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

Exercises 1

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Exercise 1.1

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).


(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 $$ {\bf1}_{A\cup B}(x) + {\bf1}_{A\cap B}(x)={\bf1}_A(x)+{\bf1}_B(x). $$

Exercise 1.2

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 \(A\cap B=[e^{-2},.5)\).

Exercise 1.3

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.

Exercise 1.4

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 \(\mathcal F\).

Exercise 1.5

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 \(\mathcal F\).

Exercise 1.6

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.

Exercise 1.7

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 \({1\over 2\sqrt x}{\bf1}_{(0,1)}(x)\).