Exercises 1

1. (a) Show that ${\bf1}_A(x){\bf1}_B(x)\ =\ {\bf1}_{A\cap B}(x) $, for any sets $A,B$.

(b) Express ${\bf1}_{A\cup B}(x) $ in terms of the indicator functions in (a).

Solution

(a) Need to check for all $x\in\Omega$. There are four possibilities (make the old Venn diagram): $x$ is in $A$ but not in $B$; $x$ is in $B$ but not in $A$; $x$ is in $A$ and in $B$; $x$ is neither in $A$ nor in $B$. By considering each of those cases one sees that: $$ {\bf1}_A(x){\bf1}_B(x)=1\quad \iff \quad{\bf1}_A(x)={\bf1}_B(x)=1\quad \iff \quad x\in A\ \hbox{and}\ x\in B. $$ (b) From the Venn diagram, again checking all the possibilities, one sees that $$ {\bf1}_{A\cup B}(x) + {\bf1}_{A\cap B}(x)={\bf1}_A(x)+{\bf1}_B(x). $$

2. Suppose $\Omega=(0,1)$, $X(\omega)=-\log\omega$ and $Y(\omega)=1-\omega$. Find the set $A\cap B$ explicitly in terms of $\omega$, if $$ A\ =\ \{X\le 2\}=\{\omega\in(0,1)\,|\, X(\omega)\le 2\},\quad B\ =\ \{Y>.5\}=\{\omega\in(0,1)\,|\, Y(\omega)>.5\}. $$

Solution

$\{\omega\in(0,1)\,|\, X(\omega)\le 2\}\ =\ \{\omega\in(0,1)\,|\, -\log\omega\le 2\}\ =\ \{\omega\in(0,1)\,|\, \omega\ge e^{-2}\}\ =\ [e^{-2},1)$. $\{\omega\in(0,1)\,|\, Y(\omega)>.5\}\ =\ \{\omega\in(0,1)\,|\, 1-\omega>.5\}\ =\ (0,.5)$. Then $A\cap B=[e^{-2},.5)$.

3. Suppose $\Omega=\{\square,\triangle, \bigcirc\}$ and let $\mathcal G$ be the set of all subsets of $\Omega$. Is $\mathcal G$ a field? A $\sigma$-field?

Solution

If $A$ is in $\mathcal G$ then $A^c=\Omega-A$ is necessarily in $\mathcal G$ too, and the same applies to $A\cup B$, $A\cap B$ for any $A,B$ that are subsets of $\Omega$. So $\mathcal G$ is a field. It is also a $\sigma$-field because the union of any number of subsets of $\Omega$ is also a subset of $\Omega$. There is no difference between a field and a $\sigma$-field when the number of elements in $\mathcal G$ is finite, as is the case here.

4. (Continued from no.3) Find the $\sigma$-field generated by $\{\{\square\}\}$.

Solution

The field (or $\sigma$-field) generated by this set of subsets of $\Omega$ must be closed for complementation, so $\{\triangle, \bigcirc\}$ must also be in it. It must be closed for unions, so $\Omega$ must be in it too. The complement of $\Omega$ is $\emptyset$. Hence, a tentative answer is: $$ \mathcal F\ =\ \{\emptyset,\Omega, \{\square\}, \{\triangle, \bigcirc\}\}. $$ One can check that the conditions required for a set of sets to be a field are satisfied by $\mathcal F$.

5. Suppose $\mathcal F$ is a $\sigma$-field, and that $A,B\in{\mathcal F}$. Show that $B\backslash A \in{\mathcal F}$ as well. (N.B. Recall that $B\backslash A=B\cap A^c$.)

Solution

We know that $A^c \in\mathcal F$, since $\mathcal F$ is a $\sigma$-field. Therefore the intersection of $B$ and $A^c$ is also in the $\sigma$-field $\mathcal F$.

6. Let $\Omega$ be any set. Check that ${\mathcal F}_0=\{\emptyset,\Omega\}$ is a $\sigma$-field.

Solution

${\mathcal F}_0$ is closed under unions, intersections and complements.

7. (a) Suppose $\Omega=(0,1)$, $\mathcal F={\mathcal B}((0,1))$ (the $\sigma$-field generated by the intervals in (0,1)) and that the probability measure $\pro$ is the Lebesgue measure restricted to $(0,1)$. This means that $$ \pro((a,b))\ =\ b-a $$ for every $0\lt a\lt b\lt 1$. Define $X(\omega)=\log(1+\omega)$. Find the distribution function and density of $X$. (b) Repeat (a) with $X(\omega)=\omega^2$.

Solution

(a) The function $\log(1+\omega)$ is increasing, so the minimum value of $X$ is $\log 1=0$ and its maximum value is $\log2$. For $x$ in between 0 and $\log2$ we have $$\eqalign{ \pro(X\le x)\ &=\ \pro(\{\omega\,|\, \log(1+\omega)\le x\})\cr & =\ \pro(\{\omega\,|\, \omega\le e^x-1\})\cr & =\ \pro((0,e^x-1))\ =\ e^x-1. } $$ Density is $e^{x}{\bf1}_{(0,\log2)}(x)$. (b) For $x\in(0,1)$, $\{\omega\,|\, \omega\le\sqrt x\}=(0,\sqrt x)$, so DF is $\sqrt x$ for those $x$; PDF is ${1\over 2\sqrt x}{\bf1}_{(0,1)}(x)$.