$$\def\q{\quad}\def\Naturals{{\bf N}}\def\Reals{{\bf R}}\def\pro{{\bf P}} \def\qpro{{\bf Q}}\def\re{{\bf Re}}\def\im{{\bf Im}}\def\esp{{\bf E}}\def\cov{{\bf Cov}\,} \def\var{{\bf Var}\,}\def\Space#1#2#3{\strut}$$

## Exercises 1

Click on each exercise to reveal/hide

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).
(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 $$A\ =\ \{X\le 2\}=\{\omega\in(0,1)\,|\, X(\omega)\le 2\},\quad B\ =\ \{Y>.5\}=\{\omega\in(0,1)\,|\, Y(\omega)>.5\}.$$
$$\{\omega\in(0,1)\,|\, X(\omega)\le 2\}\ =\ \{\omega\in(0,1)\,|\, -\log\omega\le 2\}\ =\ \{\omega\in(0,1)\,|\, \omega\ge e^{-2}\}\ =\ [e^{-2},1)$$. $$\{\omega\in(0,1)\,|\, Y(\omega)>.5\}\ =\ \{\omega\in(0,1)\,|\, 1-\omega>.5\}\ =\ (0,.5)$$. 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?
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 $$\{\{\square\}\}$$.
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 $$B\backslash A=B\cap A^c$$.)
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.
$${\mathcal F}_0$$ 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 $$X(\omega)=\omega^2$$.
(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 $${1\over 2\sqrt x}{\bf1}_{(0,1)}(x)$$.