From Stochastic Processes by Ross
On page 75 (in the proof of lemma 2.3.4 on the bottom of page 74), the author writes at the bottom of the page:
$P(Y_1 + \dots + Y_k < \tau_k, k = 1, \dots, n | \tau_n = u, Y_1 + \dots Y_n = y) = E[1 - \frac{Y_1 + \dots + Y_{n-1}}{\tau_n} | \tau_n = u, Y_1 + \dots + Y_n = y]$
I can't figure out why this is the case, anyone have any ideas?
It seems the author is proving this by induction, but I can't figure out how these two things are equal. He shows that:
$P(Y_1 + \dots Y_k < \tau_k, k = 1, \dots, n |Y_1 + \dots + Y_{n-1}, \tau_n u, Y_1 + \dots + Y_n) = 1-\frac{Y_1 + \dots + Y_{n-1}}{\tau_n}$
but I can't figure out how to prove that
$E[P(Y_1 + \dots Y_k < \tau_k, k = 1, \dots, n |Y_1 + \dots + Y_{n-1}, \tau_n = u, Y_1 + \dots + Y_n) \space | \space \tau_n = u, Y_1 + \dots + Y_n = y]$ $= P(Y_1 + \dots + Y_k < \tau_k, k = 1, \dots, n | \tau_n = u, Y_1 + \dots Y_n = y)$
The lemma restated here is:
Let $\tau_1, \dots, \tau_n$ denote the ordered values from a set of $n$ independent uniform $(0,1)$ random variables and let $Y_i, \dots, Y_n$ be iid nonnegative random variables that are also independent of $\{\tau_1, \dots, \tau_n\}.$ Then $P(Y_1 + \dots Y_k < \tau_k, k = 1, \dots, n | Y_1 + \dots + Y_n) = 1-y/t$ for $0 < y < t$ and $0$ otherwise.
Figured it out.
If $g(B) = P(A|B, C=c)$ then $E(g(B)|C=c)= P(A|C=c)$.