Probability Spaces Random Variables and Expectation — Examples

The geometric series gives $\mathbb{P}(\mathbb{N})={\sum }_{n=0}^{\mathrm{\infty }}{2}^{-(n+1)}=1,$ so the displayed weights define a probability measure as in [L1]. Since every subset of $\mathbb{N}$ is measurable, the coordinate map $X$ is a real random variable by [L2].

For each $n\in \mathbb{N}$, the fibre of $\{n\}$ is the singleton $\{n\}$, so the law definition gives ${\mathbb{P}}_X(\{n\})=\mathbb{P}(\{n\})=2^{-(n+1)}\mathrm{.}$

Applying change of variables from [L1] to the identity function on $\mathbb{N}\subseteq \mathbb{R}$ gives $\mathbb{E}[X]={\sum }_{n=0}^{\mathrm{\infty }}n{2}^{-(n+1)}=1.$

Steps 1.1, 1.2, and 2.1 verify the measure, the law, and the expectation.

By direct interval computation, $\mathbb{P}(X\le x)=0(x<0),\mathbb{P}(X\le x)=x(0\le x\le 1),\mathbb{P}(X\le x)=1(x\ge 1),$ which is exactly the displayed $F_X$.

Direct integration gives $\mathbb{E}[X]={\int }_0^{1}xdx=\frac{1}{2},\mathbb{E}[X^2]={\int }_0^1{x}^{2}dx=\frac{1}{3}\mathrm{.}$

The tail is $\mathbb{P}(X>t)=1-t$ for $0\le t\le 1$ and $0$ for $t>1$, so ${\int }_0^{\mathrm{\infty }}\mathbb{P}(X>t)dt={\int }_0^1(1-t)dt=\frac{1}{2}=\mathbb{E}[X],$ in agreement with [L2].

Steps 1.1, 1.2, and 2.1 verify the CDF, moments, and tail integral.

Because $X\ge 0$, one has $F_X(x)=0$ for $x<0$. For $x\ge 0$, $F_X(x)=1-\mathbb{P}(X>x)=1-e^{-\lambda x}\mathrm{.}$ So the displayed formula is the CDF of $X$.

The tail integral from [L3] gives $\mathbb{E}[X]={\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}dt=\frac{1}{\lambda }\mathrm{.}$

By [L2], that distribution function determines the law uniquely, and its interval increments are those of the exponential distribution with parameter $\lambda$.

Steps 1.1, 1.2, and 2.1 recover the law and its expectation from the tail.

Each displayed function is nondecreasing, right-continuous, tends to $0$ at $-\mathrm{\infty }$, and tends to $1$ at $+\mathrm{\infty }$. Therefore [L1] gives a probability law for each of them.

The jumps identify the atoms. For $F_{\mathrm{d}}$, the jumps are $1\mathrm{/}2$ at $0$ and $1\mathrm{/}2$ at $1$, so the law is purely discrete. For $F_{\mathrm{c}}$ there are no jumps, hence no atoms. For $F_{\mathrm{m}}$ there is a jump of size $1\mathrm{/}2$ at $0$ and no jump on $(0,1]$, so the law is mixed.

Thus the three distribution functions realize discrete, continuous, and mixed behavior without any implication that a density must exist in general.

By [L1], $\mathbb{E}[g(X)]={\int }_{\mathbb{R}}gd{\mathbb{P}}_X\mathrm{.}$ Because the law is supported on $\{-1,0,2\}$, this is the weighted sum $g(-1)\frac{1}{4}+g(0)\frac{1}{2}+g(2)\frac{1}{4}=\frac{2}{4}+\frac{1}{2}+\frac{5}{4}=\frac{9}{4}\mathrm{.}$

The nonlinear expectation is therefore computed entirely from the law on the target space.

Apply [L1] to the convex function $t\mapsto e^t$ and to $Y$. This gives $e^{\mathbb{E}[Y]}\le \mathbb{E}[e^Y]\mathrm{.}$

Applying the same theorem to $Z=\log X$ and the convex function $t\mapsto e^t$ gives $e^{\mathbb{E}[\log X]}\le \mathbb{E}[e^{\log X}]=\mathbb{E}[X],$ which is equivalent to $\mathbb{E}[\log X]\le \log \mathbb{E}[X]$.

For the two-point law, step 1.2 becomes $e^{p\log a+(1-p)\log b}\le pa+(1-p)b,$ that is, $a^p{b}^{1-p}\le pa+(1-p)b\mathrm{.}$

Steps 1.1, 1.2, and 2.1 give the exponential-moment and weighted AM-GM forms of Jensen.

