### 金融代写|金融数学代写Financial Mathematics代考|MATH3090

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## 金融代写|金融数学代写Financial Mathematics代考|Properties of conditional expectation

Theorem $2.1$ gives the existence of conditional expectation but does not allow us to carry out explicit calculations except in very specific cases. We thus give a series of properties that are sufficient for carrying out calculation in the majority of cases. Most of these properties are similar to those of conventional expectation, hence the name.
PROPOSITION 2.3.- Let $\mathcal{G}$ be a sub- $\sigma$-algebra of $\mathcal{F}$. We then have $\mathbb{E}[1 \mid \mathcal{G}]=1$ a.s.
PROOF.- The function $X=1$ is constant, thus $\mathcal{G}$-measurable (all constants are measurable for all $\sigma$-algebras as they are ${\emptyset, \Omega}$-measurable). Let $A \in \mathcal{G}$. We then have $\mathbb{E}\left[X \mathbb{1}{A}\right]=\mathbb{E}\left[1 \mathbb{1}{A}\right]$; thus, we indeed have $\mathbb{E}[X \mid \mathcal{G}]=\mathbb{E}[1 \mid \mathcal{G}]=1$ a.s.

Proposition 2.4.- Let $X \in L^{1}(\Omega, \mathcal{F}, \mathbb{P})$ and let $\mathcal{G}$ be a sub- $\sigma$-algebra $\mathcal{F}$. We then have
$$\mathbb{E}[\mathbb{E}[X \mid \mathcal{G}]]=\mathbb{E}[X]$$
PROOF.- It is known that $\mathbb{E}[X \mid \mathcal{G}]$ is the single, $\mathcal{G}$-measurable random variable such that for any $A \in \mathcal{G}$ we have $\mathbb{E}\left[\mathbb{E}[X \mid \mathcal{G}] \mathbb{1}{A}\right]=\mathbb{E}\left[X \mathbb{1}{A}\right]$. Taking $A=\Omega$, we obtain
$$\mathbb{E}[\mathbb{E}[X \mid \mathcal{G}]]=\mathbb{E}\left[\mathbb{E}[X \mid \mathcal{G}] \mathbb{1}{\Omega}\right]=\mathbb{E}\left[X \mathbb{1}{\Omega}\right]=\mathbb{E}[X] .$$
PROPOSITION $2.5$ (Linearity of conditional expectation).-Let $X \in L^{1}(\Omega, \mathcal{F}, \mathbb{P})$ and let $\mathcal{G}$ be a sub- $\sigma$-algebra of $\mathcal{F}$. For any $\alpha \in \mathbb{R}, Y \in L^{1}(\Omega, \mathcal{F}, \mathbb{P})$, we have
$$\mathbb{E}[\alpha X+Y \mid \mathcal{G}]=\alpha \mathbb{E}[X \mid \mathcal{G}]+\mathbb{E}[Y \mid \mathcal{G}] \text { a.s. }$$
Proof.- On the one hand, since the two random variables $\mathbb{E}[X \mid \mathcal{G}]$ and $\mathbb{E}[Y \mid \mathcal{G}]$ are $\mathcal{G}$-measurable, their linear combination $\alpha \mathbb{E}[X \mid \mathcal{G}]+\mathbb{E}[Y \mid \mathcal{G}]$ is also $\mathcal{G}$-measurable. On the other hand, by fixing $A \in \mathcal{G}$ and using the linearity of the expectation and the characteristic of the conditional expectation (statement 2 in Theorem 2.1), we obtain
\begin{aligned} \mathbb{E}\left[(\alpha \mathbb{E}[X \mid \mathcal{G}]+\mathbb{E}[Y \mid \mathcal{G}]) \mathbb{1}{A}\right] &=\mathbb{E}\left[\alpha \mathbb{E}[X \mid \mathcal{G}] \mathbb{1}{A}+\mathbb{E}[Y \mid \mathcal{G}] \mathbb{1}{A}\right] \ &=\alpha \mathbb{E}\left[\mathbb{E}[X \mid \mathcal{G}] \mathbb{1}{A}\right]+\mathbb{E}\left[\mathbb{E}[Y \mid \mathcal{G}] \mathbb{1}{A}\right] \ &=\alpha \mathbb{E}\left[X \mathbb{1}{A}\right]+\mathbb{E}\left[Y \mathbb{1}{A}\right] \ &=\mathbb{E}\left[(\alpha X+Y) \mathbb{1}{A}\right] \end{aligned}
hence the result.
PROPOSITION $2.6$ (Monotonicity in conditional expectation).-Let $X$ be a random variable in $L^{1}(\Omega, \mathcal{F}, \mathbb{P})$ and let $\mathcal{G}$ be a sub- $\sigma$-algebra of $\mathcal{F}$. Then, for any $Y \in L^{1}(\Omega, \mathcal{F}, \mathbb{P})$ such that $X \leq Y$ a.s. on $a$
$$\mathbb{E}[X \mid \mathcal{G}] \leq \mathbb{E}[Y \mid \mathcal{G}] \text { a.s. }$$
ProOF.- Let $\epsilon>0$ and $A_{e}={\mathbb{E}[X \mid \mathcal{G}]-\mathbb{E}[Y \mid \mathcal{G}] \geq \epsilon}$. We have $A_{e} \in \mathcal{G}$ since $\mathbb{E}[X \mid \mathcal{G}]-\mathbb{E}[Y \mid \mathcal{G}]$ is $\mathcal{G}$-measurable. Therefore,
$$\mathbb{E}\left[\mathbb{E}[X-Y \mid \mathcal{G}] \mathbb{1}{A{e}}\right]=\mathbb{E}\left[(X-Y) \mathbb{1}{A{e}}\right] \leq 0 \text { a.s. }$$
Consequently, it follows
$$0 \geq \mathbb{E}\left[\mathbb{E}[X-Y \mid \mathcal{G}] \mathbb{1}{A{e}}\right] \geq \in \mathbb{P}\left(A_{e}\right),$$
which implies that $\mathbb{P}\left(A_{e}\right)=0$.

