Steven R. Dunbar
Department of Mathematics
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Stochastic Processes and

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General Binomial Trees

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_______________________________________________________________________________________________ ### Rating

Mathematicians Only: prolonged scenes of intense rigor.

_______________________________________________________________________________________________ ### Section Starter Question

__________________________________________________________________________ ### Key Concepts

1. Given the probability measure $ℙ$ on paths, and the Radon-Nikodym derivative $\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}$, the probability measure $ℚ$ is the product $\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}ℙ$.
2. Note that the Radon-Nikodym derivative is deﬁned on paths, and is $\mathsc{ℱ}$-measurable, so it is also a random variable.
3. Let ${\zeta }_{t}=\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}$. Then
${\zeta }_{t}=𝔼ℙ\left[\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}\phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}{\mathsc{ℱ}}_{t}\right]$

for every $t$. This says that the expectation, knowing the information up to time $t$ with respect to $ℙ$, unpicks $\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}$ in just the right way.

4. If we want to know $𝔼ℚ\left[F\left({X}_{t}\right)\phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}{\mathsc{ℱ}}_{s}\right]$, then we would need the amount of change from time $s$ to time $t$. That is just ${\zeta }_{t}∕{\zeta }_{s}$, which is change up to time $t$ with the change up to time $s$ removed. In other words
$𝔼ℚ\left[F\left({X}_{t}\right)\phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}{\mathsc{ℱ}}_{s}\right]={\zeta }_{s}^{-1}𝔼ℙ\left[{\zeta }_{t}F\left({X}_{t}\right)\phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}{\mathsc{ℱ}}_{s}\right].$

__________________________________________________________________________ ### Vocabulary

1. A ﬁltration is an increasing sequence of $\sigma$-algebras on a measurable space.
2. A stochastic process ${X}_{t}$ is said to be adapted to the ﬁltration if each ${X}_{t}$ is an integrable random variable that is measurable with respect to the corresponding $\sigma$-algebra ${\mathsc{ℱ}}_{t}$.
3. The likelihood ratio as $\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}$, called the Radon-Nikodym derivative of $ℚ$ with respect to $ℙ$. Given the probability measure $ℙ$ on paths, and the Radon-Nikodym derivative $\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}$, the probability measure $ℚ$ is the product $\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}ℙ$.
4. If $ℙ\left[B\right]=0\phantom{\rule{3.26288pt}{0ex}}⇒\phantom{\rule{3.26288pt}{0ex}}ℚ\left[B\right]=0$ we say $ℚ$ is absolutely continuous with respect to $ℙ$ and write $ℚ\ll ℙ$.
5. Two probability measures on the space $\Omega$ with $\sigma$-algebra $\mathsc{ℱ}$ are equivalent if for any set $B\in \mathsc{ℱ}$, $ℙ\left[B\right]>0$ if and only if $QrobB>0$. This is, the probability measures are equivalent if $ℙ\ll ℚ$ and $ℚ\ll ℙ$.
6. If there is a $B$ with $ℙ\left[B\right]=0$ and $ℚ\left[B\right]=1$ we say $ℚ$ is completely singular with respect to $ℙ$. Note: $ℙ\left[{B}^{C}\right]=1$ and $ℚ\left[{B}^{C}\right]=0$ so then $ℙ$ is completely singular with respect to $ℚ$.

__________________________________________________________________________ ### Mathematical Ideas

#### General Binomial Trees

##### Path Probabilities and Filtrations

The purpose of this section is to set up a general formulation for understanding binomial processes. Ultimately the general formulation provides intuition for ﬁltrations, Radon-Nikodym derivatives, and Brownian motion

Consider the multi-step binomial process with probability $1∕2$ on each branch represented in Figure 1. For comparison, also consider the multi-step binomial process with probability $2∕3$ on each branch representing heads and probability $1∕3$ on each branch representing tails in Figure 2. Figure 1: A binomial tree with probability $1∕2$ on each branch. Figure 2: A skewed binomial tree with probabilities $2∕3$ and $1∕3$ on each branch.

Instead of branch probabilities, it is possible to assign the probability of each node in the tree. This is represented in Figure 3. This assignment gives the path probability measure instead. Given the branch probabilities, the path probabilities can be easily constructed. Given the path probabilities the branch probabilities can be recovered (if the probabilities are not $0$ or $1$.) Figure 3: A general binomial tree.

