November 15, 2023 • 17 mins
In this engaging episode of "The Magnificence of Mathematics," Eddie Kingston takes you on a journey through the fascinating world of probability. From the roll of dice to the complexities of real-world events, understand how probability shapes our understanding of chance and uncertainty. Eddie breaks down key concepts and theories, making them accessible and exciting. Whether you're a statistician at heart or just curious about how probability impacts our daily decisions, this episode has something for everyone.
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Episode Transcript
Available transcripts are automatically generated. Complete accuracy is not guaranteed.
(00:00):
Hello everyone, welcome to the Magnificence of Mathematics. I'm your host, Eddie Kingston.
(00:15):
Today we'll be talking about my favorite subfield of math, probability. You're most
likely already intuitively familiar with probability, say when you're flipping a coin. You know
that there's at least roughly a 50% chance of the coin landing on heads, and a 50% chance
of the coin landing on tails. But here's a head scratcher. Is it true that events that
(00:36):
have a 0% probability of happening can still happen? Obviously not, right? Anything with
a 0% chance of happening can't happen. You can't randomly choose dragon in a game of
rock paper scissors. Well, it might surprise you that the answer to the question I just
posed is in fact, yes. Let's build up to this. Suppose you randomly pick a number either
(00:58):
1 or 2 with equal probability. That is, the probability of picking the number 1 is 1 half,
and the probability of picking the number 2 is 1 half. Now suppose you randomly pick
a number either 1, 2, or 3 with equal probability. Then the probability of picking the number
1, or 2, or 3 is 1 third, which is smaller than 1 half. If you randomly pick a number
(01:23):
1, 2, 3, or 4 with equal probability, the probability of picking the number 1 is 1 over
4, which again is smaller than 1 third and 1 half. At this point, you might have noticed
two things. First, the probability of picking the number 1 from the numbers 1, 2, 3, all
the way up to some positive whole number n is 1 over n, since there are n numbers to
(01:46):
choose from. Second, this probability gets smaller as n gets larger. Now, what if you
wanted to choose a number from the set of all positive integers 1, 2, all the way up
to infinity? Since there are an infinite amount of numbers to choose from, the probability
of choosing the number 1, or any number for that matter, is 1 over infinity, which is
(02:09):
just 0. Wait, huh? How could it be 0 when we're still choosing a number? Well, if the
probability in question had been anything other than 0, there would be some number n
that describes the amount of numbers we're choosing from. But we just said that we're
choosing from the infinite list of positive integers, so this can't be the case. Therefore,
(02:31):
we literally have a 0% chance of choosing any positive integer whatsoever, but we still
do choose a number. Argument I just gave is an albeit hand-wavy proof by contradiction,
and the underlying idea behind all of this is the idea of a limit, which is one of the
fundamental building blocks of calculus and beyond. In a similar vein, it's also true
(02:55):
that events with 100% probability don't necessarily happen. Suppose you want to avoid choosing
the number 1 from the numbers 1, 2, and 3. The probability of avoiding the number 1 is
2 thirds, because you have two favorable outcomes out of three possible outcomes. If you want
to avoid choosing the number 1 from 1, 2, 3, and 4, that probability is 3 fourths, which
(03:19):
is greater than 2 thirds, because you have three favorable outcomes out of four possible
outcomes. Then for any positive integer n, the probability of avoiding the number 1 from
the numbers 1, 2, up to n is n minus 1 divided by n, which gets larger as n gets larger.
You might notice that this probability gets closer and closer to 1, but never actually
(03:44):
reaches it unless you have the infinite set of all positive integers, and want to avoid
choosing the number 1. So the probability of avoiding the number 1, or any number for
that matter, is 1, but we still don't avoid picking a number altogether. How cool is that?