For $X=a{\mathbf{1}}_A$, one has $\mathbb{E}[X]=a\mathbb{P}(A)$ and $\{X\ge a\}=A$, so $\mathbb{P}(X\ge a)=\mathbb{P}(A)=\frac{\mathbb{E}[X]}{a}\mathrm{.}$

For $Y=\pm a$ with equal probabilities, $\mathbb{E}[Y]=0$ and $\mathrm{Var}(Y)=a^2$, while $\mathrm{\mid }Y-\mathbb{E}[Y]\mathrm{\mid }=a$ always. Hence $\mathbb{P}(\mathrm{\mid }Y-\mathbb{E}[Y]\mathrm{\mid }\ge a)=1=\frac{\mathrm{Var}(Y)}{a^2}\mathrm{.}$

Thus both inequalities admit equality, even though other laws can make the bounds much weaker.

Define $\mathbb{P}(A):={\sum }_{n\in A}{p}_n(A\subseteq \mathrm{\Omega })\mathrm{.}$ The total mass is ${\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)}={\sum }_{n=1}^{\mathrm{\infty }}(\frac{1}{n}-\frac{1}{n+1})=1.$ If $(A_j{)}_{j\ge 0}$ is pairwise disjoint, regrouping the nonnegative series gives $\mathbb{P}\text{\hspace{0.17em}⁣}\left({\bigcup }_{j\ge 0}{A}_j\right)={\sum }_{j\ge 0}{\sum }_{n\in A_j}{p}_n={\sum }_{j\ge 0}\mathbb{P}(A_j)\mathrm{.}$ Thus $\mathbb{P}$ is a probability measure on the power set of $\mathrm{\Omega }$, and $X(n)=n$ is a measurable real random variable.

Put $X_N:=X{\mathbf{1}}_{\{1,\dots ,N\}}$. The functions $X_N$ increase pointwise to $X$. By [L1], [L2], and monotone convergence, $\mathbb{E}[X]={\lim }_{N\to \mathrm{\infty }}\mathbb{E}[X_N]={\lim }_{N\to \mathrm{\infty }}{\sum }_{n=1}^{N}n\frac{1}{n(n+1)}={\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{n+1}=+\mathrm{\infty }\mathrm{.}$ Thus the expectation exists only as an extended value, not as a finite real number, exactly as [L1] allows.

Steps 1.1 and 2.1 refute the claim that every random variable has finite expectation.

The CDF of $X$ is $F_X(x)=\{\begin{array}{ll}0,& x<0,\\ 1-p,& 0\le x<1,\\ 1,& x\ge 1.\end{array}$ So the law has atoms at $0$ and $1$ by [L2].

If this law were given by a Lebesgue density $f$, then every singleton would have probability ${\int }_{\{a\}}f(x)dx=0.$ In particular $\mathbb{P}(X=0)=0$, contradicting $\mathbb{P}(X=0)=1-p>0$.

Therefore this distribution function has no Lebesgue density.

The functions $X$ and $Y$ differ only at the single point $0$. That set is Lebesgue-null, so [L1] gives $X=Y$ almost surely.

They are not pointwise equal, because $X(0)=0$ while $Y(0)=1$. They are both bounded and hence integrable, so [L2] also gives $\mathbb{E}[X]=\mathbb{E}[Y]=0.$

Therefore almost-sure equality is strictly weaker than pointwise equality.

A chord subtending central angle $\mathrm{\Theta }$ has length $2\sin (\mathrm{\Theta }\mathrm{/}2)$. It is longer than $\sqrt{3}$ exactly when $\sin (\mathrm{\Theta }\mathrm{/}2)>\sqrt{3}\mathrm{/}2$, that is, when $\mathrm{\Theta }>2\pi \mathrm{/}3$. Under the uniform-angle model this has probability $(\pi -2\pi \mathrm{/}3)\mathrm{/}\pi =1\mathrm{/}3.$

A chord is longer than $\sqrt{3}$ exactly when its midpoint is within distance $1\mathrm{/}2$ of the center. Along a fixed radius, the favorable segment has length $1\mathrm{/}2$ inside the unit segment, so the midpoint-on-a-radius model gives probability $1\mathrm{/}2$. Over the whole disk, the favorable region is the disk of radius $1\mathrm{/}2$, so the midpoint-in-the-disk model gives area ratio $(\pi (1\mathrm{/}2{)}^2)\mathrm{/}\pi =1\mathrm{/}4.$

The three different answers $1\mathrm{/}3$, $1\mathrm{/}2$, and $1\mathrm{/}4$ come from three different probability models for the same geometric question, exactly as [L1] predicts.