For stochastic processes, a ﬁltration represents information about the process available up to and including each time $t$ through the sets of sample paths that satisfy all measurable conditions up to time $t$. A ﬁltration has more information as time increases. That is, the set of measurable events for the stochastic process stays the same or increases as more conditions from the evolution of the stochastic processes become available.

Deﬁnition. A ﬁltration is an increasing sequence of $\sigma$-algebras on a measurable space $\Omega$. That is, given a measurable space ($\Omega ,\mathsc{ℱ}$), a ﬁltration is a sequence of $\sigma$-algebras ${\left\{{\mathsc{ℱ}}_{t}\right\}}_{t\ge 0}$ with ${\mathsc{ℱ}}_{t}\subseteq \mathsc{ℱ}$ for each $t$ and ${t}_{1}\le {t}_{2}\phantom{\rule{3.26288pt}{0ex}}⇒\phantom{\rule{3.26288pt}{0ex}}{\mathsc{ℱ}}_{{t}_{1}}\subseteq {\mathsc{ℱ}}_{{t}_{2}}$.

A stochastic process ${X}_{t}$ is said to be adapted to the ﬁltration if each ${X}_{t}$ is an integrable random variable that is measurable with the respect to the corresponding $\sigma$-algebra ${\mathsc{ℱ}}_{t}$.

Example. For the multi-stage binomial process represented in Figure 3 the sample space, denoted

is the set of all possible inﬁnite sequences of $0$s and $1$s representing all possible outcomes of the composite experiment.

The $\sigma$-algebra denoted ${\Omega }_{n}$ is the set of all possible sequences of $n$ 0’s and 1’s representing all possible outcomes of the composite experiment. We denote an element of ${\Omega }_{n}$ as $\omega =\left({\omega }_{1},\dots ,{\omega }_{n}\right)$, where each ${\omega }_{k}=0$ or $1$. That is, ${\Omega }_{n}={\left\{0,1\right\}}^{n}$.

A ﬁltration adapted to the multi-stage binomial process is

$\begin{array}{llll}\hfill {\mathsc{ℱ}}_{0}& =\left\{\varnothing ,\Omega \right\}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\mathsc{ℱ}}_{1}& =\left\{\varnothing ,\left(0\right),\left(1\right),\Omega \right\}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\mathsc{ℱ}}_{2}& =\left\{\varnothing ,\left(0\right),\left(1\right),\left(00\right),\left(01\right),\left(10\right),\left(11\right),\Omega \right\}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\mathsc{ℱ}}_{3}& =\left\{\varnothing ,\left(0\right),\left(1\right),\left(00\right),\left(01\right),\left(10\right),\left(11\right),\left(000\right),\left(001\right),\left(010\right),\left(011\right),\left(100\right),\left(101\right),\left(110\right),\left(111\right),\Omega \right\}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

In general ${\mathsc{ℱ}}_{n}$ is a set of ${2}^{n+1}$ sets. This ﬁltration represents the sets of sample paths that satisfy all measurable conditions in the binomial process up to time $t$.

Next put a probability measure $ℙ$ on the ﬁltration as the product of the probabilities along each branch of the binomial tree. So for example at the third level of the tree, the probabilities would be

 $\left(000\right)$ ${p}_{30}{p}_{20}{p}_{10}$ ${\pi }_{000}$ $\left(001\right)$ ${p}_{30}{p}_{20}{p}_{11}$ ${\pi }_{001}$ $\left(010\right)$ ${p}_{30}{p}_{21}{p}_{10}$ ${\pi }_{010}$ $\left(011\right)$ ${p}_{30}{p}_{21}{p}_{11}$ ${\pi }_{011}$ $\left(100\right)$ ${p}_{31}{p}_{20}{p}_{10}$ ${\pi }_{100}$ $\left(101\right)$ ${p}_{31}{p}_{20}{p}_{11}$ ${\pi }_{101}$ $\left(110\right)$ ${p}_{31}{p}_{21}{p}_{10}$ ${\pi }_{110}$ $\left(111\right)$ ${p}_{31}{p}_{21}{p}_{11}$ ${\pi }_{111}$