Want another counterintuitive way probability works? You might have heard of the famous
(04:05):
Monty Hall problem inspired by Let's Make a Deal host Monty Hall, who ran the show from
1963 until 1987. If you're unfamiliar with the show or the problem itself, imagine you're
put in front of three doors. The host Monty Hall tells you that behind one of the doors
is a brand new car. Behind the other two doors are goats. Your task is to pick the door the
(04:27):
car is hiding behind. Let's say you pick door 1. Monty then opens door 3, revealing a goat.
Now Monty gives you a choice. Do you want to stick with door 1, or change your door
to door 2? The problem is, given all this information, would it be in your best interest
to switch, or to stick with your original choice? Your knee-jerk reaction might be that
(04:49):
it doesn't matter whether you stick with your first choice or switch, since now there's
only two doors the car could be behind and it's a 50-50 chance. But it turns out that
that's wrong. There is a 2 in 3 chance that switching will net you the car. To explain
this, it might be easier to think about the game as if there were 100 doors to choose
from, 99 goats and 1 car. If you pick door 36, and Monty opens every door except door
(05:14):
36 and door 67, at this point you might be inclined to think, yeah, I'm switching. There's
only a 1 in 100 chance I was right to begin with. I probably guessed wrong, the car is
actually behind door 67. And you're right. Given the other 98 doors that have opened,
the probability that the door you first chose was right all along is only 1 out of 100,
(05:36):
meaning the probability that door 67 was the right door is 99 out of 100. If we scale this
back to the original problem, the probability that the door you first chose was right all
along is 1 out of 3, meaning the probability that the car is behind the other unopened
door is 2 out of 3. If you think about randomly choosing 1 door out of 3 or 100 or however
(05:58):
many as an experiment, this paradox became a thing because when thinking about probabilities
and what would be better to do, people thought about the experiment as if the start of the
game didn't happen. They thought of the experiment as having started at the point where they
have to choose to either stick with their door or change it, which is where they get
50% from. But if you think about the experiment as starting when you first choose a door,
(06:21):
then you only have a 1 in 3 or 1 in 100 or 1 in however many chance of choosing the right
door on the first try. This is all well and good, but where did any of this come from?
Why study probability in the first place? Probability came from the desire to quantify
uncertainty, in particular when it comes to gambling. In the mid 1600s, Gerolamo Cardano
(06:44):
showed the use of defining odds as the ratio of the number of favorable outcomes to the
number of unfavorable outcomes. So for example, 1 to 1 odds means there is 1 favorable outcome
and 1 unfavorable outcome, and around the same time, French mathematicians Pierre de
Fermat and Blaise Pascal wrote to each other and laid the foundations of probability as
(07:05):
we know it. In particular, Blaise Pascal is famous for what's known as Pascal's Triangle,
which tells you how many ways you can pick a certain number of items from a larger set
of items. The reason it's called a triangle is because you can write out these numbers
as if you were building a side of a pyramid, i.e. a triangle from the top down. For example,
(07:25):
the top of the triangle, called row 0, is just 1 because there is only one way to pick
0 items from 0 items. That's just doing nothing. Similarly, the row below, row 1, has a 1 on
the left and a 1 right next to it, because again, there's only one way to pick 0 items
from a set of 1 item, which is to just do nothing, and one way to pick 1 item from that
(07:48):
set, which is just picking that item. Now for row 2, there's a 1 on either end because
there's only one way to pick 0 items from a set of 2, but not doing anything, and only
one way to pick 2 items from a set of 2 by picking both items, and a 2 in the middle
because there's 2 ways to pick 1 item from a set of 2, picking one thing or the other.
(08:10):
For row 3, there's once again a 1 on either end, one way to pick nothing and one way to
pick everything, and 2 3s in the middle. 3 ways to pick 1 object, and 3 ways to pick
2 objects. That is, 3 ways to leave 1 object out. Let's recap. For row 0, we have a 1.
