Let our probability space $(\Omega, \mathcal B, \lambda)$ be the unit interval $\Omega=[0,1]$ with the Borel subsets $\mathcal B$ and the Lebesgue measure $\lambda$. Is it possible to find an infinite sequence of independent and identically distributed random variables $X_1$, $X_2$, $\dots$ of any given distribution supported on this probability space? Surprisingly, yes, and the proof relies on a version of Cantor’s diagonal argument. Proof It is sufficient to find a sequence of independent and uniformly distributed random variables $U_1$, $U_2$, $\dots$, on $[0,1]$. Then if $F$ is any cumulative distribution function, the sequence $X_n = F^{-1}(U_n)$ has the desired property. Thus we proceed to construct the required sequence $U_1$, $U_2$, $\dots$ In order to apply Cantor’s diagonal argument, we first need to write an arbitrary point $a \in [0,1]$ as an infinite binary fraction

Here we illustrate three of the famous zero-one laws for convergent sequences of random variables. Their names are: the Borel-Cantelli lemma, the second Borel-Cantelli lemma, and Kolmogorov’s zero-one law. They are going to help us study the nuances in the relationship between almost sure convergence and complete convergence. Suppose we are given an infinite sequence of random variables $$ \tag{1}{X_1, X_2, \dots} $$ defined on some probability space $(\Omega, \mathcal A, \operatorname P)$. We list three different modes in which the sequence may be convergent to zero. $(i)$ The sequence $(1)$ converges to zero in probability if for every $\epsilon>0$ $$ \lim_{n\rightarrow \infty} \operatorname P(|X_n|>\epsilon) = 0. $$ $(ii)$ The sequence $(1)$ converges to zero almost surely if for every $\epsilon>0$ $$ \lim_{n\rightarrow \infty} \operatorname P({|X_n|> \epsilon} \cup {|X_{n+1}|> \epsilon} \cup \cdots) = 0.

Let $X=(X_1, X_2)$ and $Y=(Y_1, Y_2)$ be two random vectors in $\operatorname{R}^2$. Suppose that $X_i$ and $Y_j$ are independent for each pair of indices. We have the following questions. Are $X$ and $Y$ independent? Are $X$ and $Y$ independent if each of them is bivariately normally distributed? Are $X$ and $Y$ independent if they are jointly normally distributed? Answer Let $X_1, X_2 \in U(0,1)$ be uniformly distributed and independent. Define the variables $$ \begin{aligned} Y_1 &= X_1 + X_2 \qquad \operatorname{mod} 1 \\ Y_2 &= X_1 - X_2 \qquad \operatorname{mod} 1, \end{aligned} $$ where the $\operatorname{mod} 1$ operation means that we identify all integer values on the real line and turn it into a circle of circumference one. Notice that $Y_1 \mid X_1 \sim U(0,1)$, hence the distribution of $Y_1$ does not depend on $X_1$.

“A generating function is a device somewhat similar to a bag. Instead of carrying many little objects detachedly, which could be embarrassing, we put them all in a bag, and then we have only one object to carry, the bag.” – George Polya, Mathematics and plausible reasoning (1954) Introduction Often in mathematics we have to deal with recurrence relations. One of the best known examples of recurrence relations is the Fibonacci numbers given by the relation $$ {\displaystyle F_{n}=F_{n-1}+F_{n-2}}, $$ with initial conditions ${\displaystyle F_{0}=1}, {\displaystyle F_{1}=1.}$ The Fibonacci sequence appeared in 1202 in the book *Liber Abaci* by the Italian mathematician Leonardo of Pisa, also known today as Fibonacci, where he tries to model the population growth of an idealized pair of newborn rabbits.