conditional expectation of disjoint events

Question

Let $B_{1},B_{2},\dots,B_{n} \in \mathcal{F}$ partition of $\Omega$

and $\mathcal{P}=\sigma(B_{1},B_{2},\dots,B_{n})$

If $P(B_j)>0 , j = 1,\dots,n$.

Then show that :

$$\mathbb{E}[X|\mathcal{P}] = \sum_{j=1}^{n}\mathbb{E}[X| B_{j}]\mathbf{1}_{B_{j}}$$ where :

$$\mathbb{E}[X|{B_i}] = \frac{\mathbb{E}[ X \mathbf{1}_{B_{j}}]}{P(B_{j})}$$

Hint : (all the non empty events of $\mathcal{P}$ have the form of $B_{j1} \bigsqcup B_{j2} \bigsqcup,\dots \bigsqcup B_{jk}$ )

With $\bigsqcup$ we denote the union of disjoint events.

My effort is :

\begin{align*} \mathbb{E}\left[X | B_{1} \bigsqcup B_{2} \bigsqcup,\dots \bigsqcup B_{j}\right] &= \frac{1}{P(B_{1} \bigsqcup B_{2} \bigsqcup \dots \bigsqcup B_{j})} \mathbb{E} \left[X \cdot \mathbf{1}_{B_{1} \bigsqcup B_{2} \bigsqcup,\dots \bigsqcup B_{j}} \right] \\ &=\frac{1}{ P\left( \bigsqcup_{j=1}^{n} B_{j} \right) }\mathbb{E}\left[ X \left[ \mathbf{1}_{B_{1}} + \mathbf{1}_{B_{2}} + \dots + \mathbf{1}_{B_{j}} \right] \right] \\ &=\frac{1}{\sum_{j=1}^{n} P(B_{j})} \mathbb{E}(X\cdot \mathbf{1}_{B_{1}}) + \mathbb{E}(X \cdot \mathbf{1}_{B_{2}}) + \dots +\mathbb{E}(X \cdot \mathbf{1}_{B_{j}}) \\ &=\sum_{j=1}^{n}\mathbb{E}[X| B_{j}]\mathbf{1}_{B_{j}} \end{align*}

But why it says $B_{j1},B_{j2},\dots,B_{jk}$

Proposal :

\begin{align*} \mathbb{E}\left[X | B_{j1} \bigsqcup B_{j2} \bigsqcup,\dots \bigsqcup B_{jk}\right] &= \frac{1}{P(B_{j1} \bigsqcup B_{j2} \bigsqcup \dots \bigsqcup B_{jk})} \mathbb{E} \left[X \cdot \mathbf{1}_{B_{j1} \bigsqcup B_{j2} \bigsqcup,\dots \bigsqcup B_{jk}} \right] & \text{by definition}\\ &=\frac{1}{ P\left( \bigsqcup_{j=1}^{n} B_{j} \right) }\mathbb{E}\left[ X \left[ \mathbf{1}_{B_{j1}} + \mathbf{1}_{B_{j2}} + \dots + \mathbf{1}_{B_{jk}} \right] \right]& \text{by mutual exclusivity} \\ &=\frac{1}{\sum_{j=1}^{n} P(B_{j})} \mathbb{E}(X\cdot \mathbf{1}_{B_{j1}}) + \mathbb{E}(X \cdot \mathbf{1}_{B_{j2}}) + \dots +\mathbb{E}(X \cdot \mathbf{1}_{B_{jk}})&\text{by linearity of expectation} \\ &=\sum_{j=1}^{n}\mathbb{E}[X| B_{j}]\mathbf{1}_{B_{j}} \end{align*}

Answer 1

Hint

What has to be proved it's that for all $F\in \mathcal P$ (i.e. for $F$ of the form $B_{i_1}\sqcup\cdots \sqcup B_{i_k}$), $$\mathbb E[X\boldsymbol 1_F]=\mathbb E\left[\sum_{j=1}^n \mathbb E[X|B_j]\boldsymbol 1_{B_j}\boldsymbol 1_F\right].$$

Try to adapt what you did in your original post to $\mathbb E[X\boldsymbol 1_F]$ and conclude.

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Edit

We can suppose WLOG that $F=B_1\sqcup\cdots B_k$ (where $k\leq n$).

Then

\begin{align*} \mathbb E[X\boldsymbol 1_F]&=\sum_{i=1}^k\mathbb E[X_i\boldsymbol 1_{B_i}]\\ &= \sum_{i=1}^k\mathbb P(B_i)\mathbb E[X_i\mid B_i]\\ &=\mathbb E\left[\sum_{i=1}^k\mathbb E[X\mid B_i]\boldsymbol 1_{B_i}\right]\\ &=\mathbb E\left[\sum_{i=1}^n\mathbb E[X\mid B_i]\boldsymbol 1_{B_i}\boldsymbol 1_F\right], \end{align*} as wished.