这是一份GLA格拉斯哥大STATS4024_1 /STATS5026_1作业代写的成功案例

from which it follows that, by monotonicity,
$$
\forall n \in \mathbb{N}: \quad E\left[Y_{n+1} \mid X=x\right] \geq E\left[Y_{n} \mid X=x\right], \quad P_{X} \text {-a.s. }
$$
Moreover,
$$
\forall B \in \mathcal{B}: \quad \int_{[X \in B]} Y_{n} d P=\int_{B} E\left[Y_{n} \mid X=x\right] d P_{X}(x)
$$
and
$$
\forall B \in \mathcal{B}: \quad \int_{[X \in B]} Y d P \geq \int_{[X \in B]} Y_{n} d P
$$
Thus
$$
E[Y \mid X=x] \geq E\left[Y_{n} \mid X=x\right], \quad P_{X} \text {-a.s. }
$$

STATS4024_1 /STATS5026_1 COURSE NOTES ：

Therefore,
$$
P([Y \in A] \mid \cdot)=P([Y \in A]), \quad P_{X} \text {-a.s. }
$$
and if $Y$ is a real-valued integrable random variable, then
$$
E[Y \mid \cdot]=E[Y], \quad P_{X}-a . s .
$$
Proof: Independence of $X$ and $Y$ is equivalent to
$$
P([X \in B] \cap[Y \in A])=P([X \in B]) P([Y \in A]) \quad \forall A \in \mathcal{B}{1}, B \in \mathcal{B} $$ or $$ \begin{aligned} \int{[X \in B]} I_{[Y \in A]}(\omega) P(d \omega) &=P([Y \in A]) \int I_{B}(x) d P_{X}(x) \
&=\int_{D} P([Y \in A]) d P_{X}(x)
\end{aligned}
$$