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For a sequence $X_1, X_2, \dots $, Let $F_n(x)$ denote the cdf of $X_n$.

Suppose our sequence is $X_n \sim N(0,n) $ then for all $x$ the point-wise limit of $F_n(x)$ is $\frac{1}{2}$.

How would one prove this?

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  • $\begingroup$ You should ask this on 'Mathematics' $\endgroup$ – Hemant Rupani Apr 18 '15 at 18:25
  • $\begingroup$ @Hermant: While this question would be fine on math.SE, it is perfectly within the scope of Cross Validated as well. $\endgroup$ – cardinal Apr 19 '15 at 1:14
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More generally, let $X$ be any random variable with distribution $F$ having unit variance. Letting $(\mu_n)$ and $(\sigma_n)$ be any sequences of numbers, set

$$X_n = \sigma_n X + \mu_n.$$

Define $F_n$ to be the distribution of $X_n$. Suppose that as $n\to\infty$, $$\sigma_n\to\infty$$ and $$\mu_n / \sigma_n\to z.$$ If $F$ is continuous in a neighborhood of $-z$, then the pointwise limit of $F_n(x)$ must be $F(-z)$.

Intuitively, this is because

  1. The distributions of the $X_n$ are becoming more and more spread out but

  2. Relative to the spreads $\sigma_n$, any fixed number $x$ is becoming ever closer to $-z$.

Thus the pointwise limit ought to be $F(-z)$.

It remains to make this intuition rigorous. From the basic definitions and a little bit of algebra, observe that

$$F_n(x) = \Pr(X_n \le x) = \Pr(\sigma_n X + \mu_n \le x) = \Pr\left(X \le \frac{x-\mu_n}{\sigma_n}\right) = F\left(\frac{x-\mu_n}{\sigma_n}\right).$$

The assumptions about the limiting values of $\mu_n/\sigma_n$ and $\sigma_n$ were made specifically to imply $\lim_{n\to \infty} (x-\mu_n)/\sigma_n = -z$. Applying the assumed continuity of $F$ in a neighborhood of $-z$ finishes the demonstration. (The details, which are straightforward, are left to the reader because this is a self-study question that asks for guidance and intuition rather than a complete answer.)

Application of this result to the standard Normal distribution $F$, $(\mu_n) = (0)$, and $\sigma_n=(n)$ answers the question as stated, because $F(-0) = 1/2$.

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  • $\begingroup$ Presumably $F$ should also have zero mean? $\endgroup$ – Dougal Apr 18 '15 at 23:48
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    $\begingroup$ @Dougal Not at all. The mean is irrelevant. Because $X$ has a variance, its mean is defined and finite, but that's all we can say about the mean in general. Your intuition can help here: when $\sigma_n$ is very large, $F_n(x)$ depends only on a value of $F$ near $\mu_n/\sigma_n \approx z$, whereas the mean also depends on the chances that $X$ is extremely large or extremely small. $\endgroup$ – whuber Apr 18 '15 at 23:49
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    $\begingroup$ Hmm, okay, I see that now – I was just thrown by $\mathbb{E} X_n \ne \mu_n$. $\endgroup$ – Dougal Apr 18 '15 at 23:51

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