Notes, solutions, and commentary on Introduction to Mathematical Statistics by Robert Hogg, Joseph Mckean, and Allen Craig.
1.1 Introduction
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1.1.1 Sample Spaces and Experiment
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Many kinds of investigations involve repeated experiment under approximately the same conditions.
All experiments terminate with an outcome.
A key property of these experiments is that the outcome of the experiment cannot be predicted with certainty.
A collection of all possible outcomes is called a Sample Space.
An experiment that can be repeated over and over under the same conditions for which we know the sample space is called a Random Experiment.
Ex 1.1.1
A coin toss is a random experiment because we know the Sample Space, C = {H, T} where H is heads and T is tails, before performing the experiment and we can repeat the experiment, that is flipping the coin over and over under approximately the same conditions.
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1.1.2 Relative Frequency
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During experiments we are often interested in chances of subsets of elements of the sample space, such subsets are called events.
For instance in the coin toss example we may be interested in the chance that the subset A = {H} occurs, that is what is the chance we get heads when flipping the coin.
If an event is repeated N times we can count the number of times f an event occurs, that is the frequency of the event, and f/N, the ratio of the frequency of the event and the number of experiments, is called the relative frequency of that event.
Relative frequencies are often erratic for small numbers of N and more stable for large numbers of N.
For large numbers of N we associate the number p to the relative frequency known as the probability, that is when the relative frequency becomes stable we consider it the probability of the event.
Ex 1.1.3
Let C denote the sample space of a red and white die, that i C = {(1,1),...,(1,6),(2,1),...,(2,6),...,(6,6)}.
Let B be the set of every ordered pair of C for which the sum of the pair is equal to seven, that is,
B = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}.
We cast the dice N=400 times and get a pair of seven 60 times, then 60 is the frequency of a pair of seven.
The relative frequency of an outcome in B is f/N = 60/400 = 0.15.
If we associate a number p with B then we say that the probability of an even B is 0.15.
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1.1.3 Interpretations of Probability
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The purpose of a theory of mathematical statistics is to provide mathematical models of random events.
Once the model of a system has been established and the theory worked out inferences can be made.
The use of relative frequency as probability is known as the Relative Frequency Approach.
The Relative Frequency Approach is a interpretation of Probability that relies on the idea that a experiment can be repeated under identical conditions which could be viewed as an idealistic condition.
Other interpretations of probability consider it to be a rational measure of belief.
Under the Rational Belief interpretation a statement like p = 2/5 for an event A means that the personal probability of an even A is equal to 2/5.
The rational belief interpretation can be viewed as the willingness of a person to bet on an outcome given a consistent measure of previous events.
Different interpretations of probability are usually equivalent, that is you get the same answers in the end, what differs are how you philosophically justify those answers.
Relative Frequency.
Say we cast the dice N/400 with the same result with identical conditions and get f/n=60/400 =0.15
Then we changed the conditions to something more or less extreme with identical conditions N/400
Will we get f/n=60/400=0.15 again ?
I'm guessing it would have to be done before we can write something like f/n=45/400, whatever the result would be. If the experiment is identical then must have a stable frequency.
I wouldn't go with a pair of seven cause I'm not sure how that works with 6 sided dice. Maybe with 4 dice, but that would at least decrease the odds when rolling a pair of 7 in a single roll, but that shouldn't matter, it's the result that determins the relative frequency.
Maybe I don't understand that part, so I'll settle for a pair of 6. Probably scaled down from the OP model. I should probably use 1 dice for starters.
This is good for short term memory, but that's the best I can do for now.
1.2 Sets
A set is a collection of objects.
The objects we are most interested in are numbers, as such the sets we are interested in are sets of numbers.
An object in a set is known as an element.
If a set is countable then that set is finite.
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1.2.1 Review of Set Theory
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Definitions : :
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Definition 1.2.1 (Component)
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The complement of an event A is the set of all elements in C which are not in A. We denote the complement of A by A^c. That is A^c = {x∈C : x∉A}.
Ex. If C = {1,2,3,4,5,6,7,8,9,10} and A = {1,2,3,4,5} then A^C = {6,7,8,9,10}
Definition 1.2.2 (Subset)
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If each element of a set A is also an element of set B, the set A is called a subset of the set B. This is indicated by writing A ⊂ B. If A ⊂ B and also B ⊂ A, the two sets have the same elements, and this is indicated by writing A = B.
Ex. If A = {1,2,3} and B = {1,2,3,4,5} then A ⊂ B because all elements in A can be found in B.
Ex. If A = {1,2,3} and B = {1,2,3} then A ⊂ B and B ⊂ A implying A = B because all elements in A are found in B and all elements in B are found in A therefore both A and B have the same elements.
Definition 1.2.3 (Union)
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Let A and B be events. Then the union of A and B is the set of all elements that are in A or in B or in both A and B. The union of A and B is denoted by A∪B.
