PROB_Q1

Quiz 1 Probability Theory 2025.pdf

Problem 1

Define the group operation to be symmetric difference.

Problem 2

$(\Omega ,\mathcal{F},\mathbb{P})$ is a probability space such that $\mathbb{P}(A)\in \{0,1\}$ for all $A\in \mathcal{F}$. Let $X:\Omega \to \mathbb{R}$ be a random variable, not necessarily discrete. Prove that there exists $c\in \mathbb{R}$ such that $\mathbb{P}(X=c)=1$.

Let $G=\{g_n=\{X\in [n,n+1)\}|n\in \mathbb{Z}\}$. Note that $G$ is countable. Now,

$$\mathbb{P}(\Omega )=\mathbb{P}\left(\bigcup _{n\in \mathbb{Z}}{g}_n\right)=\sum _{n\in \mathbb{Z}}P(g_n).$$

Since $\mathbb{P}(\Omega )=1$, it must be the case that $P(g_n)=1$ for exactly one $n$. Consider the interval $i_1=[n,n+1]$. If $P\left(X=n+\frac{1}{2}\right)=1$, we are done. Else, either $P\left(X\in \left[n,n+\frac{1}{2}\right]\right)=1$ or $P\left(X\in \left[n+\frac{1}{2},n+1\right]\right)=1$. Repeat the steps on the interval whose probability is $1$. We will either get a point $c$ such that $P(X=c)=1$ and stop, or keep going and obtain a sequence of nested intervals $S=\{i_1,i_2,\dots \}$. We know from the nested interval property that their intersection will be a singleton, $c$. Also, if we consider the set $\{\{X\in i_1\},\{X\in i_2\},\dots \}$ in $\Omega$, it forms a countable descending sequence. Let their intersection be $B$. We know that the probability of their intersection is the limit of their probabilities, which is $1$. Thus, $P(B)=1$. Clearly, $\{X=c\}=B$, since if $b\notin B$, $X(b)\ne c$, and there exists an interval in $S$ containing $c$ but not containing $X(b)$. Thus, $\mathbb{P}(X=c)=P(B)=1$.

Problem 3

Let $(\Omega ,\mathcal{A},\mathbb{P})$ be a probability space. Let $(A_n)$ be a sequence of events in $\mathcal{A}$. Note that

$$\underset{n\to \infty }{\mathrm{limsup}}{A}_n=\bigcap _{n=1}^{\infty }\left(\bigcup _{k=n}^{\infty }{A}_k\right),$$ $$\underset{n\to \infty }{\mathrm{liminf}}{A}_n=\bigcup _{n=1}^{\infty }\left(\bigcap _{k=n}^{\infty }{A}_k\right).$$

Fatou’s Lemma

Theorem 1(Lemma).

$(\Omega ,\mathcal{A},P)$ is a probability space, and $(A_n)$ is a sequence of events in $\mathcal{A}$. Then,

$$\mathbb{P}(\underset{n\to \infty }{\mathrm{liminf}}{A}_n)\le \underset{n\to \infty }{\mathrm{liminf}}\mathbb{P}(A_n)\le \underset{n\to \infty }{\mathrm{limsup}}\mathbb{P}(A_n)\le \mathbb{P}(\underset{n\to \infty }{\mathrm{limsup}}{A}_n).$$

$$\mathbb{P}(\underset{n\to \infty }{\mathrm{liminf}}{A}_n)=\mathbb{P}\left(\bigcup _{n=1}^{\infty }\left(\bigcap _{k=n}^{\infty }{A}_k\right)\right)=\underset{n\to \infty }{\lim }\mathbb{P}\left(\bigcap _{k=n}^{\infty }{A}_k\right)=\sup \{\mathbb{P}(A_n\cap A_{n+1}\cap \dots )|n\in \mathbb{Z}\},$$

where this property has been used. Results from here have also been used.

