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Suppose $X_n$ are independent Gaussian random variables with mean $\mu_n$ and variance $\sigma_n^2$.

Prove that $\{X_n\}$ is stochastically bounded iff $\mu_n$ and $\sigma_n^2$ are bounded sequences.

[If the sequences are bounded then the random variables are stochastically bounded. This is probably the easier side and I could solve it. May I have some hints for the other direction?]

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If $X$ is a standard normal random variable, then $$ \mathbb{P}(X>x)\sim \frac{1}{\sqrt{2\pi}x}e^{-x^2/2}$$ as $x\to\infty$.

Therefore if $X$ is Gaussian with mean $\mu$ and variance $\sigma^2$, then $$ \mathbb{P}(X>x)\sim \frac{\sigma}{\sqrt{2\pi}(x-\mu)}e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$ as $x\to\infty$. I think this should be enough to prove the direction you want.

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  • $\begingroup$ I just tried that but I am struggling with the case, where both $\mu_n$ and $\sigma_n^2$ are unbounded along a subsequence. Could you please elaborate? probably I am missing something. $\endgroup$ – the number 17 Feb 21 '16 at 7:06
  • $\begingroup$ I'd deal with the case when $\mu_n$ is unbounded (regardless of what $\sigma_n$ is doing) looking at something like $\mathbb{P}(|X-\mu_n|<\sigma_n)$. Otherwise $\{\mu_n\}$ is bounded, and the formula in my answer should handle the case when $\sigma_n$ is unbounded. $\endgroup$ – carmichael561 Feb 21 '16 at 7:47
  • $\begingroup$ To prove that $\mu_n$ is bounded it is enough to write that $P(X_n>\mu_n) =1/2$. To deduce further that $\sigma_n$ is bounded, it suffices to write that the probability $P(X_n > \mu_n +\sigma_n)$ is independent of $n$. $\endgroup$ – zhoraster Feb 21 '16 at 10:25
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Evidently $\Bbb P[X_n>x]=1-\Phi((x-\mu_n)/\sigma_n)$, and if $\{X_n\}$ is stochastically bounded then this quantity must tend to $0$ as $x\to+\infty$, uniformly in $n$. This is the same as $\lim_{x\to +\infty}\inf_n[(x-\mu_n)/\sigma_n]=+\infty$. In particular, given $N>0$, there exists $x_0>0$ such that if $x\ge x_0$ then $x\ge \mu_n+N\sigma_n$ for all $n$. This implies that $\{\mu_n\}$ is bounded above (because $\sigma_n\ge 0$); similar consideration of the left tails $\Bbb P[X_n\le x]$ shows that $\{\mu_n\}$ is bounded below, by $K$, say. Consequently, if $x_0$ corresponds to $N=1$ as above, then $x_0\ge\mu_n+\sigma_n\ge K+\sigma_n$, so $\{\sigma_n\}$ is bounded above (and trivially, below by $0$).

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