User:Albinomonkey/Sandbox/Q2b

I got this:

$$\mathbb{E}\{X|Z\} = \sum_{j=0}^{\infty}\mathbb{E}\{X|B_j\}\mathbb{I}_{B_j}(\omega)$$

...which is

$$\mathbb{E}\{X|Z\} = \sum_{j=0}^{\infty}\left(j+1-\frac{1}{e-1}\right)\mathbb{I}_{B_j}(\omega)$$ $$ = \sum_{j=0}^{\infty}j\mathbb{I}_{B_j}(\omega) + \left(1-\frac{1}{e-1}\right)\sum_{j=0}^{\infty}\mathbb{I}_{B_j}(\omega)$$

The first term in this is just the Z, ie if X is in $$B_j = [j,j+1)\,$$, you get j. The second sum will give you 1 since the $$B_j$$'s are disjoint and their union is the whole of $$\mathbb{R}^+$$.

So you get $$Z + \left(1-\frac{1}{e-1}\right)$$, and you can put the second bit in a calculator...