The probabilities on paths, ${\pi }_{0},{\pi }_{1},{\pi }_{10},{\pi }_{11},{\pi }_{01},\dots ,{\pi }_{000},\dots ,$ contain all the information about the probabilities along each path. For example at the third level, knowing the $8$ probabilities ${\pi }_{000},\dots ,{\pi }_{111}$ and the $4$ probabilities ${\pi }_{11},{\pi }_{10},{\pi }_{01},{\pi }_{00}$ is suﬃcient to ﬁnd the $8$ branch probabilities ${p}_{10},{p}_{11},\dots ,{p}_{31}$, so long as none of ${\pi }_{00},\dots ,{\pi }_{11}$ are $0$ or $1$. Figure 4: Another general binomial tree.

Now suppose we had a diﬀerent measure $ℚ$ on the paths with branch probabilities ${q}_{ij}$ summarized in path probabilities ${\varphi }_{ijk\dots }$ and so on. As in the previous case, it is suﬃcient to specify the path probabilities, since the branch probabilities can be determined from that path information, so long as each ${\varphi }_{ijk\dots }$ are not $0$ or $1$.

The likelihood ratios ${\varphi }_{ijk\dots }∕{\pi }_{ijk\dots }$provide a way of mapping between the two measures $ℙ$ and $ℚ$. Write the likelihood ratio as $\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}$, called the Radon-Nikodym derivative of $ℚ$ with respect to $ℙ$. Given the probability measure $ℙ$ on paths, and the Radon-Nikodym derivative $\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}$, the probability measure $ℚ$ is the product $\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}ℙ\left[\cdot \right]$. Note also that since the Radon-Nikodym derivative $\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}$ is deﬁned on paths, it is a random variable on the space $\Omega$. Even more, the Radon-Nikodym derivative is ${\mathsc{ℱ}}_{t}$-adapted. Figure 5: A general binomial tree with the Radon-Nikodym derivative.

##### Two Probability Measures

The case when some branch probability ${p}_{ij}$ is $0$ or $1$ needs brief consideration. For example, if ${p}_{21}=0$ then ${\pi }_{111}$ and ${\pi }_{110}$ are both $0$ and no further information about paths through ${\pi }_{111}$ or ${\pi }_{110}$ is recoverable. Since all paths past the node $\left(11\right)$ have probability $0$, that is to say, are impossible, the probabilities ${p}_{21}$ and ${p}_{20}$ are irrelevant anyway, so this is not a serious loss of information. If the probability measure is provided only for paths along which ${\pi }_{ijk\dots }\ne 0$ then we can recover the non-zero information about the corresponding branch probabilities.

Considering the probability measures $ℙ$ and $ℚ$ and the Radon-Nikodym derivative $\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}$ in the case when some ${p}_{ij}$ is $0$ leads to an important deﬁnition. Then some path probabilities ${\pi }_{ijk\dots }$ are $0$, that is those paths are impossible. Suppose for the moment that with respect to the probability $ℚ$ those same paths do not have probability $0$. Then the Radon-Nikodym derivative $\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}$ is not deﬁned along those paths, and it is not possible to recover $ℚ$ from $ℙ$. Restricting to the non-zero probability paths with respect to $ℙ$ is not adequate, because that still loses $ℚ$ information about the paths that are $ℙ$-impossible but are $ℚ$-possible.

Deﬁnition. If $ℙ\left[B\right]=0\phantom{\rule{3.26288pt}{0ex}}⇒\phantom{\rule{3.26288pt}{0ex}}ℚ\left[B\right]=0$ we say $ℚ$ is absolutely continuous with respect to $ℙ$ and write $ℚ\ll ℙ$.

Note that an equivalent deﬁnition of absolute continuity is that $ℚ\left[B\right]>0\phantom{\rule{3.26288pt}{0ex}}⇒\phantom{\rule{3.26288pt}{0ex}}ℙ\left[B\right]>0$.

Deﬁnition. Two probability measures on the space $\Omega$ with $\sigma$-algebra $\mathsc{ℱ}$ are equivalent if for any set $B\in \mathsc{ℱ}$, $ℙ\left[B\right]>0$ if and only if $QrobB>0$. This is, the probability measures are equivalent if $ℙ\ll ℚ$ and $ℚ\ll ℙ$.

