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I have a succession of random variables $X_n$ on $\Omega=[0,1]$ with $X_n=(1-\omega)^n$.

I have to prove the convergence almost sure and/or in law in these case:

  1. $\mathbb P=\delta_{0}$
  2. $\mathbb P=\frac12 \delta_{0}+\frac12 \delta_{1}$
  3. $\mathbb P=$ Lebesgue measure on $[0,1]$

I dont' know how do.

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1 Answer 1

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For 1), recall that $\mathbb P=\delta_0$ just means that $\mathbb P(\{0\})=1$. So $X_n(0) = (1-0)^n = 1$, and $\mathbb P(X_n=1)=1$. Hence trivially $X_n\stackrel{a.s.}{\longrightarrow}1$ and $X_n\stackrel{d}{\longrightarrow}1$

For 2), recall that $\mathbb P=\frac12\delta_0 + \frac12\delta_1$ means $\mathbb P(\{0\})=\mathbb P(\{1\})=\frac12$. So $X_n(0) = 1$ and $X_n(1)=0$, and $\mathbb P(X_n=0)=\mathbb P(X_n=1)=\frac12$. Since the $X_n$ all have the same distribution, $X_n\stackrel{a.s.}{\longrightarrow}X_1$ and $X_n\stackrel{d}{\longrightarrow}X_1$.

For 3), recall that Lebesgue measure on $[0,1]$ is just a uniform distribution over $[0,1]$, that is, $\mathbb P([a,b])=b-a$ for $0\leqslant a\leqslant b\leqslant 1$. Observe that $$(1-\omega)^n\stackrel{n\to\infty}{\longrightarrow}\begin{cases} 0,& \omega\ne 0\\1,&\omega=0.\end{cases}$$

Since $\mathbb P(\{0\})=0$, it follows that $$\mathbb P\left(\lim_{n\to\infty} X_n=0 \right)=1, $$ i.e. $X_n\stackrel{a.s.}{\longrightarrow}0$ (and therefore $X_n\stackrel{d}{\longrightarrow}0$).

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