acotamiento de una martingala secuencia?

Question

Deje $F_n(t)=\frac{1}{n} \sum_{k=1}^n 1_{X_k \leq t}$ ser el emprical función de distribución del yo.yo.d. variables aleatorias $X_k\sim U[0,1]$. Para definir $t\in [0,1)$ $$M_n(t)=\frac{F_n(t)-t}{1-t}.$$ I have shown that this is a martingale. If anybody is interested, I'm posting it. Now I want to show that if we put $M_n(t)=1$ for $t\geq 1$, then $M_n$ is not UI, but bounded in $L^1$. But then, I am not sure what $\lim_{t\rightarrow 1} M_n(t)$ es, me puedes ayudar con eso? Supongo, tengo que demostrar que no es IU, ¿verdad? Además, estoy muy pegado con el acotamiento, ninguna de las desigualdades he encontrado, llevan a ninguna parte. Alguna idea?

Muchas gracias!

Answer 1

Sobre el acotamiento en $\mathcal{L}_{1}$: en Primer lugar, tenga en cuenta que $F_{n}(t)\in[0,1]$ . Consideramos $$E\left|M_{n}(t)\right|=E\left(\frac{F_{n}(t)-t}{1-t}\right)^{+}+E\left(\frac{F_{n}(t)-t}{1-t}\right)^{-},$$ where $\a la izquierda(x\right)^{+}=\max\left\{ 0,x\right\}$ and $\left(x\right)^{-}=\max\left\{ 0,-x\right\}$. Of course, for $t\geq 1$ the boundedness is trivial since then $M_{n}(t)=1$, so we focus on $t\in[0,1)$. On this interval I have shown, that $M_{n}$ is a martingale, which gives us $E\left(\frac{F_{n}(t)-t}{1-t}\right)^{+}-E\left(\frac{F_{n}(t)-t}{1-t}\right)^{-}=E\left[M_{n}(t)\right]=E\left[M_{0}(t)\right]=0$. Thus we have $E\left(\frac{F_{n}(t)-t}{1-t}\right)^{+}=E\left(\frac{F_{n}(t)-t}{1-t}\right)^{-}$, so we can focus on $E\left(\frac{F_{n}(t)-t}{1-t}\right)^{+}$. We exclude the special case, that the fraction is zero, because then the boundedness is trivially true as well. We notice that for $\frac{F_{n}(t)-t}{1-t}$ to belong to the positive part, we need to have that either either both, nominator and denominator, are positive or both of them are negative. Since we are considering $t\in[0,1)$, we have that both have to be positive. Since $F_{n}(t)\in[0,1]$ we obtain $$E\left(\frac{F_{n}(t)-t}{1-t}\right)^{+}\leq E\left(\frac{1-t}{1-t}\right)=1.$$ Since $E\left(\frac{F_{n}(t)-t}{1-t}\right)^{+}=E\left(\frac{F_{n}(t)-t}{1-t}\right)^{-}$ then also the negative part is bounded by 1 and we conclude, that for all $t\geq 0$ it holds that $E\left|M_{n}(t)\right|\leq 2.$

Answer 2

Tenga en cuenta que $M_n(t)=1-\frac{1-F_n(t)}{1-t}$ y $1-F_n(t)\geqslant0$ $\mathbb E(F_n(t))=t$ por lo tanto $$\mathbb E(|M_n(t)|)\leqslant1+\frac{1-\mathbb E(F_n(t))}{1-t}=2.$$ Por otro lado, $\mathbb E(M_n(t))=0$ por cada $t\lt1$ $M_n(t)=1$ por cada $t\geqslant T_n$$T_n=\max\{X_k\mid 1\leqslant k\leqslant n\}$. Desde $T_n\lt1$ casi seguramente, $M_n(t)\to M^*_n=1$ cuando $t\to1$, $t\lt1$.

Si $(M_n(t))_{0\leqslant t\lt1}$ fue de interfaz de usuario, $\mathbb E(M^*_n)=1$ sería el límite de $\mathbb E(M_n(t))$ cuando $t\to1$, $t\lt1$, es decir, $0$. Desde $1\ne0$, $(M_n(t))_{0\leqslant t\lt1}$ no es de IU.