Take the 2-minute tour ×

I'm stuck with a part of a homework problem and need some clarification. Let $\Omega$ be the sample space for flipping $n$ fair coins, i.e. the set of all $n$-tuples of $E$ and $N$, eagle and number, the $\sigma$-algebra $\cal{F}$ of all subsets of $\Omega$ and $P(A)=\frac{|A|}{2^n}$.

Let $X_k$ denote the number of the eagles in the first $k$ flips, i.e. if $\omega=(x_1,...,x_n)$, then $X_k(\omega)=|\{1\leq i\leq k:x_i=E\}|$. Describe $\cal{F}_{X_k}$. How many elements does it have? Give an example of an element in $\Omega$, but not in $\cal{F}_{X_k}$. Show that that $E[X_n |\cal{F}_{X_k}]= X_k+\frac{n-k}{2}$

Wouldn't $\cal{F}_{X_k}$ be a set of all $k$-tuples, hence $2^k$ elements? But then no element of $\cal{F}_{X_k}$ would be in $\Omega$, so that seems not correct. But what is $\cal{F}_{X_k}$ then? Or is this just a trick question?

For the second part, this is the expectation of flipping $n$ coins after the result of the first $k$ coins is known; then we are left with $\frac{n-k}{2}$ eagles out of $n-k$ further coins. But how do I prove this formally?

Yours, Marie

share|improve this question

2 Answers 2

up vote 1 down vote accepted

Let $\Psi=\{E,N\}$, then $$\mathcal F_k=\{A_I^k\times\Psi^{n-k}\mid I\subseteq\{0,1,\ldots,k\}\}, $$ where, for every $I\subseteq\{0,1,\ldots,k\}$, $$ A_I^k=\{(x_n)_{1\leqslant n\leqslant k}\in\Psi^k\mid x_1+\cdots+x_k\in I\}. $$ For example, $A_\varnothing^k=\varnothing$ hence $A_\varnothing^k\times\Psi^{n-k}=\varnothing$, and $A_{I_k}^k=\Psi^k$ for $I_k=\{0,1,\ldots,k\}$ hence $A_{I_k}^k\times\Psi^{n-k}=\Omega$. And $\mathcal F_k$ is equipotent to the set of subsets of $\{0,1,\ldots,k\}$, hence the size of $\mathcal F_k$ is...

Introduce $Y_k:\Omega\to\{0,1\}$, $\omega=(x_i)_{1\leqslant i\leqslant n}\mapsto\mathbf 1_{x_k=E}$. Then the value of $\mathrm E(X_n\mid\mathcal F_k)$ is a consequence of the fact that, for every $1\leqslant i\leqslant k\lt j\leqslant n$, $$ \mathrm E(Y_i\mid\mathcal F_k)=Y_i,\qquad\mathrm E(Y_j\mid\mathcal F_k)=1/2. $$

share|improve this answer
So $\cal{F}_{X_k}=\cal{F}_k$, right? –  Marie. P. Mar 7 '12 at 19:54
Right. Answer modified. –  Did Mar 7 '12 at 20:05

${\cal F}_{X_k}$ is the set of all subsets $A$ of $\Omega$ such that the value of $X_k$ determines whether an outcome is in $A$. Thus $A$ could be the set of $n$-tuples $\omega$ such that the number of E's in $(\omega_1, \ldots, \omega_k)$ is ....

The question asking for an example may be a mistake, or at least poorly worded. Elements of $\Omega$ are $n$-tuples, but elements of ${\cal F}_{X_k}$ are sets of $n$-tuples. It could be that they want a single $\omega \in \Omega$ such that the set $\{\omega\}$ is not in ${\cal F}_{X_k}$, or perhaps they meant a member of $\cal F$ instead of a member of $\Omega$.

share|improve this answer

Your Answer


