# How to find density process in this binomial model

Suppose we have the binomial model, i.e. time horizon $T=2$, a process $S_k$ with $k=0,1,2$, filtration generated by $S$. The Probability space consists of the following elements: $\{uu,ud,du,dd\}$, where $u$ stands for up and $d$ for down. S evolves like this: $S_0=100$, then at time $k=1$, $S_1$ can take the two values $V_u$ or $V_d$ with probability $p_u$ or $p_d$. At $k=2$, $S_2$ can take the values $V_{uu}$ with probability $p_up_{uu}$, $V_{ud}$ with $p_up_{ud}$, $V_{du}$ with $p_dp_{du}$ and $V_{dd}$ with $p_dp_{dd}$. Hence $p_{ij}$ with $i,j\in\{u,d\}$ is the conditional probability given time $k=1$ ending in $V_{ij}$.

Now suppose that I found also an equivalent measure $Q$ which can also be described in the same way, i.e. we have $q_j>0$ and $q_{ij}>0$ with $i,j\in\{u,d\}$. Then We define the density process $Z$ as

$$Z_k:=E[\frac{dQ}{dP}|\mathcal{F}_k] \mbox{ and }D_k:=\frac{Z_k}{Z_{k-1}}$$

where $dQ/dP$ refers to the Randon Nikodym derivative. Why is it true that

$$\frac{dQ}{dP}\{(i,j)\}=\frac{Q({i,j})}{P(i,j)}=\frac{q_iq_{ij}}{p_ip_{ij}}$$

The second equality is clear, just pluggin in. It is the first one, which bothers me. It seems like coming from the definition, however it is not absolutely clear for me. Thanks for your help

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