For row 1, we have a 1 and a 1, which would sandwich the 1 from row 0 if row 0 was brought
(08:34):
down. For row 2, we have a 1, 2, and 1, with the 2 sandwiched between the 2 1s from row
1. Row 3 reads 1 3 3 1, and so on. If you write these out, or just look up the triangle,
you might notice that if you take any two adjacent numbers in a certain row, and look
at the number in the row below that's sandwiched between those two numbers, the sandwiched
(08:57):
number is the sum of the two numbers directly above it. So if you look at the left side
of row 2 and row 3, you see a 1 and 2 in row 2, a 1 on the far left in row 3, and a 3 between
the 1 and the 2 in row 2, and 1 plus 2 is 3. Another way these are read is in the choose
notation. So row 2 can be read as 2 choose 0, which equals 1, 2 choose 1, which equals
(09:21):
2, and 2 choose 2, which equals 1. 2 choose 1, for example, can be interpreted as, from
a set of two objects, choose one of them. This triangle is crucial for beginning to
understand the field of combinatorics, which is what mathematicians fancily call counting.
If you were to take a beginning college level probability class, you'd likely encounter
(09:42):
basic combinatorics at the beginning of the class, and be able to answer questions like,
if you have a hand of 5 standard playing cards, what's the probability of having 2 aces? To
answer this, you want 2 of the 4 aces from the deck in your hand, so there's 4 choose
2, which is 6 ways to choose the aces for your hand. Now you don't want any more aces,
(10:04):
so you have to pick 3 non-ace cards out of the remaining 48 possible cards. So you have
48 choose 3, which equals 17,296 ways to pick the remaining cards. Multiplying these together
means you have 103,776 favorable outcomes. Since you have 52 cards, and you choose 5
(10:25):
of them for your hand, you have 52 choose 5, which equals 2,598,960 total possible outcomes.
So the answer we're looking for is the number of favorable outcomes divided by the number
of possible outcomes, giving us 103,776 divided by 2,598,960, which comes out to be slightly
(10:51):
less than 4%. If you've ever played poker or powerball or something, this is how the
probability of certain hands or certain amounts of numbers matched are calculated. If I haven't
broken your brain by this point, I want to talk a bit about how mathematicians think
of probability. Two of the famous mathematicians to get the ball rolling are André Marcot
and André Kolmogorov in the early 1900s. Marcot was largely responsible for the development
(11:16):
of stochastic processes, which are sequences of random events that happen over time. Going
back to our coin flipping example, suppose you play a game where you win $1 every time
the coin lands on heads, and you lose $1 every time the coin lands on tails. The amount of
money you win or lose is a stochastic process. You can map out how the game goes over time,
(11:36):
and you'll get a different map every time you play the game, assuming you play enough
of course. Stochastic processes can be used to model all kinds of things like wait times
and even stock prices. Kolmogorov established the foundations of modern probability via
measure theory. What is measure theory, you may ask? Well, a measure is how mathematicians
generalize the notions of measuring things as you know them, like areas, volumes, and
(12:00):
masses. To define a measure, a measure space, and a probability space, we need three key
ingredients. The first ingredient is a set, which is a collection of distinct objects,
say numbers. So let's just say we have the set containing the numbers 1, 2, and 3. These
are our elements in our set. Let's call this set X. The second ingredient is what's known
(12:22):
as a sigma algebra. This is a collection of subsets of our original set, meaning a collection
of sets that comprise either part of or all of our original set. This sigma algebra satisfies
three important properties. The first is that the entire set X is an element of our sigma
algebra. The second property is that the sigma algebra is closed under complements, meaning
(12:47):
if any subset of X is in our sigma algebra, then so is every element in X outside of that
subset. For example, if the set containing just the number 1 is in our sigma algebra,
then so is the set containing the numbers 2 and 3. An immediate consequence of this
second property is that the set containing nothing, or the so-called empty set, is in
(13:08):
our sigma algebra. Finally, the third property is that if you have a countable collection
of subsets in our sigma algebra, meaning you can assign each subset a number, then the
union of those subsets is also in the sigma algebra. This just means you're able to combine
elements from two or more distinct subsets into one big subset.
