5
$\begingroup$

Let $(Y_1, \ldots, Y_n)$ be a $[0,1]^n$-valued random vector and $U_1, \ldots, U_n$ independent random variables, uniformly distributed on $[0,1]$, and independent of $(Y_1, \ldots, Y_n)$. For some fixed $j \in \{1, \ldots, n\}$ consider the random variable $X := \mathbb{1}_{\{U_j \leq Y_j\}}$. Then I want to show that $$ \mathbb{E}[X \vert \sigma(Y_1, \ldots, Y_n)] = \mathbb{E}[X \vert \sigma(Y_j)]. $$ To prove this, I thought I can use the result, that if $\mathcal{F}$ is independent of $\sigma(\mathcal{G},\sigma(X))$, then $$ \mathbb{E}[ X \vert \sigma(\mathcal{F},\mathcal{G}) ] = \mathbb{E}[X \vert \mathcal{G}]. $$ Hence we set $\mathcal{G} : = \sigma(Y_j)$ and $\mathcal{F} : = \sigma(Y_i \colon i \in \{1, \ldots, n\} \setminus \{j\} )$. Then $\sigma(\mathcal{F},\mathcal{G}) = \sigma(Y_1, \ldots, Y_n)$. Now it remains to show that $\mathcal{F}$ is independent of $$ \sigma(\mathcal{G},\sigma(X)) = \sigma(Y_j,\mathbb{1}_{\{ U_j \leq Y_j \}}). $$ But I only know that $U_j$ is independent of $\mathcal{F}$. I don't know anything about $Y_j$ and $Y_i$ for $i \neq j$. How can I safe this?

$\endgroup$
7
$\begingroup$

If $V$ and $W$ are two independent vectors (say of dimension $m$ and $k$) and $f\colon\mathbb R^m\times \mathbb R^k\to\mathbb R $ is measurable, then $$\mathbb E\left[f\left(V,W\right)\mid \sigma\left(W\right) \right]=g\left(W\right) $$ where $g$ is defined by $$ g\left(w_1,\dots,w_k\right)=\mathbb E\left[f\left(V,w_1,\dots,w_k\right) \right].$$ Apply this to the following cases:

  1. $m=1$, $k=n$, $f\left(v,w_1,\dots,w_n \right) =\mathbf 1\left\{v\leqslant w_j \right\}$, $V=U_j$ and $W=\left(Y_1,\dots,Y_n\right)$.
  2. $m=k=1$, $f\left(v,w \right) =\mathbf 1\left\{v\leqslant w\right\}$, $V=U_j$ and $W= Y_j$.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.