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3 votes
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Finding $\lim_{\,n\to\infty} \, \sum_{k=n}^{3n} \binom{k-1}{n-1}\left(\frac{1}{3}\right)^n\left(\frac{2}{3}\right)^{k-n}$

Note that if $X \sim \mathrm{NB}(1/3, n)$, where $X$ counts the number of trials required to see $n$ successes (each with probability $p = 1/3$), then each term in your sum is the pmf of $X$. In ...
dmk's user avatar
  • 2,386
3 votes
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An example application where convergence in probability is useful or valuable

Convergence in probability can be highly valuable precisely because it is weaker than almost sure convergence. In many practical senarios, we can only establish convergence in probability, whereas ...
Hirofumi Shiba's user avatar
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
  • 425
1 vote
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Uniform Tightness of a Sequence of Probability Measures in $\mathbb R^{\mathbb N}$

For reference, another proof of the fact that OP is referring to can be found in Bogachev Vol. II, Example 8.2.16, which however does not answer OP's question. OP is asking to show that in this case, ...
Snoop's user avatar
  • 15.8k
1 vote
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If a random variable converges to zero in probability what can we say about its almost sure boundedness?

Neither $1.)$ implies $3.)$ nor the other way round. For example $X_n(x)=1/\sqrt{nx}$ on $[0,1]$ with the Lebesgue measure. These random variables are all unbounded. However, $X_n\rightarrow 0$ in ...
Severin Schraven's user avatar
1 vote
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Probability of the limit equals the limit of the probability?

Yes. For any sequence of events $\{A_n\}_{n=1}^{\infty}$ we have $$ \cap_{m=n}^{\infty} A_m \subseteq A_n \subseteq \cup_{m=n}^{\infty}A_m \quad \forall n $$ Thus $$P[\cap_{m=n}^{\infty}A_m]\leq P[A_n]...
Michael's user avatar
  • 24.9k

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