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A common proof for this proposition is the following: Fix $\epsilon>0$ and $M_o>0$. There exists an N such that $P(|X_n|\geq M_o) \leq \epsilon$ for all $n\geq N$. For the rest of the $X_n$ where $n<N$ (finitely many) we find $M_n$ such that $$P(|X_n|\geq M_n) \leq \epsilon$$ We then take the maximum of all the $M_n$ $n=0,1,2,..N-1$ and we get the constant for tightness or bounded in probability. My confusion is with proving the second statement of the proof. I have seen some cite Portmanteau lemma as justification and others mention points of continuity of the cdf. Thank you.

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    $\begingroup$ Hey! Welcome to MSE. You may want to rewrite your question so that what you are asking is clear and you can show what you have done and what specifically you are stuck with (you are talking about a proof but no proof shown). Also, avoid using links or images as some devices will not open them correctly. $\endgroup$
    – William M.
    Commented Feb 7, 2022 at 17:16
  • $\begingroup$ @WilliamM. I edited the question. Hopefully it is better now. $\endgroup$ Commented Feb 7, 2022 at 19:08
  • $\begingroup$ Any random variable is finite a.s. Then, the functions $\mathbf{1}_{|X_n| > k}$ ($n$ is fixed) converge to zero a.s. as $k \to \infty.$ Since these functions are bounded (and we are on a probability space), we see that for each $n$ there exists a constant $M_n$ such that $P(|X_n| > M_n) = \int \mathbf{1}_{|X_n| > M_n} dP \leq \varepsilon.$ Obviously, $\sup_{n \in \mathbf{N}} M_n$ can be infinity, yet the supremum of finetely many numbers is finite. $\endgroup$
    – William M.
    Commented Feb 7, 2022 at 19:24
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    $\begingroup$ @WilliamM. Thank you for responding. This may be my mistake but I meant literally the second sentence of the proof was not clear to me which ends with: for all $n\geq N$. Your response was still a nice perspective on the last part of the proof. $\endgroup$ Commented Feb 7, 2022 at 20:06
  • $\begingroup$ So $X_n \to X$ weakly. There exists $M > 0$ such that $P(|X| > M) \geq 1- \varepsilon.$ Since $\liminf P(|X_n| > M) \geq P(|X| > M) \geq 1 - \varepsilon,$ we get that $P(|X_n| \leq M) \leq 2\varepsilon$ for all $n$ large. $\endgroup$
    – William M.
    Commented Feb 7, 2022 at 20:11

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If I understand correctly, you are wondering why $X_{n}$ converging implies that for all $\epsilon > 0$ and $M_{0} > 0$, there exists $N$ such that $P(|X_{n}| \geq M_{0}) \leq \epsilon$ for all $n \geq N$.

Well, first of all, this is actually not true. Consider $X_{n} = 2$ and $M_{0} = 1$. Then no matter how large we select $N$, we will not get $P(|X_{n}| \geq M_{0}) < \epsilon$. What I think you meant is instead that, for all $\epsilon > 0$, there exists $M_{0} > 0$ and $N$ such that $$P(|X_{n}| \geq M_{0}) \leq \epsilon$$ for all $n \geq N$.

To find this $M_{0}$, we consider $X$, the variable that $X_{n}$ converges to in distribution. Then, since $X$ is necessarily tight, for all $\delta > 0$ there exists $M_{0} > 0$ with $$ P(|X| \geq M_{0}) \leq \delta. $$ Take $\delta = \epsilon / 2$ and fix this $M_{0}$. Since $X_{n} \to X$, we can find $N$ such that \begin{equation} | P(|X_{n}| \geq M_{0}) - P(|X| \geq M_{0}) | \leq \delta \end{equation} for all $n \geq N$ (this is apparently justified by the Portmanteau Lemma, although it seems like it would follow by definition). As such, by the triangle inequality, $$ P(|X_{n}| \geq M_{0}) \leq P(|X| \geq M_{0}) + | P(|X_{n}| \geq M_{0}) - P(|X| \geq M_{0}) | \leq 2 \delta = \epsilon $$ for all $n \geq N$.

To be clear, I've adapted this from the proof in van der Vaart's Asymptotic Statistics (chapter 2.1, theorem 2.4 - Prohonov's theorem).

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  • $\begingroup$ I apologize for the late reply. This is exactly what I was looking for. $\endgroup$ Commented Mar 19, 2022 at 3:45
  • $\begingroup$ @UniformIntegrability great! You accept the answer by clicking on the check mark under the upvote/downvote arrows. $\endgroup$ Commented Mar 20, 2022 at 18:22

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