Deﬁnition. If there is a $B$ with $ℙ\left[B\right]=0$ and $ℚ\left[B\right]=1$ we say $ℚ$ is completely singular with respect to $ℙ$. Note that $ℙ\left[{B}^{C}\right]=1$ and $ℚ\left[{B}^{C}\right]=0$ so $ℙ$ is completely singular with respect to $ℚ$ and we write

$ℙ\perp ℚ$

If

• $ℙ$ and $ℚ$ are probability measures on common $\sigma$-algebra $\mathsc{ℱ}$; and
• $ℚ$ is absolutely continuous with respect to $ℙ$ then there is a function
$L\left(x\right)=\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}.$

and furthermore for any $\mathsc{ℱ}$-measurable function $F$

$𝔼ℚ\left[F\left(X\right)\right]=𝔼ℙ\left[F\left(X\right)L\left(X\right)\right]=𝔼ℙ\left[F\left(X\right)\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}\right].$

Note that the Radon-Nikodym derivative is deﬁned on paths, and is $\mathsc{ℱ}$-measurable, so it is also a random variable. For simplicity, let ${\zeta }_{t}=\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}$. Then

${\zeta }_{t}=𝔼ℙ\left[\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}\phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}{\mathsc{ℱ}}_{\bigsqcup }\right]$

for every $t$. This says that the expectation, knowing the information up to time $t$, with respect to $ℙ$ accumulates $\frac{\phantom{\rule{0.3em}{0ex}}dℚ}{\phantom{\rule{0.3em}{0ex}}dℙ}$ in just the right way. The process ${\zeta }_{t}$ represents the amount of change of measure so far up to time $t$ along the current path. If we want to know $𝔼ℚ\left[F\left({X}_{t}\right)\right]$, it would be $𝔼ℙ\left[{\zeta }_{t}F\left({X}_{t}\right)\right]$. If we want to know $𝔼ℚ\left[F\left({X}_{t}\right)\phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}{\mathsc{ℱ}}_{s}\right]$, then we would need the amount of change from time $s$ to time $t$. That is just ${\zeta }_{t}∕{\zeta }_{s}$, which is change up to time $t$ with the change up to time $s$ removed. In other words

 $𝔼ℚ\left[F\left({X}_{t}\right)\phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}{\mathsc{ℱ}}_{s}\right]={\zeta }_{s}^{-1}𝔼ℙ\left[{\zeta }_{t}F\left({X}_{t}\right)\phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}{\mathsc{ℱ}}_{s}\right].$ (1)

Example.

The point of the example is to apply the deﬁnitions and notations to the pair of binomial trees with $ℙ$ deﬁned by $ℙ\left[H\right]=\frac{1}{2}$ and $ℙ\left[T\right]=\frac{1}{2}$ and $ℚ\left[H\right]=\frac{2}{3}$ and $ℚ\left[T\right]=\frac{1}{3}$. This has minimal mathematical content, but is a good illustration of deﬁnitions and notation.

Calculating conditional probabilities with the simple deﬁnition from elementary probability remains easy on a point-by-point basis.

$\begin{array}{llll}\hfill & 𝔼ℚ{\left[{1}_{\left[{X}_{3\ge 1}\right]}\phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}{\mathsc{ℱ}}_{2}\right]}_{X=0}=ℚ\left[{X}_{3}=1\phantom{\rule{0.3em}{0ex}}|HT,HT\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}=ℚ\left[HHH\phantom{\rule{0.3em}{0ex}}|HT,TH\right]=\frac{ℚ\left[HHH\right]}{ℚ\left[HT,TH\right]}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}=\frac{8∕27}{4∕9}=\frac{2}{3}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The Radon-Nikodym derivative along this tree is

${\zeta }_{t}={\left(\frac{4}{3}\right)}^{\frac{t+{X}_{t}}{2}}\cdot {\left(\frac{2}{3}\right)}^{\frac{t-{X}_{t}}{2}}.$

Now calculate the same conditional probabilities with formula (1) in Tables 1, 2,3.

