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Let $X_1, X_2 . . . $ be iid $N(0, 1)$ RVs. Consider the sequence of RVs {$\overline X_n$}, where $\overline X_n=n^{-1}\sum_{i=1}^nX_i$. Let $F_n$ be the CDF of $\overline X_n$, $n = 1,2,...$ Find $\lim_{x\to \infty}F_n(x)$. Is this limit a CDF?

My attempt: I know that the sum of $n$ standard normal random variables will have a normal distribution with mean $0$ and variance $n$. How does the $n^{-1}$ factor into it? Once I find that, I am guessing it will still be normal and then I could take the limit of an appropriate normal CDF. I'm inclined to say that this will also be a CDF, but I need to find out what the first CDF is in order to test right continuity and the other axioms.

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Recall that $$ \text{Var}[X]=\mathbb{E}[X^2]-(\mathbb{E}[X])^2. $$ So, for any random variable $X$ and any $\alpha\in\mathbb{R}$, $$ \begin{align*} \text{Var}[\alpha X]&=\mathbb{E}[\alpha^2X^2]-(\mathbb{E}[\alpha X])^2\\ &=\alpha^2\mathbb{E}[X^2]-\alpha^2(\mathbb{E}[X])^2\\ &=\alpha^2\text{Var}[X]. \end{align*} $$ So, the net effect of multiplying a random variable by $\frac{1}{n}$ is that the variance is multiplied by $\frac{1}{n^2}$.

Further, you can show (by transforming the density function, for instance) that if $X$ is $N(0,1)$, then $\alpha X$ is also normal, with mean $0$ and variance $\alpha^2$.

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    So then the variance of $\overline X_n$ would be $1/n$? – 2017-01-20
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    Yep. This is why you would normally see the term $\frac{1}{\sqrt{n}}$ instead of $\frac{1}{n}$ in the central limit theorem -- that scaling factor would make the variance $1$ instead. – 2017-01-20
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    I see. So the distribution function would remain normal, but how do I obtain the limit? In my class we said that the normal distribution has a PDF, not a DF. We didn't get into the "error function" or anything like that. – 2017-01-20
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    I assumed that by DF you meant Density Function, but that's apparently not the case. Are you talking about the Cumulative Distribution Function? – 2017-01-20
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    Yes I am. Sorry for the confusion. I will fix that in the question now. – 2017-01-20
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    Well, that being the case: think about $\overline{X}_n$ when $n$ is really large. It is normal, but it has an extremely small variance. What does this tell you about $P(\overline{X}_n0$? What about when $x<0$? Further, note that $P(\overline{X}_n\leq 0)=\frac{1}{2}$ for all $n$. – 2017-01-20
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    The denominator would go to $0$ as $n$ approaches $\infty$. Then the first probability would be $1$, and the second would be $0$. Why is the last one $1/2$? – 2017-01-20
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    All of the random variables in your sequence are symmetric and have mean $0$. So, whether they are positive or negative is a coin flip. (And I wouldn't say the denominator tends to 0 -- it isn't clear what denominator you're talking about. But the variance tends to $0$, and so you could easily show (say, by Chebyshev's inequality) that you are correct about the probabilities in the limit. – 2017-01-20
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    What does the $1/2$ have to do with the CDF in question? $\lim_{x\to \infty}F_n(x)$ would be $1$ for $x>0$ and $0$ for $x<0$? – 2017-01-20
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    You aren't supposed to take a limit in $x$. That variable is fixed. You are taking a limit in $n$. – 2017-01-20
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    Then how do I use Chebyshev's? I have: $P(|\overline X_n-0|>\epsilon)\le var(\overline X_n)/\epsilon^2$ where the $0$ in the absolute value is the mean. – 2017-01-20
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    You know the variance, and it tends to zero. – 2017-01-20
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    Ah, so then that probability is always $0$, then $P(\overline X_n$0$ when $x<0$ but for $x>0$ I would need to do $1-0$? – 2017-01-20
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    Yep. And in the one case not covered -- the case $x=0$ -- you can compute the limit directly as $\frac{1}{2}$. – 2017-01-20
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    And then finally that satisfies all the properties of a CDF, so this limit is indeed a CDF. Thank you so much! – 2017-01-20
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    Actually, no. Remember the right continuity condition for CDFs; this function is not right-continuous at $x=0$. – 2017-01-20
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    My book says that it should be a CDF though – 2017-01-20
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    Not sure what to tell you on that one, but it isn't a CDF. – 2017-01-20
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    I thought about this some more and I was wondering: could it be that the $0$ case should share the limit of 1? – 2017-01-21
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    Nope. It is the pointwise limit of a sequence that is constant $\frac{1}{2}$. – 2017-01-21