1
$\begingroup$

Consider the probability space $(\Omega_3,\mathcal{F}_3,\mathbb{P})$ where the outcome space $\Omega_3$ is all sequences of three coin tosses, the $\sigma$-algebra $\mathcal{F}_3$ is all subsets of $\Omega_3$, and the probability measure is generated by a fair coin. Suppose that $Y_n=Y_{n-1}+X_n$ where $X_n=1$ if the $n^{th}$ toss is heads and $X_n=-1$ if the $n^{th}$ toss is tails. Define $$ V_n = \left\{ \begin{array}{lr} Y_n & if \;Y_m\geq0\;for\;0\leq m \leq n\\ 0 &otherwise \end{array} \right.$$ 1. What are the elements of the $\sigma$-algebra $\sigma(Y_2)$?

2.Compute the generalized conditional expectation $\mathbb{E}(V_3\;|\;\sigma(Y_2))$.

I've been messing with this problem but am not entirely sure I know what I'm doing. Can someone give a solution? Here is what I came up with. Define the sets $$A_{HH}=\{HHH,HHT\},\;A_{HT\cup TH}=\{HTH,HTT,THH,THT\},\;A_{TT}=\{TTT,TTH\}$$ It seems to me that $\sigma(Y_2)$ is all complements and unions of $$\{\emptyset,A_{HH},A_{HT\cup TH},A_{TT},\Omega_3\}$$ As far as the generalized conditional expectation I get $$\mathbb{E}(V_3\;|\;A_{HH})=\frac{1}{2}4+\frac{1}{2}2=3,$$ and similarly $\;\mathbb{E}(V_3\;|\;A_{HT \cup TH})=1,\;\mathbb{E}(V_3\;|\;A_{TT})=0$.

$\endgroup$
0
$\begingroup$

Disclaimer (Jun 21'16): This answer was recently downvoted, twice. Needless to say, it is perfectly correct, and it answers the question. The downvotes might be due to extra-mathematical reasons. Happy reading!

Yes. And all this, since you are asked $E(V_3\mid Y_2)$, should be summarized as $$E(V_3\mid Y_2)=\tfrac12Y_2(Y_2+1).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.