So in our example, since we said the set containing 1 and the set containing 2 and 3 are both
(13:33):
in the sigma algebra, the set containing 1, 2, and 3 is in the sigma algebra, but we do
that anyway since that's just the original set X, which we established was in the sigma
algebra via the first property.
Okay, let's take a step back. So far we have two ingredients, a set X and a collection
of subsets of X, called a sigma algebra, satisfying three different properties. The third and
(13:57):
final ingredient is what's known as a measure. A measure is a function that takes in a subset
of X from our sigma algebra and outputs either a real number, negative infinity, or positive
infinity. You can assign values to these sigma algebra elements in any way as long as it
obeys three rules. First, the measure of any subset has to be greater than or equal to
(14:20):
zero. Second, the measure of the empty set is zero. Third, the measure of the union of
disjoint subsets, meaning subsets that don't share any elements, is equal to the sum of
the measures of the disjoint subsets. So say you have two subsets, the intervals zero to
one inclusive and the interval one to two inclusive. If you picture these two intervals
(14:43):
on a number line, you might say the distance between zero and one is one and the distance
between one and two is one. So the distance between zero and two is just one plus one
equals two. So these three ingredients, a set, a sigma algebra, and a measure, come
together to create what's known as a measure space. This general notion of a measure space
(15:03):
is important in analysis, which you can think of as an extension and or generalization of
calculus. A probability space is just a measure space where the measure of the whole set X
is equal to one. We can think of probabilities as measures and the set of all possible outcomes
of an experiment as our set X. For example, flipping a coin once has two possibilities,
(15:25):
heads or tails, and we can ignore the possibility of the coin landing on its side and say the
probability of getting either heads or tails is one or 100%. The assignment of probability
to certain events are usually determined by distributions, think bell curves if you've
ever seen those. I'll talk more about distributions next time. Komagoro was able to use this framework
(15:46):
to establish the foundations of all modern probability, but to talk about it much further
requires ideas from calculus, which I'll talk about in upcoming episodes. Thank you
for tuning into this episode of the Magnificence of Mathematics. In the next episode or two,
I'll talk more about statistics, my time in grad school, and advice for undergrads
and grad students in general. Hope to see you then. 175 00:16:20,720 -->00:17:28,080 Algid Productions LLC Outro. Thank you for listening!
Hello everyone, welcome to the Magnificence of Mathematics. I'm your host, Eddie Kingston.
(00:15):
Today we'll be talking about my favorite subfield of math, probability. You're most
likely already intuitively familiar with probability, say when you're flipping a coin. You know
that there's at least roughly a 50% chance of the coin landing on heads, and a 50% chance
of the coin landing on tails. But here's a head scratcher. Is it true that events that
(00:36):
have a 0% probability of happening can still happen? Obviously not, right? Anything with
a 0% chance of happening can't happen. You can't randomly choose dragon in a game of
rock paper scissors. Well, it might surprise you that the answer to the question I just
posed is in fact, yes. Let's build up to this. Suppose you randomly pick a number either
(00:58):
1 or 2 with equal probability. That is, the probability of picking the number 1 is 1 half,
and the probability of picking the number 2 is 1 half. Now suppose you randomly pick
a number either 1, 2, or 3 with equal probability. Then the probability of picking the number
1, or 2, or 3 is 1 third, which is smaller than 1 half. If you randomly pick a number
(01:23):
1, 2, 3, or 4 with equal probability, the probability of picking the number 1 is 1 over
4, which again is smaller than 1 third and 1 half. At this point, you might have noticed
two things. First, the probability of picking the number 1 from the numbers 1, 2, 3, all
the way up to some positive whole number n is 1 over n, since there are n numbers to
(01:46):
choose from. Second, this probability gets smaller as n gets larger. Now, what if you
wanted to choose a number from the set of all positive integers 1, 2, all the way up