Ex. If A = {1,2,3,4,5} and B = {6,7,8,9,10} then A∪B = {1,2,3,4,5,6,7,8,9,10}
Definition 1.2.4 (Intersection)
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Let A and B be events. Then the intersection of A and B is the set of all elements that are in both A and B. The intersection of A and B is denoted as A∩B.
Ex. If A = {1,2,3,4,5,6,7} and B = {6,7,8,9,10,11,12} then A∩B = {6,7}
Definition 1.2.5 (Disjoint)
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Let A and B be events. Then A and B are disjoint if A∩B = ∅.
Ex. If A = {1,2,3,4,5} and B = {6,7,8,9,10} then A∩B = ∅ because A and B share no element B.
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Properties of Sets : :
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Properties 1.2.1 (Distributive Laws)
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
A ∪ (B ∩ C) = (A u B) ∩ (A u C)
Properties 1.2.3 (DeMorgan's Law)
(A ∩ B)^C = A^C ∪ B^C
(A ∪ B)^C = A^C ∩ B^C
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Quantifiers : :
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The Universal Quantifier ∀ is a symbolic representation of the phrases 'for every' or 'for all'.
In ∀xP(x), where P(x) is some proposition, the universal quantifier states that the proposition is true if and only if it is true for any value you substitute for x.
Ex. ∀x(x² >= 0) states that whatever value for x you plug into the expression x² the result will be greater than or equal to zero.
The Existential Quantifier ∃ is a symbolic representation of the phrase "there exists".
In ∃xP(x), where P(x) is some proposition, the existential quantifier states that the proposition is true if and only if there is at least one value of x that makes P(x) true.
Ex. ∃x(x>=x²) is true given x=0 is a solution.
Notice that quantifiers decide the domain of what makes something true or not, sometimes we care about if something is true for any and all cases while other times we care if it is true sometimes.
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Indexed Sets : :
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Many problems that utilize the notion of set are not finite, as such we need set theoretical objects that allows use to deal with non-finite operations or at the least very large ones.
Definition 1.2.8 (Indexed Sets)
If I is a set, called an index set, where i ∈ I we can associate a set Aᵢ. The set {Aᵢ:i∈I} is called an Indexed Set.
To represent unions greater than two we use the following notation,
∪Aᵢ = {x|x∈Aⱼ ∃j∈I}
Notice the use of the existential quantifier, this appropriately describes the union of the indexed set Aᵢ because the condition of membership is to be in at lest one of the sets Aᵢ.
To represent intersections greater than two we use the following notation,
∩Aᵢ = {x|x∈Aⱼ ∀j∈I}
Notice the use of the universal quantifier, this appropriately describes the intersection of the indexed set Aᵢ because the condition of membership is to be in all of the sets Aᵢ.
Ex. i = {1,2,3}, A₁ = {0,1,2,3,4,5}, A₂ = {-3, -2, -1, 0, 1, 2}, A₃ = {-4, -2, 0, 2, 4}
∪ᵢ Aᵢ = {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5} (that is all elements in at least one of the sets)
∩ᵢ Aᵢ = {0, 2} (that is elements in all sets)
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Sequences of Sets : :
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Definition 1.2.9 (Increasing Sequence)
A sequence is called increasing if aₙ <= aₙ₊₁ ∀n ∈ N
Definition 1.2.10 (Decreasing Sequence)
A sequence is called decreasing if aₙ >= aₙ₊₁ ∀n ∈ N
A sequence is called Monotone if it is decreasing or increasing.
Sequences of sets, denoted as {Aₙ} are sometimes monotone.
{Aₙ} is increasing if Aₙ ⊂ Aₙ₊₁ ∀n∈N
For an increasing {Aₙ} we define lim Aₙ = ∪ Aₙ
{Aₙ} is decreasing if Aₙ₊₁ ⊂ Aₙ ∀n∈N
For a decreasing {Aₙ} we define lim Aₙ = ∩ Aₙ
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1.2.2 Set Functions : :
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Many functions used in calculus are functions that map real numbers into real numbers.
We are also concerned with functions that map sets into real numbers.
Functions that map sets to real numbers are called functions of a set.
Ex. Let C = R, the set of real numbers. For a subset A in C, let Q(A) be equal to the number of points in A that correspond to positive integers. Then Q(A) is a set function of the set A. Thus if A = {x:0<x<5} then Q(A)=4; if A={-2,-1}, then Q(A)=0; and if A={x:-∞<x<6} then Q(A)=5.
Relative Frequency.
Say we cast the dice N/400 with the same result with identical conditions and get f/n=60/400 =0.15
Then we changed the conditions to something more or less extreme with identical conditions N/400
I don't know what you mean by N/400 given N is the number of times an experiment is performed.
Will we get f/n=60/400=0.15 again ?
I'm guessing it would have to be done before we can write something like f/n=45/400, whatever the result would be. If the experiment is identical then must have a stable frequency.
