Probability Theory 13 | Independence for Random Variables

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This is my video series about Probability Theory. I hope that it will help everyone who wants to learn about it.
This video is about probability theory, also known as stochastics, stochastic processes or statistics. I keep the title in this general notion because I want cover a lot of topics with the upcoming videos.

Here we talk about independence again, namely about independent random variables. First we discuss two random variables and then infinitely many random variables.

#ProbabilityTheory
#Analysis
#Calculus
#Mathematics

I hope that this helps students, pupils and others. Have fun!

(This explanation fits to lectures for students in their first and second year of study: Mathematics for physicists, Mathematics for the natural science, Mathematics for engineers and so on)

[undisclosedmusic]

I love the series! It would be awesome if you could also go into some advanced uses of probability theory in information theory, e.g. divergences, distances and things like couplings. Thanks!

[sinx]

Why is it sufficient to look at the preimages of only the half-bounded intervals? I don't see why it isn't necessary to look at all (measurable) subsets of R

[leeu-geen]

I would love to see more German videos please ❤
Thank you soo much for all u do . u r really one of the best !!

[debasishdas]

Great videos. Do you have a book reference for more details? I like your style so if you have a favourite book this will be great.

[novicadakovic]

Is the last definition mutual independence of random variables?

[jimallysonnevado]

I am having a bit of confusion in trying to show the equality at 7:31. I just want to verify if this is correct:

P(preim_X((-inf, x]) intersect preim_Y((-inf, y]))
= P({(w_1, w_2) | f(w_1)<=x} intersect {(w_1, w_2) | g(w_2)<=y})
=P({(w_1, w_2) | f(w_1)<=x and g(w_2)<=y})
I just combined it into a single set with two conditions. I think this is a valid step because an intersection of two sets can be thought of combining both sets into one and putting the conditions together under an "and" connector.
=P({w_1| f(w_1)<= x} x {w_2| g(w_2)<= y})

This is a part where I'm not sure if what I'm doing is legal or not. I made it into a Cartesian Product of two sets. I think the reasoning here is that since the function f, and g depend solely on w_1, and w_2 respectively, we can think of the tuple (w_1, w_2) as belonging to omega_1 x omega_2 and w_1 can be any element of Omega_1 as long as f(w_1)<=x and likewise w_2 can also be any element of Omega_2 as long as g(w_2)<=y.

From here we can use the properties of product of probability spaces namely, P(A_1 x A_2) = P_1(A_1)P_2(A_2).
Thus,

P({w_1| f(w_1)<= x} x {w_2| g(w_2)<= y})
= P_1({w_1| f(w_1)<= x}) P_2({w_2| g(w_2)<= y})
The index is due to Omega_1 and Omega_2 have different probability measures.
=P(preim_X((-inf, x]) P(preim_Y((-inf, y]))
Here is my other confusion. I dropped the index, because technically P_1({w_1| f(w_1)<= x}) measures the probability of preimage under f not X. But since f is just X but only depends on the first entry, we can say that the pre-image of X is just all tuples (w_1, w_2) where w_2 is arbitrary member of omega_2 and is only restricted in the first entry by f. Thus, the probability
P_1({w_1| f(w_1)<= x})
=P_1({w_1| X(w_1, w_2)<= x} )
=P_1({w_1| X(w_1, w_2)<= x} ) P_2(Omega_2)
This is where I just multiplied by 1 since P_2(Omega_2)=1
=P({(w_1| X(w_1, w_2)<= x} x Omega_2)
This is where I combined it as a Cartesian product and used the Properties of Product Spaces.
=P(preim_X((-inf, x])
This is precisely the preimage of X because w_2 can be arbitrary element from Omega_2 since X does not depend on Omega_2.
The same reasoning goes for showing P_2({w_2| g(w_2)<= y}) = P(preim_Y((-inf, y]).

I'm not really sure if assuming that w_2 can be arbitrary element of Omega_2 is legit or not in terms of the random variable X and likewise can w_1 be arbitrary element of Omega_1 or not in terms of the random variable Y.

[IsomerSoma]

Immer eine gute Wiederholung nach der Vorlesung.

[shahidkamal]

Hi, how many lectures will it take to complete the whole probability theory . And how frequently can we expect the later lectures to be uploaded

[whatitmeans]

Always when the independence topic is presented, it is intuitively explained through the two dices example, which simply show what independent events are... but sincerely, trying to think of them through Venn Diagrams makes it a lot harder: actually, from the first definition it is a requirememt that the intersection exists to be independent (so somehow disjoint events cannot be independent?.. like the two dice by itself don't stand this, but they will do if they are drop jointly?), and so, from the Left hand side you show that it is actually a relation about the proportion of the sizes of both sets that it is covered by their intersection, proportion which will be telling if they are independent or not, and this is totally counterintuitive: Why independence is related to these sizes' proportions?... So, just any different proportion could drop the independence of the events?... Could the sets be "slightly independent"?... or random noise could make independent events looks correlated? Or otherwise, dependent events look independent because of random correlations?... thinking about points in the real line, I feel like choosing just the "small" set of points that will be fulfilling independence is like integrating from one point to itself, like choosing a measure-zero set of points in a continium of possibilities... that is why I think Venn Diagrams could be quite complicated on probability topics and somehow teachers hide this from students.... other example, when teaching superposition of events, we say that if events are disjoint, their probability are added, and if they are connected, a substraction of their intersection is needed (with the formula that extend it for many sets), but always are show only the two sets examples, which are disjoint or they have only one intersections... but actually, for three and more sets, the possible different combinations and possible intersections grows dramatically as is beatyfully explained on the youtube video "How many ways can circles overlap? - Numberphile".... hope you see it and comment.
Also, I hope you extend your explanations with these caveats.

[jfg]

I love the series! Super useful and super clear, thanks!

My question about this video: is it always possible to combine the d/dx and the d/dy and express the derivative in terms of just z?