# Where is the mistake using trivial formulas for conditional expectation?

I want to calculate a conditional expectation and I do not see where my mistake is. I'm solving the following exercise (namely, 1a):

(source)

Here $S^1,S^2$ are stochastic processes, which model for example a stock price. At time (day) $0$, $S^1_0=100$, which means that the price of stock one is $100$ unit of money. Then the stock price can go up, with probability $p_u=\frac{2}{3}$ and go down with prob. $p_d=\frac{1}{3}$. Hence at day $1$ it can take the prices $S^1_1=104$ or $S^1_1=98$ and so on.

The exercise want to construct an equivalent martingale measure for $S^1$. To have an equivalent martingale for $S^1$ I need to find transition probabilities $q_u,q_d=1-q_u$ (u=up, d=down) and $q_{u,u},q_{u,d}=1-q_{u,u},q_{d,u},q_{d,d}=1-q_{d,u}$ such that $S^1=(S^1_k), k=0,1,2$ is a $Q^1$-martingale. The equivalence of the measure follows immediately if $q_j,q_{j,i}>0$, $i,j\in\{u,d\}$. The filtration is generated by $S^1$, i.e. $\mathcal{F}_0=\sigma(S^1_0)$ which is trivial and $\mathcal{F}_1=\sigma(S^1_1)$. To be a martingale under the measure $Q^1$ (which is characterized by transition probabilities $q_j,q_{j,i}$ $i,j\in\{u,d\}$ we have to solve the equations:

$$E_{Q^1}[S^1_1]=100$$ $$E_{Q^1}[S^1_2|\mathcal{F}_1]=S^1_1$$

the first one is easy to solve and gives $q_u=\frac{1}{3}$ and $q_d=\frac{2}{3}$. Now for the second equation I want to use that $\mathcal{F}_1$ is generated by $\sigma(A_1,A_2)$, where $A_1=\{S^1_1=104\},A_2=\{S^1_1=98\}$. Then I know that

$$E_{Q^1}[S^1_2|\mathcal{F}_1]=\sum_{j=1}^2\frac{E[S^1_2\mathbf1_{A_j}]}{Q^1[A_j]}\mathbf1_{A_j}$$

hence for writting this out gives two equations:

$$\frac{1}{Q^1[A_1]}(116.48\cdot q_{u,u}+(1-q_{u,u})\cdot 99.84)=104$$ $$\frac{1}{Q^1[A_2]}(101.92\cdot q_{d,u}+(1-q_{d,u})\cdot 96.04)=98$$

where clearly $\frac{1}{Q^1[A_1]}=\frac{1}{q_u}$ and $\frac{1}{Q^1[A_2]}=\frac{1}{q_d}$. I would get the right result without the $\frac{1}{Q^1[A_j]}$ in the front of the equations. But I do not see why they do not have to be there. Right result: $q_{u,u}=\frac{1}{4}$, $q_{d,u}=\frac{1}{3}$. It would be very helpful, if someone could point out, where I my mistake is exactly.

As mentioned in the comment, I used the following theorem for calculating the conditional expectation:

Let $(\Omega,\mathcal{F},P)$ be a prob. space, $A_i\in \mathcal{F}$, for $1\le i\le N\le\infty$ pairwise disjoint measurable sets with $P(A_i)>0$ and $\bigcup_{i=1}^N A_i=\Omega$. Let $\mathcal{A}=\sigma(A_i;1\le i\le N)$. Let $X\in L^1(\Omega,\mathcal{F},P)$, then $$E[X|\mathcal{A}]=\sum_{i=1}^N\frac{E[X\mathbf1_{A_i}]}{P(A_i)}\mathbf1_{A_i}$$

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Please make your post self-contained instead of relying on some off-site information. Why should people spend some time to help you if you cannot be bothered to spend some time to ask your question properly? –  Did Jan 4 '13 at 20:07
@did please let me know what kind of off-site information you mean. I spent a lot of time solving this problem and also posting this question. If something should not be clear, please let me know what exactly. But I do not see, where my post should rely on some off-site information. –  user20869 Jan 4 '13 at 20:16
@did if it is the equation for $E_{Q^1}[S^1_2|\mathcal{F}_1]$, I will add this. But I thought this is a standard example in probability theory for cond. expectation –  user20869 Jan 4 '13 at 20:19
To begin with: $S^i_j$, $q_i$, $q_{i,j}$, $Q^i$, $100$, $104$, $98$, $116.48$, etc., etc., etc. (Note that your second comment is mildly insulting, but that I love the upvote on your first comment, so sweet...) –  Did Jan 4 '13 at 21:00
@did I added the details, and thanks to Zev Chonoles, who insert the picture of the pdf. I did not add the details first, because it seemed to be an "overkill" to me, since they are all explained in the link, I posted. I hope, that everything is fine now. About my second comment: It was not at all my intention to insult someone. If this should be the case, I apologize. Since I need help on this question this would be rather counterproductive. –  user20869 Jan 5 '13 at 9:46

The formula that you use is correct, however you make a mistake when computing $\mathsf E_{\mathsf Q^1}[S_2^1\cdot 1_{A_1}]$. To see this, let us write explicitly everything that we have, i.e. $\Omega = \{uu,ud,du,dd\}$ and you further partition it as $A_1 = \{uu,ud\}$ and $A_2 = \Omega\setminus A_1$. You thus have $$\mathsf E_{\mathsf Q^1}[S_2^1\cdot 1_{A_1}] = S_2^1(uu)\mathsf Q^1(uu)+S_2^1(ud)\mathsf Q^1(ud).$$ Note now that $\mathsf Q^1(uu) = q_u\cdot q_{uu}$ since this it the probability that the price will go up both times. Instead, you compute the expectation assuming that $\mathsf Q^1(uu) = q_{uu}$ which is not correct. That's why the terms with $q_u$ do not cancel and your obtain an incorrect answer in the end.
after a while, there comes up another small question. Am I right, that $Q[A_1]=q_u$? Because: $Q[A_1]=q_u(q_{uu}+q_{ud})=q_u$ since $q_{uu}+q_{dd}=1$. And if we continue drawing the picutre, which means trading $T=10$ days. Then the probability of for example : $\{uudduud\}$ which is the event of being at day $7$ at the position $\{uudduud\}$ would be $q_uq_{uu}q_{uud}q_{uudd}q_{uuddu}q_{uudduu}q_{uudduud}$, am I right? Thanks again for your help –  user20869 Jan 22 '13 at 14:33
@hulik: yes, as these are conditional transition probabilities, so to apply this formula for 3 steps you have $$\mathsf P(X_1 = u,X_2 = u,X_3 = d) = \mathsf P(X_3 = d|X_1 = u,X_2 = u)\cdot \mathsf P(X_2 = u|X_1 = u)\cdot \mathsf P(X_1 = u)$$ –  Ilya Jan 22 '13 at 14:37