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Toss a coin three times, so event space $\Omega=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$. We win $\$1$ if we flip a Head and lose $\$1$ for a Tail. Let $\mathbb{P}(H) = p$ and $\mathbb{P}(T) = q$. The change in our wealth after flip $i$ is the r.v.

$$X_i = \cases{+1 \text{ if }H \\ -1 \text{ if } T}$$

Our wealth after turn $i$ is:

$$ S_i = S_0 + X_1 + \dots X_i $$

There are three questions (these are not homework problems, rather revision) and I have some questions about their solutions.

  1. Find $\mathbb{P}(S_3|S_1)$

This is the probability of $S_3$ occuring given that $S_1$ occurs, and by definition:

$$ \mathbb{P}(S_3|S_1) = \frac{\mathbb{P}(S_1 \cap S_3)}{\mathbb{P}(S_1)} $$

I think the answer should be $\frac{1}{2}$ thinking of the $H/T$ outcome as paths on a binary tree. Can someone provide a more algebraic solution?

  1. Find $\mathbb{E}(S_3|S_1)$

We interpret this as our expected wealth after $3$ flips given $S_1$. This is just: $S_1 + p^2(2) + 2pq(0) - q^2(2) = S_1$ iff $p=q$. I think this is okay.

  1. Given that $F_0 = \{\phi,\Omega\}$ what are $F_1, F_2$ and $F_3$ where $F_i$ is the smallest event space that we can identify from complete knowledge of $F_j$ for $1\leq j\leq i$.

I am unfamiliar with filtrations, and have only a vague sense of what they actually are. So is $F_1$ is the event space we can identify from complete knowledge of the first coin flip? What does it mean to identify an event space from complete knowledge?

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The meaning of the phrase the probability of $S_3$ occuring given that $S_1$ occurs is unclear since $S_i$ are random variables, not events. – Did Jun 19 '12 at 7:29
So there is no single value we can associate with $P(S_3|S_1)$ in this case? – anthus Jun 19 '12 at 8:14
Personally, I never use expressions like $P(S_3\mid S_1)$ and I do not know what they could mean. In your view, what could be a definition of $P(S_3\mid S_1)$ as a real number? – Did Jun 19 '12 at 8:34
up vote 2 down vote accepted

$S_1$ and $S_3$ are random variables rather than events, so when reading $\Pr(S_3|S_1)$ in your first question you may want to say something like $$\Pr(S_3=s+2|S_1=s) = \frac14$$ $$\Pr(S_3=s|S_1=s) = \frac12$$ $$\Pr(S_3=s-2|S_1=s) = \frac14.$$

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Thanks Henry, this is much clearer. Could you comment on the filtration, $F = (F_1,F_2,F_3)$? What is $F_1$ for instance. I know that $F_1 \subset F_2$ and that these are subsets of the event space. But the meaning of $F_i$ is not clear to me. – anthus Jun 19 '12 at 8:03
$\mathcal{F_1}$ will be the sigma algebra based on $\{HHH,HHT,HTH,HTT\}$ and $\{THH,THT,TTH,TTT\}$ since you can only distinguish the first event, while $\mathcal{F_2}$ will be the sigma algebra based on $\{HHH,HHT\},$ $\{HTH,HTT\}$, $\{THH,THT\}$ and $\{TTH,TTT\}$, and $\mathcal{F_3}$ will be the power set of your $\Omega$.… might help – Henry Jun 19 '12 at 21:23

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