If i rolled two die again I probably would not get 60/400 and therefore the probability of obtaining a sum of 7 from a my dice roll probably is not 0.15.
This is true regardless of exact ideal conditions during a repeated experiment.
Think in terms of a coin toss. If I flipped a coin 1 time and it came up heads then f/n = 1/1 = 1 and 1 implies that the coin should come up heads every single time but we know that is silly. If the coin is fair then we know the probability of heads is 0.5. The greater the number N the closer the actual probability of heads will be 0.5.
Look how the relative frequency (probability) of heads changes as we increase the number N.
N = 10
N = 50
N= 10000
I wouldn't go with a pair of seven cause I'm not sure how that works with 6 sided dice. Maybe with 4 dice, but that would at least decrease the odds when rolling a pair of 7 in a single roll, but that shouldn't matter, it's the result that determins the relative frequency.
Maybe I don't understand that part, so I'll settle for a pair of 6. Probably scaled down from the OP model. I should probably use 1 dice for starters.
This is good for short term memory, but that's the best I can do for now.
P₆ = {(5,1),(4,2),(3,3),(1,5),(2,4)}, so there are actually only 5 ways to get a sum of 6 while there is 6 ways to get a sum of 7, hence betting on seven is technically better.
There is a trade off between adding more dice because despite there now being a greater number of ways to obtain any given sum there is also a much larger set of possible outcomes as a whole. I won;t do the calculations now but I wonder how probabilities of different sums of pairs scales with increasing dice.
it's the result that determins the relative frequency.
if we consider f/N where it is the number of occurrences of an even and the number of samples that determines the relative frequency.
Infinite samples (N -> ∞) is what gives us our absolute probability of an event, that is a coin flip is 0.5 heads and 0.5 tales.
Relative Frequency.
Say we cast the dice N/400 with the same result with identical conditions and get f/n=60/400 =0.15
Then we changed the conditions to something more or less extreme with identical conditions N/400
I don't know what you mean by N/400 given N is the number of times an experiment is performed.
Yeah this > N=400
Will we get f/n=60/400=0.15 again ?
I'm guessing it would have to be done before we can write something like f/n=45/400, whatever the result would be. If the experiment is identical then must have a stable frequency.
If i rolled two die again I probably would not get 60/400 and therefore the probability of obtaining a sum of 7 from a my dice roll probably is not 0.15.
This is true regardless of exact ideal conditions during a repeated experiment.
Think in terms of a coin toss. If I flipped a coin 1 time and it came up heads then f/n = 1/1 = 1 and 1 implies that the coin should come up heads every single time but we know that is silly. If the coin is fair then we know the probability of heads is 0.5. The greater the number N the closer the actual probability of heads will be 0.5.
Is that with identical conditions as well ?
Look how the relative frequency (probability) of heads changes as we increase the number N.
N = 10
N = 50
N= 10000
I wouldn't go with a pair of seven cause I'm not sure how that works with 6 sided dice. Maybe with 4 dice, but that would at least decrease the odds when rolling a pair of 7 in a single roll, but that shouldn't matter, it's the result that determins the relative frequency.
Maybe I don't understand that part, so I'll settle for a pair of 6. Probably scaled down from the OP model. I should probably use 1 dice for starters.
This is good for short term memory, but that's the best I can do for now.
P₆ = {(5,1),(4,2),(3,3),(1,5),(2,4)}, so there are actually only 5 ways to get a sum of 6 while there is 6 ways to get a sum of 7, hence betting on seven is technically better.
Alice I don't get it. ( Those graphs make life easier ) But the part I don't get...
.
Wait I see, had to count them. Probability 6 P₆={(5,1),(4,2),(3,3),(1,5),(2,4)} each adds up to 6. And each are the possible outcomes. I was thinking 6 sided dice, but this isn't the same.
Oh wait, the B was assigned, but that isn't the case this time. ( I'm writing my thoughts to show you how amature I am are getting this. I'm not quick at it. )
There is a trade off between adding more dice because despite there now being a greater number of ways to obtain any given sum there is also a much larger set of possible outcomes as a whole. I won;t do the calculations now but I wonder how probabilities of different sums of pairs scales with increasing dice.
Got it.
In the previous result of f/n=60/400=0.15 ( let's say stable ) would we be able to simply use addition and make it
f/n=240/1600=0.60 ? Or is that cheating ? ( I have my doubt about this, maybe there's something I don't see )
it's the result that determins the relative frequency.if we consider f/N where it is the number of occurrences of an even and the number of samples that determines the relative frequency.
Infinite samples (N -> ∞) is what gives us our absolute probability of an event, that is a coin flip is 0.5 heads and 0.5 tales.
Is there a machine that can say, flip a coin with the exact identical conditions ( down to the coin's exact starting point) and achieve the same result at various frequencies ? ?
This took long to write, so much thinking. It's easier to get it done faster eh ?