 75 and the SE is 10. Also, we observe
that in 25 draws the sum ranges from 0 to 150 (all 0 through all
6).
What are the chances that the sum will be between 50 and 100?
To answer this question we observe that:
50 - 75 = -25 = -2.5 x 10 and 100 – 75 = 25 = 2.5 x 10
So that 50 and 100 are 2.5 times SEs away from the expected
value. We say that 25 is 2.5 standard units.
Expected Value and Standard Error
We can apply the following approximation that will be better
justified by the use of the Central Limit Theorem:
• 68% of the draws will be within one standard unit of the
expected value.
• 95% of the draws will be within two standard units of the
expected value.
• 99% of the draws will be within 2.5 standard units of the
expected values.
For our example the ranges we get for one, two and 2.5
standard units are 75 ± 10, 75 ± 20 and 75 ± 25.
Expected Value and Standard Error
Example: Suppose there are 10,000 independent plays on a
roulette wheel in a casino. Suppose all plays are of $1 on red at
each play. What are the chances that the casino will win more
than $250 from these plays?
The box model for this problem is
The expected value of the casino's net gain is the average:
$20 − $18
≈ $0.05
38
times the number of plays 10000 x $0.05 = $500.
So the casino expects to win $500 a month.
Expected Value and Standard Error
We now need to calculate the SD. We have 20 deviations of 0.95
and 18 deviations of -1.05 from the average. So
and being conservative we can approximate this number to 1. So
the SE is approximately 10000 = 100.
$250 is 2.5 SE units from the expected gain of $500, so we see
that there is 99% chance that the net gain for the casino will be
between $250 and $750.
Probability Histograms
Consider the box:
Then, the chances of
obtaining a ticket with a 1
are 4/7, the chances of a
3 are 1/7 and the chances
of a 4 are 2/7. We can
display that information
graphically in a probability
histogram.
Each box is centered at a number and its area corresponds to the
probability of that number. The sum of the areas of the boxes is
equal to one. This histograms are used to represent chance and
not the frequency of data.
Probability Histograms
Empirical
histograms based
on the
frequencies of
observed
outcomes of an
experiment
converge to the
corresponding
probability
histograms, as
can be seen by
the example of
rolling two dice.
Probability Histograms
Consider now
taking the product
of the two dice.
The convergence
is also true then,
although the
probability
histogram is much
more irregular
than the one
obtained for the
sum. The
regularity is a
general feature
related to the
sum.
Probability Histograms
Consider the problem
of tossing a fair coin a
certain number of
times n. We can
obtain the probability
histogram for each n.
We observe that
the probability
histogram of the
number of tails converges to a very
regular curve (the
NORMAL) as the
number of tosses is
increased.
Normal Approximation
We can approximate the probability histogram of the sum of
heads in a large number of coin tosses using the normal curve.
Example: A coin is tossed 100 times, what is the probability of
getting:
a) exactly 50 heads?
b) between 45 and 55 heads inclusive?
c) between 45 and 55 heads exclusive?
Solution:
a) We can look at the probability histogram for this case. We
observe that the chances corresponding to 50 are equal to the
area of the box that has a base from 49.5 to 50.5. The area of
this box is 7.96%.
Normal Approximation
What about an approximation using the normal curve?
First step is to calculate the mean and standard deviation.
Consider a box model where there is a zero for the tail and 1 for
the head,
We draw a ticket from this box 100 times
The expected number of heads is
100 × ½ = 50 and the standard error is given by the square root
law 100 × 1/ 2 = 5. Now we have to convert to standard units:
49.5 − 50
50.5 − 50
= −0.01 and
= 0.01
5
5
So the normal approximation consists of the area under the
normal curve for the interval (-0.1,0.1). According to the normal
table, this is equal to 7.97%.
Normal Approximation
b) The probability of getting between 45 and 55 heads is equal
to the areas of the rectangles between 45 and 55 in the
probability histogram. This is approximated by the area under the
normal curve for the interval (44.5,55.5). In standard units this
corresponds to the interval (-1.1,1.1), which has a probability of
72.87% according to the table.
c) This time the probability is given by the areas of the
rectangles between 46 and 54, which is approximately the area
under the curve corresponding to the interval (45.5,54.5), this is
the interval (-0.9,0.9) in standard units, which has a probability
of 63.19%.
Very often it is not specified if the end points are included or not.
In that case we consider the approximation using the given
interval. So, for the previous example, we would have (45,55)
that is converted to (-1,1) in standard units and yields 68.27%
probability.
When can we use the normal
approximation
Consider the box
. Then the probability histogram
for the tickets in the box is far from being normal.
Nevertheless, if we consider the
experiment of drawing tickets from
the box and sum the results over
and over again, then the probability
histogram of the sum will be
approximated by the normal curve.
What if we consider the product of the tickets? In that case the
probability histogram will not be approximated by a normal curve,
no matter how many draws from the box we take.
The Central Limit Theorem
In general it is true that the probability histogram of the sum of
draws from a box of tickets will be approximated by the normal
curve. This is a mathematical fact that can be expressed and
proved as a theorem.
The reason why the CLT is used as an approximation for
distributions of lists of numbers is that it often happens that the
uncertainty in the data can be thought of as the sum of several
sources of randomness.