$$\underset{n\to \infty }{\mathrm{liminf}}\mathbb{P}(A_n)=\underset{n\to \infty }{\lim }\inf \{\mathbb{P}(A_n),\mathbb{P}(A_{n+1}),\dots \}=\sup \{\inf \{\mathbb{P}(A_n),\mathbb{P}(A_{n+1}),\dots \}|n\in \mathbb{Z}\}$$

Now, $\mathbb{P}(A_n\cap A_{n+1}\cap \dots )\le \mathbb{P}(A_k)$ for all $k\ge n$. Thus, it is a lower bound for $\{\mathbb{P}(A_n),\mathbb{P}(A_{n+1}),\dots \}$ and hence must be less than or equal to its infimum. Thus, we have

$$\mathbb{P}(A_n\cap A_{n+1}\cap \dots )\le \inf \{\mathbb{P}(A_n),\mathbb{P}(A_{n+1}),\dots \}.$$

This holds for all $n$. Thus, it must be that $\sup \{\mathbb{P}(A_n\cap A_{n+1}\cap \dots )|n\in \mathbb{Z}\}\le \sup \{\inf \{\mathbb{P}(A_n),\mathbb{P}(A_{n+1}),\dots \}|n\in \mathbb{Z}\}$.

Hence the first inequality holds.

The second inequality follows from the definition of lim inf and lim sup.

For the third inequality, we have

$$\underset{n\to \infty }{\mathrm{limsup}}\mathbb{P}(A_n)=\inf \{\sup \{\mathbb{P}(A_n),\mathbb{P}(A_{n+1}),\dots \}|n\in \mathbb{Z}\}$$ $$\mathbb{P}(\underset{n\to \infty }{\mathrm{limsup}}{A}_n)=\inf \{\mathbb{P}(A_n\cup A_{n+1}\cup \dots )|n\in \mathbb{Z}\},$$

Since $\mathbb{P}(A_n\cup A_{n+1}\cup \dots )$ is an upper bound for $\{\mathbb{P}(A_n),\mathbb{P}(A_{n+1}),\dots \}$, $\sup \{\mathbb{P}(A_n),\mathbb{P}(A_{n+1}),\dots \}\le \mathbb{P}(A_n\cup A_{n+1}\cup \dots )$. Thus, the fourth inequality holds.

Borel-Cantelli lemma

Theorem 2(Lemma).

Let $(B_n)$ be a sequence of events such that ${\sum }_{n\in \mathbb{N}}{B}_n<\infty$. Then, $P({\mathrm{limsup}}_{n\to \infty }{B}_n)=0$. In other words, the probability of infinitely many $B_n$ occurring is $0$.

$\mathbb{P}(\lim {\sup }_{n\to \infty }(B_n))=\inf \{\mathbb{P}(B_n\cup B_{n+1}\cup \dots )|n\in \mathbb{Z}\}$. We know that

$$\mathbb{P}(B_n\cup B_{n+1}\cup \dots )\le \mathbb{P}(B_n)+\mathbb{P}(B_{n+1})+\cdots <\infty .$$

So, $\inf \{\mathbb{P}(B_n\cup B_{n+1}\cup \dots )|n\in \mathbb{Z}\}\le \inf \{\mathbb{P}(B_n)+\mathbb{P}(B_{n+1})+\dots )|n\in \mathbb{Z}\}$. Since ${\lim }_{n\to \infty }(\mathbb{P}(B_n)+\mathbb{P}(B_{n+1})+\dots )=0$, $\inf \{\mathbb{P}(B_n)+\mathbb{P}(B_{n+1})+\dots )|n\in \mathbb{Z}\}=0$.

Second Borel-Cantelli lemma

A partial converse of the first lemma.

Theorem 3(Lemma).

If $(E_n)$ is a sequence of independent events such that ${\sum }_{n\in \mathbb{N}}P(E_n)=\infty$, then $P(\lim {\sup }_{n\to \infty }{E}_n)=1$. In other words, If the events $E_n$ are independent and the sum of the probabilities of the $E_n$ diverges to infinity, then the probability that infinitely many of them occur is 1.