to infinity? Since there are an infinite amount of numbers to choose from, the probability
of choosing the number 1, or any number for that matter, is 1 over infinity, which is
(02:09):
just 0. Wait, huh? How could it be 0 when we're still choosing a number? Well, if the
probability in question had been anything other than 0, there would be some number n
that describes the amount of numbers we're choosing from. But we just said that we're
choosing from the infinite list of positive integers, so this can't be the case. Therefore,
(02:31):
we literally have a 0% chance of choosing any positive integer whatsoever, but we still
do choose a number. Argument I just gave is an albeit hand-wavy proof by contradiction,
and the underlying idea behind all of this is the idea of a limit, which is one of the
fundamental building blocks of calculus and beyond. In a similar vein, it's also true
(02:55):
that events with 100% probability don't necessarily happen. Suppose you want to avoid choosing
the number 1 from the numbers 1, 2, and 3. The probability of avoiding the number 1 is
2 thirds, because you have two favorable outcomes out of three possible outcomes. If you want
to avoid choosing the number 1 from 1, 2, 3, and 4, that probability is 3 fourths, which
(03:19):
is greater than 2 thirds, because you have three favorable outcomes out of four possible
outcomes. Then for any positive integer n, the probability of avoiding the number 1 from
the numbers 1, 2, up to n is n minus 1 divided by n, which gets larger as n gets larger.
You might notice that this probability gets closer and closer to 1, but never actually
(03:44):
reaches it unless you have the infinite set of all positive integers, and want to avoid
choosing the number 1. So the probability of avoiding the number 1, or any number for
that matter, is 1, but we still don't avoid picking a number altogether. How cool is that?
Want another counterintuitive way probability works? You might have heard of the famous
(04:05):
Monty Hall problem inspired by Let's Make a Deal host Monty Hall, who ran the show from
1963 until 1987. If you're unfamiliar with the show or the problem itself, imagine you're
put in front of three doors. The host Monty Hall tells you that behind one of the doors
is a brand new car. Behind the other two doors are goats. Your task is to pick the door the
(04:27):
car is hiding behind. Let's say you pick door 1. Monty then opens door 3, revealing a goat.
Now Monty gives you a choice. Do you want to stick with door 1, or change your door
to door 2? The problem is, given all this information, would it be in your best interest
to switch, or to stick with your original choice? Your knee-jerk reaction might be that
(04:49):
it doesn't matter whether you stick with your first choice or switch, since now there's
only two doors the car could be behind and it's a 50-50 chance. But it turns out that
that's wrong. There is a 2 in 3 chance that switching will net you the car. To explain
this, it might be easier to think about the game as if there were 100 doors to choose
from, 99 goats and 1 car. If you pick door 36, and Monty opens every door except door
(05:14):
36 and door 67, at this point you might be inclined to think, yeah, I'm switching. There's
only a 1 in 100 chance I was right to begin with. I probably guessed wrong, the car is
actually behind door 67. And you're right. Given the other 98 doors that have opened,
the probability that the door you first chose was right all along is only 1 out of 100,
(05:36):
meaning the probability that door 67 was the right door is 99 out of 100. If we scale this
back to the original problem, the probability that the door you first chose was right all
along is 1 out of 3, meaning the probability that the car is behind the other unopened
door is 2 out of 3. If you think about randomly choosing 1 door out of 3 or 100 or however
(05:58):
many as an experiment, this paradox became a thing because when thinking about probabilities
and what would be better to do, people thought about the experiment as if the start of the
game didn't happen. They thought of the experiment as having started at the point where they
have to choose to either stick with their door or change it, which is where they get
50% from. But if you think about the experiment as starting when you first choose a door,
(06:21):
then you only have a 1 in 3 or 1 in 100 or 1 in however many chance of choosing the right
door on the first try. This is all well and good, but where did any of this come from?
