I know from standard textbooks that "Given the measurable functions $X_i:(\Omega,\mathcal{F})\rightarrow(\Omega_i,\mathcal{A}_i)$, the $\sigma$-algebra generated by a set of random variables $(X_i; i\in I)$ is given by \begin{equation} \sigma\big(X_i;~i\in I\big)=\sigma\left(\bigcup_{i\in I}\sigma(X_i)\right)=\sigma\left(\bigcup_{i\in I} X_i^{-1}(\mathcal{A}_i)\right)". \end{equation} I could understand this idea conceptually, but when this becomes to the real examples I am a bit confused. For example, suppose that $I$ is a discrete time set $I=\mathbb{N}$ and $X_i:\Omega\rightarrow\mathbb{R}$ denotes the number of heads in $i$th draw of a coin. Now it is clear that e.g. \begin{align} \sigma(X_1)=\big\{\emptyset,\{T\},\{H\},\Omega\big\} ~~etc.. \end{align} My questions are as follows:

  1. What would then be the exact elements of $\sigma(X_1,X_2,...,X_k)=:\mathcal{F}_1^k$ for some $k\in\mathbb{N}$? (say when $k=2$?)
  2. Should I distinguish the events H and T in $i$th draw from those in $j$th draw? ($i\ne j$)
  3. Does this idea hold also for the random vector? That is, $\sigma(X_1,X_2)=\sigma(Y)$ where $Y$ is a random vector defined as $Y=(X_1,X_2)^{T}$ (with $T$ being transpose), and $I$ is now obviously $\{1,2\}$.

Many thanks in advance, John.

  • 2
    $\begingroup$ I would not use $T$ for tails and for $\Bbb N$ at the same time. Also, $$ \sigma(X_i) = \{\emptyset, \{\omega:X_i(\omega) = T\},\{\omega:X_i(\omega) = H\},\Omega\} $$ which in particular for $i=1$ is not what you wrote. That answers your question 2. Also, think first of the case $I = \{1,2\}$ for your original problem of coin tossing, and imagine that $\Omega = \{HH,HT,TH,TT\}$. Can you write down now $\mathcal F_1$ and $\mathcal F_2$? What about $I = \{1,2,3\}$? Finally, your 3rd question does not seem to make sense, but first let's do these exercises. $\endgroup$
    – Ilya
    Nov 26 '14 at 18:41
  • 2
    $\begingroup$ Many thanks for your comment and help Ilya. (I guess you meant $\sigma(X_i)=\{\emptyset, \{w:X_i(w)=0\}, \{w:X_i(w)=1\},\Omega\}$ right?) I see. So in this case $\sigma(X_1)=\{\emptyset, \{TT,TH\}, \{HH,HT\},\Omega\}$ and $\sigma(X_2)=\{\emptyset,\{HT,TT\},\{TH,HH\},\Omega\}$. So it follows that $\sigma(X_1,X_2)=\{\emptyset, \{TT,TH\}, \{HH,HT\}\{HT,TT\},\{TH,HH\},\Omega\}$ (which is itself a $\sigma$-algebra in this case). Similarly for the case $I=\{1,2,3\}$. Thanks very much. $\endgroup$
    – John
    Nov 27 '14 at 16:45
  • 1
    $\begingroup$ Indeed, that's the case - I guess now it shall be easier to approach the original problem. Just tell me if anything else is unclear $\endgroup$
    – Ilya
    Nov 27 '14 at 16:47
  • $\begingroup$ Just a quick question, do we then consider $\Omega=\mathbb{R}^k$ (where $k$ is the cardinality of $I$) in the continous case? $\endgroup$
    – John
    Nov 27 '14 at 16:49
  • $\begingroup$ (By the way, in the third question, I meant transpose; sorry for the typo; hope the question now make sense to you.) $\endgroup$
    – John
    Nov 27 '14 at 16:50

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