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2 votes
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show that if $(\sqrt{n}(Y_n - \theta) \overset{d}{\to} N(0, 1))$ then $(Y_n \overset{P}{\to} \theta)$

According to Slutsky's theorem, if $Z_n \xrightarrow{d} Z$ and $a_n \xrightarrow{p} a$, then $a_n Z_n \xrightarrow{d} a Z$. Here, $Z_n=\sqrt{n}\left(Y_n-\theta\right)$ and $Z \sim N(0,1)$, and $\frac{...
bruno's user avatar
  • 399
1 vote
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probability convergence as compactness for a.s. convergence

First, the mentioned result actually implies that there is no "topology of almost sure convergence." Indeed, let $(X_n)$ be a sequence of random variables that converges in probability to a ...
Michael Greinecker's user avatar
1 vote

MIT Statistic For Applications course Question 1

It's Markov's inequality. thanks Matthew Towers.
tom tom's user avatar
  • 29
1 vote
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Convergence of positive random vector

Let $(\Omega,\mathcal{A},\mathbb{P})$ be our probability space. For $1\leqslant p<\infty$, let $L_p$ denote the space of (equivalence classes of) random variables $X:\Omega\to \mathbb{R}$ (that is, ...
user469053's user avatar
  • 2,522

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