Why study probability in the first place? Probability came from the desire to quantify
uncertainty, in particular when it comes to gambling. In the mid 1600s, Gerolamo Cardano
(06:44):
showed the use of defining odds as the ratio of the number of favorable outcomes to the
number of unfavorable outcomes. So for example, 1 to 1 odds means there is 1 favorable outcome
and 1 unfavorable outcome, and around the same time, French mathematicians Pierre de
Fermat and Blaise Pascal wrote to each other and laid the foundations of probability as
(07:05):
we know it. In particular, Blaise Pascal is famous for what's known as Pascal's Triangle,
which tells you how many ways you can pick a certain number of items from a larger set
of items. The reason it's called a triangle is because you can write out these numbers
as if you were building a side of a pyramid, i.e. a triangle from the top down. For example,
(07:25):
the top of the triangle, called row 0, is just 1 because there is only one way to pick
0 items from 0 items. That's just doing nothing. Similarly, the row below, row 1, has a 1 on
the left and a 1 right next to it, because again, there's only one way to pick 0 items
from a set of 1 item, which is to just do nothing, and one way to pick 1 item from that
(07:48):
set, which is just picking that item. Now for row 2, there's a 1 on either end because
there's only one way to pick 0 items from a set of 2, but not doing anything, and only
one way to pick 2 items from a set of 2 by picking both items, and a 2 in the middle
because there's 2 ways to pick 1 item from a set of 2, picking one thing or the other.
(08:10):
For row 3, there's once again a 1 on either end, one way to pick nothing and one way to
pick everything, and 2 3s in the middle. 3 ways to pick 1 object, and 3 ways to pick
2 objects. That is, 3 ways to leave 1 object out. Let's recap. For row 0, we have a 1.
For row 1, we have a 1 and a 1, which would sandwich the 1 from row 0 if row 0 was brought
(08:34):
down. For row 2, we have a 1, 2, and 1, with the 2 sandwiched between the 2 1s from row
1. Row 3 reads 1 3 3 1, and so on. If you write these out, or just look up the triangle,
you might notice that if you take any two adjacent numbers in a certain row, and look
at the number in the row below that's sandwiched between those two numbers, the sandwiched
(08:57):
number is the sum of the two numbers directly above it. So if you look at the left side
of row 2 and row 3, you see a 1 and 2 in row 2, a 1 on the far left in row 3, and a 3 between
the 1 and the 2 in row 2, and 1 plus 2 is 3. Another way these are read is in the choose
notation. So row 2 can be read as 2 choose 0, which equals 1, 2 choose 1, which equals
(09:21):
2, and 2 choose 2, which equals 1. 2 choose 1, for example, can be interpreted as, from
a set of two objects, choose one of them. This triangle is crucial for beginning to
understand the field of combinatorics, which is what mathematicians fancily call counting.
If you were to take a beginning college level probability class, you'd likely encounter
(09:42):
basic combinatorics at the beginning of the class, and be able to answer questions like,
if you have a hand of 5 standard playing cards, what's the probability of having 2 aces? To
answer this, you want 2 of the 4 aces from the deck in your hand, so there's 4 choose
2, which is 6 ways to choose the aces for your hand. Now you don't want any more aces,
(10:04):
so you have to pick 3 non-ace cards out of the remaining 48 possible cards. So you have
48 choose 3, which equals 17,296 ways to pick the remaining cards. Multiplying these together
means you have 103,776 favorable outcomes. Since you have 52 cards, and you choose 5
(10:25):
of them for your hand, you have 52 choose 5, which equals 2,598,960 total possible outcomes.
