# Conditioned expectation

I would like to understand the proposed solution of this example.

I can't understand the $E[N_k|N_{k-1}]$ expression in the second line. Shouldn't that be, with total probability:

$$E[N_k|N_{k-1}] = p\ (1 + N_{k-1}) + (1 - p)\ (1 + E[N_k])$$

Where is the p-multiplicant before $(N_{k - 1} + 1)$ and why can he (Roos, Introduction to the Probability Models, p. 113) alternate between random variable ($N_{k - 1}$) and expectation ($E[N_k]$) in the same expression? The derived expression is correct, by the way. I verified it with a simulation and a more straightforward algebraic approach.

• I think you are using the 10th edition by Sheldon Ross? The final recursive equation of $M_k$ is right, though the derivation back then was problematic at best. In the 11th edition it is corrected (p107 to p108). – Lee David Chung Lin Aug 8 '17 at 9:34
• Your total probability line is missing one term: $E[N_k|N_{k-1}] = p\ (1 + N_{k-1}) + (1 - p)\ ({\color{red}{N_{k-1}}}+1 + E[N_k])$, which actually simplifies to the line you see in Ross' text. – Lee David Chung Lin Aug 8 '17 at 9:43
• Sheldon Ross often sacrifices rigor, and unfortunately often clarity, for what he believes is intuitive. That RHS $N_{k-1}+1 +(1-p)E[N_k]$ can be understood this way: given $N_{k-1}$, there's gonna be $+1$ no matter what, and if the next trial is a success then that's the end of it, while if the next trial is a failure we start over and have an additional $E[N_k]$. – Lee David Chung Lin Aug 8 '17 at 9:52
• @LeeDavidChungLin Thank you. I do use 10th edition, I checked the solution in the 11th edition via Google Books. Also I do miss that term in my expression. How about mixing variates and expectations? Shouldn't there expectations only in our expression? – BoLe Aug 8 '17 at 9:58
• The conditional expectation $E[N_k|N_{k-1}]$ in the LHS is also a random variable which is $\sigma(N_{k-1})$ measurable, therefore it is not surprising that the expression in the RHS is also in terms of $N_{k-1}$ – BGM Aug 8 '17 at 14:46