Let $(X_n)_{n\geq 1}$ be a sequence of random variables such that $X_n\sim N(\mu_n,\sigma_n)$ for all $n\geq 1$. Then i'm trying to deduce that if $(X_n)_{n\geq 1}$ is tight in the sense that $$ \forall \varepsilon>0\,\exists r>0: \sup_{n\geq 1}P(|X_n|>r)<\varepsilon, $$ then $\sup_{n\geq 1} (|\mu_n|+\sigma_n^2)<\infty$. My approach so far has been to assume that $\sup_{n\geq 1} (|\mu_n|+\sigma_n^2)=\infty$ and then try to show that $(X_n)$ cannot be tight. Under this assumption me must have that either $\sup_n |\mu_n|=\infty$ or $\sup_n \sigma_n^2=\infty$. In either case I'm not sure how to show that for $r>0$ the following $$ P(|X_n|>r)=\int_{(-\infty,-r]\cup [r,\infty)} \frac{1}{\sqrt{2\pi \sigma_n^2}}e^{-(x-\mu_n)^2/\sigma_n^2} dx $$ is not arbitrarily small. Is this the right approach?

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    $\begingroup$ Your logic is wrong. You are going to prove that if $\sup_n (|\mu_n| + \sigma_n^2) = + \infty$ then $(X_n)$ is not tight, which is true, but not what you want. $\endgroup$ – D. Thomine Mar 5 '12 at 10:01
  • $\begingroup$ Thanks, I made a mistake writing it up, and i've edited it. I want to show that $(X_n)$ being tight implies $\sup_n (|\mu_n|+\sigma_n^2)<\infty$. $\endgroup$ – Stefan Hansen Mar 5 '12 at 10:07

Event if it is not necessary, I would rather distinguish two cases: the case $\sup_n |\mu_n| = + \infty$, and the case $\sup_n \sigma_n = + \infty$

1) If $\sup_n |\mu_n| = + \infty$: note that $\mu_n$ is the median of $X_n$. Hence, $\mathbb{P} (|X_n| > |\mu_n|) \geq 1/2$. For all $M> 0$, there exists a $n$ such that $|\mu_n| \geq M$, and for this value of $n$ we get $\mathbb{P} (|X_n| > M) \geq 1/2$. hence, for all $M \geq 0$, we have:

$$\sup_n \ \mathbb{P} (|X_n| > M) \geq 1/2,$$

which contradicts the tightness. Actually, one can also prove that, in this case, $\sup_n \ \mathbb{P} (|X_n| > M) = 1$.

2) If $\sup_n \sigma_n = + \infty$: note that the density of the law of a normally distributed random variable of mean deviation $\sigma$ is bounded above by $(\sigma \sqrt{2 \pi})^{-1}$. Hence, for all $m > 0$ and all $n$,

$$\mathbb{P} (|X_n| \leq M) \leq \int_{-M}^M \frac{1}{\sigma_n \sqrt{2 \pi}} \ dx = \frac{M}{\sigma_n} \sqrt{\frac{2}{\pi}},$$

so that:

$$\sup_n \ \mathbb{P} (|X_n| > M) \geq \sup_n \left( 1- \frac{M}{\sigma_n} \sqrt{\frac{2}{\pi}} \right) = 1.$$

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    $\begingroup$ Excellent answer. Thanks alot! $\endgroup$ – Stefan Hansen Mar 5 '12 at 14:29
  • $\begingroup$ @ Didier Pau: Thank you for the remark, I edited the post. I had indeed forgotten one half of the gaussian. $\endgroup$ – D. Thomine Mar 5 '12 at 14:48

Let $M:=\sup_{n\geq 1}|\mu_n|+\sigma_n^2$. Let $n\geq 1$ and $r>0$. Then \begin{align} P(|X_n|>r)&=\frac 1{\sqrt{2\pi \sigma_n^2}}\int_r^{+\infty}\exp\left(-\frac{(x-\mu_n)^2}{2\sigma_n^2}\right)dx\\ &=\frac 1{\sqrt{2\pi \sigma_n^2}} \sigma_n\int_{\frac{r-\mu_n}{\sigma_n}}^{+\infty}\exp\left(-\frac{t^2}2\right)dt\mbox{ substitution }t=\frac{x-\mu_n}{\sigma_n}\\ &=\frac 1{\sqrt{2\pi}}\int_{\frac{r-\mu_n}{\sigma_n}}^{+\infty}\exp\left(-\frac{t^2}2\right)dt \end{align} and $\frac{r-\mu_n}{\sigma_n}\geq \frac{r-M}{\sigma_n}\geq\frac{r-M}{\sqrt M}$ so for all $n\geq 1$ and for all $r>0$ $$P(|X_n|>r)\leq \frac 1{\sqrt{2\pi}}\int_{\frac{r-M}{\sqrt M}}^{+\infty}\exp\left(-\frac{t^2}2\right)dt,$$ which gives tightness since the last integral doesn't depend on $n$ and converges to $0$ as $r\to\infty$.

  • $\begingroup$ Thanks for your reply, but unfortunately I wrote by mistake that I was trying to prove $\sup_n |\mu_n|+\sigma_n^2<\infty$ implies $(X_n)$ being tight in the original post. I've now edited the original post, and it's actually the other way around i'm trying to show. Sorry for the inconvenience. $\endgroup$ – Stefan Hansen Mar 5 '12 at 10:18
  • $\begingroup$ Yes I realize now you edited your post when I was typing. I will think on your problem now. $\endgroup$ – Davide Giraudo Mar 5 '12 at 10:19

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