 Node Expectation ${X}_{2}$ Prob HH $\frac{2}{3}\cdot 1+\frac{1}{3}\cdot 1$ 2 1 HT $\frac{2}{3}\cdot 1+\frac{1}{3}\cdot 0$ 1 $\frac{2}{3}$ TH $\frac{2}{3}\cdot 1+\frac{1}{3}\cdot 0$ 1 TT $\frac{2}{3}\cdot 0+\frac{1}{3}\cdot 0$ 0 0

Table 1: $𝔼ℚ\left[{1}_{\left[{X}_{3\ge 1}\right]}\phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}{\mathsc{ℱ}}_{2}\right]$

 Node Expectation ${X}_{2}$ Prob HH $\frac{1}{2}\cdot 1\cdot {\left(\frac{4}{3}\right)}^{3}+\frac{1}{2}\cdot 1\cdot {\left(\frac{4}{3}\right)}^{2}\left(\frac{2}{3}\right)$ $2$ ${\left(\frac{4}{3}\right)}^{2}$ HT $\frac{1}{2}\cdot 1\cdot {\left(\frac{4}{3}\right)}^{2}\left(\frac{2}{3}\right)+\frac{1}{2}\cdot 0\cdot \left(\frac{4}{3}\right){\left(\frac{2}{3}\right)}^{2}$ 1 $\frac{1}{2}{\left(\frac{4}{3}\right)}^{2}\frac{2}{3}$ TH $\frac{1}{2}\cdot 1\cdot {\left(\frac{4}{3}\right)}^{2}\left(\frac{2}{3}\right)+\frac{1}{2}\cdot 0\cdot \left(\frac{4}{3}\right){\left(\frac{2}{3}\right)}^{2}$ 1 $\frac{1}{2}{\left(\frac{4}{3}\right)}^{2}\frac{2}{3}$ TT $\frac{1}{2}\cdot 0\cdot \left(\frac{4}{3}\right){\left(\frac{2}{3}\right)}^{2}+\frac{1}{2}\cdot 0\cdot {\left(\frac{2}{3}\right)}^{3}$ 0 $0$

Table 2: $𝔼ℙ\left[{1}_{\left[{X}_{3}\ge 1\right]}{\zeta }_{3}\phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}{\mathsc{ℱ}}_{2}\right]$

 ${X}_{2}$ $𝔼ℚ\left[{1}_{\left[{X}_{3\ge 1}\right]}|{\mathsc{ℱ}}_{2}\right]$ $𝔼ℙ\left[{1}_{\left[{X}_{3}\ge 1\right]}{\zeta }_{3}|{\mathsc{ℱ}}_{2}\right]$ ${\zeta }_{2}$ Prob 2 $1$ ${\left(\frac{4}{3}\right)}^{2}$ ${\left(\frac{4}{3}\right)}^{2}$ 1 1 $\frac{2}{3}$ $\left(\frac{1}{2}\right){\left(\frac{4}{3}\right)}^{2}\frac{2}{3}$ $\left(\frac{4}{3}\right)\left(\frac{2}{3}\right)$ $\frac{2}{3}$ 0 $0$ $0$ ${\left(\frac{2}{3}\right)}^{2}$ 0

Table 3: Summary of $𝔼ℚ\left[F\left({X}_{t}\right)\phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}{\mathsc{ℱ}}_{s}\right]={\zeta }_{s}^{-1}𝔼ℙ\left[{\zeta }_{t}F\left({X}_{t}\right)\phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}{\mathsc{ℱ}}_{s}\right]$.

#### Sources

This section is adapted from: “Chapter 2, Discrete Processes” in Financial Calculus by M. Baxter, A. Rennie, Cambridge University Press, Cambridge, 1996, .

_______________________________________________________________________________________________ ### Problems to Work for Understanding

1. On the general binomial tree, prove that
${\zeta }_{t}=𝔼ℙ\left[\frac{\phantom{\rule{0.3em}{0ex}}dℙ}{\phantom{\rule{0.3em}{0ex}}dℚ}\phantom{\rule{0.3em}{0ex}}|\phantom{\rule{0.3em}{0ex}}{\mathsc{ℱ}}_{\bigsqcup }\right]$

holds for $t=1$, $2$, and $3$.

__________________________________________________________________________ ### References

   M. Baxter and A. Rennie. Financial Calculus: An introduction to derivative pricing. Cambridge University Press, 1996. HG 6024 A2W554.

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