Let $Y(0)$ a random variable uniformly distribuited on $[0,1]$ and define $X(n)=2^{n}(1-Y(n))$ where $Y(n)$ is uniformly distribuited on $[Y_{n-1},1]$. Show that $X(n)$ is a martingale.
I am not sure that $X(n)$ is a Markov process, so I tried to compute the first conditional expectation so
\begin{equation} \begin{array} aE(X(1)\,|\,\mathcal{F}_0)&=2E((1-Y(1))\,|\,Y(0))\\ &=2E(u\,|\,Y(0)) \end{array} \end{equation} where $u$ is uniformly distribuited in $[0,Y_0]$. Then \begin{equation} \begin{array} 2E(u\,|\,Y(0)=y)&=\int_0^y \frac{u}{y}du\\ &=\frac{y}{2} \end{array} \end{equation}
my question are: $E(u\,|\,Y(0))=E(u\,|\,Y(0)=y)1_{\{Y(0)=y\}}$? and is $X$ a Markov process?
Notice how $Y_{n+1}|X_0,X_1,\ldots ,X_n \sim \mathcal{U}\Big[1-2^{-n}X_n,1\Big]$ and so $$\begin{eqnarray*}\mathbb{E}\left(X_{n+1}|X_0,X_1,\dots,X_n\right) &=& \mathbb{E}\left(2^{n+1}(1-Y_{n+1})|X_0,X_1,\dots,X_n\right) \\ &=& 2^{n+1}\left(1-\mathbb{E}\left(Y_{n+1}|X_0,X_1,\dots,X_n\right)\right) \\ &=& 2^{n+1}\Big(1-\frac{1-2^{-n}X_n+1}{2}\Big) \\ &=&X_n\end{eqnarray*} $$ Hence $\{X_i\}_{i\geq 0}$ is a martingale. No integration necessary.