## 金融代写|金融数学代写Financial Mathematics代考|Geometric interpretation

We have directly defined conditional expectation via the Radon-Nykodym theorem for integrable variables. A more geometric construction is possible if we restrict ourselves to square integrable random variables. Conditional expectation may then be interpreted as an orthogonal projection.

Let us recall that a Hilbert space is a complete vector space $H$ endowed with a scalar product. Let us also recall the definition of the orthogonal projection on a sub-space of a Hilbert space. Let $F$ be a closed sub-space of a Hilbert space $H$. Then, for any $x \in H$, there exists a unique $y \in F$, written as $y=\pi F(x)$, which satisfies one of the following equivalent conditions
1) $\forall z \in F,=0$.
2) $\forall z \in F,|x-y|_{H} \leq|x-z|_{H}$.
The vector $y=\pi_{F}(x)$ is called the orthogonal projection of $x$ on $F$.
Let us now establish the relation between projection and conditional expectation in the Hilbert space $L^{2}(\Omega, \mathcal{F}, \mathbb{P})$ endowed with the scalar product $\langle X, Y>=\mathbb{E}[X Y]$.
THEOREM 2.5.-Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, and
$$L^{2}(\mathcal{G})=L^{2}(\Omega, \mathcal{G}, \mathbb{P})=\left{Y \mathcal{G}-\text { measurable; } \mathbb{E}\left[Y^{2}\right]<\infty\right}$$
for any sub- $\sigma$-algebra $\mathcal{G}$ of $\mathcal{F}$. We then have the following properties:

1) $L^{2}(\mathcal{G})$ is a Hilbert space for the scalar product
$$=\mathbb{E}[X Y]$$
2) For any sub- $\sigma$-algebra $\mathcal{H}$ of $\mathcal{G}$, we have $L^{2}(\mathcal{H})$ is a closed sub-space of $L^{2}(\mathcal{G})$.
3) The orthogonal projection of $X$ on $L^{2}(\mathcal{H})$ is $\mathbb{E}[X \mid \mathcal{H}]$.
PROOF.- The first two statements directly follow from the definition of $L^{2}(\mathcal{G})$. For the third statement, let $Z \in L^{2}(\mathcal{H})$. We have
$$\mathbb{E}[Z X \mid \mathcal{H}]=Z \mathbb{E}[X \mid \mathcal{H}]$$
and therefore
$$\mathbb{E}[Z \mathbb{E}[X \mid \mathcal{H}]]=\mathbb{E}[\mathbb{E}[Z X \mid \mathcal{H}]]=\mathbb{E}[Z X]$$
or again
$$\forall Z \in L^{2}(\mathcal{H}),=0$$
hence the result.
REMARK 2.3.- The fact that the conditional expectation of $X$ given $\mathcal{H}$ is the orthogonal projection of $X$ on $L^{2}(\mathcal{H})$ makes it possible to interpret $\mathbb{E}[X \mid \mathcal{H}]$ as the best approximation of $X$ (in the norm $L^{2}$ ) given the information contained in the $\sigma$-algebra $\mathcal{H}$.

COROLLARY 2.2.-Let $X, Y \in L^{2}(\Omega, \mathcal{F}, \mathbb{P})$. Then, $X-\mathbb{E}[X \mid Y]$ and $Y$ are non-correlated.

PrOOF.- Using the definition of the orthogonal projection, we have $\mathbb{E}[(X-$ $\mathbb{E}[X \mid Y]) Y]=0$ and $\mathbb{E}[X-\mathbb{E}[X \mid Y]]=0$; therefore, $\operatorname{Cov}(X-\mathbb{E}[X \mid Y], Y)=0$.

