Expected value $\mathbb{E}(Y_n)$ in conditional case Let $\lbrace Y_n \rbrace_{n \in \mathbb{N_0}}$ be a family of random variables and $\mathbb{P}(Y_0=0)=1$. 
$Y_n$ is conditional on $\lbrace Y_1=y_1,...,Y_{n-1}=y_{n-1} \rbrace$ unifmorly distributed on the interval $(y_{n-1}, y_{n-1}+1$) for all $n \in \mathbb{N}$.
What is the expected value $\mathbb{E}(Y_n)$?
My idea was:
(If continuous): If $f_X(x)$ and $f_Y(y)$ are the probability density functions of $X$ and $Y$ and $f_{X|Y=y}(x)$ is the PDF, then the PDF $f_X(x)=\int_{-\infty}^{\infty}f(x,y)dy$ is given with 
$f(x,y)=f_{X|Y=y}(x)f_Y(y)$
So $\mathbb{E}[X]=\int \mathbb{E}[X|Y=y]f_Y(y)dy$
How to resume from here?

In general if $X$ has uniform distribution on interval $\left[a,b\right]$
then $\mathbb{E}X=\frac{1}{2}\left(a+b\right)$.
This tells us immediately that: $$\mathbb{E}\left[Y_{n}\mid Y_{n-1}\right]=Y_{n+1}+\frac{1}{2}$$
and consequently: $$\mathbb{E}Y_{n}=\mathbb{E}\left(\mathbb{E}\left[Y_{n}\mid Y_{n-1}\right]\right)=\mathbb{E}\left(Y_{n-1}+\frac{1}{2}\right)=\frac{1}{2}+\mathbb{E}Y_{n-1}$$
Knowing that $\mathbb{E}Y_{0}=0$ it is now not difficult anymore
to find with e.g. induction that: $$\mathbb{E}Y_{n}=\frac{1}{2}n$$