So the answer we're looking for is the number of favorable outcomes divided by the number
of possible outcomes, giving us 103,776 divided by 2,598,960, which comes out to be slightly
(10:51):
less than 4%. If you've ever played poker or powerball or something, this is how the
probability of certain hands or certain amounts of numbers matched are calculated. If I haven't
broken your brain by this point, I want to talk a bit about how mathematicians think
of probability. Two of the famous mathematicians to get the ball rolling are André Marcot
and André Kolmogorov in the early 1900s. Marcot was largely responsible for the development
(11:16):
of stochastic processes, which are sequences of random events that happen over time. Going
back to our coin flipping example, suppose you play a game where you win $1 every time
the coin lands on heads, and you lose $1 every time the coin lands on tails. The amount of
money you win or lose is a stochastic process. You can map out how the game goes over time,
(11:36):
and you'll get a different map every time you play the game, assuming you play enough
of course. Stochastic processes can be used to model all kinds of things like wait times
and even stock prices. Kolmogorov established the foundations of modern probability via
measure theory. What is measure theory, you may ask? Well, a measure is how mathematicians
generalize the notions of measuring things as you know them, like areas, volumes, and
(12:00):
masses. To define a measure, a measure space, and a probability space, we need three key
ingredients. The first ingredient is a set, which is a collection of distinct objects,
say numbers. So let's just say we have the set containing the numbers 1, 2, and 3. These
are our elements in our set. Let's call this set X. The second ingredient is what's known
(12:22):
as a sigma algebra. This is a collection of subsets of our original set, meaning a collection
of sets that comprise either part of or all of our original set. This sigma algebra satisfies
three important properties. The first is that the entire set X is an element of our sigma
algebra. The second property is that the sigma algebra is closed under complements, meaning
(12:47):
if any subset of X is in our sigma algebra, then so is every element in X outside of that
subset. For example, if the set containing just the number 1 is in our sigma algebra,
then so is the set containing the numbers 2 and 3. An immediate consequence of this
second property is that the set containing nothing, or the so-called empty set, is in
(13:08):
our sigma algebra. Finally, the third property is that if you have a countable collection
of subsets in our sigma algebra, meaning you can assign each subset a number, then the
union of those subsets is also in the sigma algebra. This just means you're able to combine
elements from two or more distinct subsets into one big subset.
So in our example, since we said the set containing 1 and the set containing 2 and 3 are both
(13:33):
in the sigma algebra, the set containing 1, 2, and 3 is in the sigma algebra, but we do
that anyway since that's just the original set X, which we established was in the sigma
algebra via the first property.
Okay, let's take a step back. So far we have two ingredients, a set X and a collection
of subsets of X, called a sigma algebra, satisfying three different properties. The third and
(13:57):
final ingredient is what's known as a measure. A measure is a function that takes in a subset
of X from our sigma algebra and outputs either a real number, negative infinity, or positive
infinity. You can assign values to these sigma algebra elements in any way as long as it
obeys three rules. First, the measure of any subset has to be greater than or equal to
(14:20):
zero. Second, the measure of the empty set is zero. Third, the measure of the union of
disjoint subsets, meaning subsets that don't share any elements, is equal to the sum of
the measures of the disjoint subsets. So say you have two subsets, the intervals zero to
one inclusive and the interval one to two inclusive. If you picture these two intervals
(14:43):
on a number line, you might say the distance between zero and one is one and the distance
between one and two is one. So the distance between zero and two is just one plus one
equals two. So these three ingredients, a set, a sigma algebra, and a measure, come
together to create what's known as a measure space. This general notion of a measure space
(15:03):
is important in analysis, which you can think of as an extension and or generalization of
calculus. A probability space is just a measure space where the measure of the whole set X
is equal to one. We can think of probabilities as measures and the set of all possible outcomes
of an experiment as our set X. For example, flipping a coin once has two possibilities,
(15:25):
heads or tails, and we can ignore the possibility of the coin landing on its side and say the
probability of getting either heads or tails is one or 100%. The assignment of probability
to certain events are usually determined by distributions, think bell curves if you've
ever seen those. I'll talk more about distributions next time. Komagoro was able to use this framework
(15:46):
to establish the foundations of all modern probability, but to talk about it much further
requires ideas from calculus, which I'll talk about in upcoming episodes. Thank you
for tuning into this episode of the Magnificence of Mathematics. In the next episode or two,
I'll talk more about statistics, my time in grad school, and advice for undergrads
and grad students in general. Hope to see you then. 175 00:16:20,720 -->00:17:28,080 Algid Productions LLC Outro. Thank you for listening!
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