## 金融代写|金融数学代写Financial Mathematics代考|Conditional expectation and independence

To conclude this chapter, let us now explore the relations between conditional expectation and independence.

PROPOSITION 2.11. – Let $X$ be an integrable random variable and $\mathcal{G}$ be a $\sigma$-algebra such that $X$ is independent of $\mathcal{G}$. Then,
$$\mathbb{E}[X \mid \mathcal{G}]=\mathbb{E}[X]$$
In particular, if $Y$ is another random variable independent of $X$, then $\mathbb{E}[X \mid Y]=\mathbb{E}[X]$ a.s.

Proof.- On the one hand, $\mathbb{E}[X]$ is a constant; therefore, it is $\mathcal{G}$-measurable. On the other hand, for any $A \in \mathcal{G}$, because $X$ is independent of $\mathcal{G}$, we have
$$\mathbb{E}\left[X \mathbb{1}{A}\right]=\mathbb{E}[X] \mathbb{E}\left[\mathbb{1}{A}\right]=\mathbb{E}\left[\mathbb{E}[X] \mathbb{1}{A}\right],$$ which implies that $\mathbb{E}[X \mid \mathcal{G}]=\mathbb{E}[X]$. COROLLARY 2.3.- Let $X \in L^{1}(\Omega, \mathcal{F}, \mathbb{P}), f$ be a bounded or positive measurable function or, more generally, a function such that $f(X)$ is integrable and let $\mathcal{G}$ be a sub- $\sigma$-algebra of $\mathcal{F}$. If $X$ is independent of $\mathcal{G}$, then $$\mathbb{E}[f(X) \mid \mathcal{G}]=\mathbb{E}[f(X)] .$$ ProOF.- If $X$ is independent of $\mathcal{G}$, then $f(X)$ is also independent of $\mathcal{G}$. It is then sufficient to apply the previous proposition to $f(X)$. Thus, an independent $\sigma$-algebra of $X$ naturally gives no information on $X$. ExAMPLE 2.7.- Simple symmetric random walk: Let $\left(X{n}\right){n \geq 1}$ be a sequence of independent random variables with the same Rademacher distribution with parameter $1 / 2$ $$\mathbb{P}\left(X{1}=1\right)=\mathbb{P}\left(X_{1}=-1\right)=1 / 2 .$$
Let $S_{0}=0$ and $S_{n}=\sum_{k=1}^{n} X_{k}$ for $n \geq 1$. Let $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$ be the natural filtration of $S=\left(S_{n}\right){n \in \mathbb{N}}$ and $n \geq 0$. Let us calculate $\mathbb{E}\left[S{n+1} \mid \mathcal{F}{n}\right]$ (all the variables are integrable): \begin{aligned} \mathbb{E}\left[S{n+1} \mid \mathcal{F}{n}\right] &=\mathbb{E}\left[S{n}+X_{n+1} \mid \mathcal{F}{n}\right] \ &=\mathbb{E}\left[S{n} \mid \mathcal{F}{n}\right]+\mathbb{E}\left[X{n+1} \mid \mathcal{F}{n}\right] \quad \text { by linearity } \ &=S{n}+\mathbb{E}\left[X_{n+1}\right], \end{aligned}
because $S_{n}$ is $\mathcal{F}{n}$-measurable and $X{n+1}$ is independent of $\mathcal{F}{n}$. We then use the fact that $\mathbb{E}\left[X{n+1}\right]=0$ to obtain $\mathbb{E}\left[S_{n+1} \mid \mathcal{F}{n}\right]=S{n}$.

## 金融代写|金融数学代写Financial Mathematics代考|Properties of conditional expectation

$$0 \geq \mathbb{E}\left[\mathbb{E}[XY \mid \mathcal{G}] \mathbb{1} {A {e}}\right] \geq \in \mathbb{P}\left(A_{e}\right),$$

## 金融代写|金融数学代写Financial Mathematics代考|Geometric interpretation

1)∀和∈F,=0.
2) ∀和∈F,|X−是|H≤|X−和|H.

L^{2}(\mathcal{G})=L^{2}(\Omega, \mathcal{G}, \mathbb{P})=\left{Y \mathcal{G}-\text { 可测量的；} \mathbb{E}\left[Y^{2}\right]<\infty\right}L^{2}(\mathcal{G})=L^{2}(\Omega, \mathcal{G}, \mathbb{P})=\left{Y \mathcal{G}-\text { 可测量的；} \mathbb{E}\left[Y^{2}\right]<\infty\right}

1) 大号2(G)是标量积的希尔伯特空间

=和[X是]
2) 对于任何子σ-代数H的G， 我们有大号2(H)是一个封闭的子空间大号2(G).
3) 的正交投影X上大号2(H)是和[X∣H].

∀从∈大号2(H),=0

ProOOF.- 使用正交投影的定义，我们有和[(X− 和[X∣是])是]=0和和[X−和[X∣是]]=0; 所以，这⁡(X−和[X∣是],是)=0.

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