# Bringing limit inside a right-continuous function.

1. Oct 17, 2012

### operationsres

1. The problem statement, all variables and given/known data

Suppose that we have $N : \mathbb{R}\cup\{-\infty,\infty\} \to [0,1]$ which is the standard normal cumulative distribution function. It is right-continuous.

What I want to evaluate is $\lim_{b\to 0^+}N(\frac{a}{b})$, where $a \in \mathbb{R}^+$, and alternatively where $a \in \mathbb{R}^-$

2. The attempt at a solution
I opened a thread yesterday on the same topic but the consequences of the fact that $N(.)$ is right-continuous wasn't answered/addressed, which is why I decided to re-open and start fresh so that we can focus on this one aspect.

I already know that $N(-\infty)$ and $N(\infty)$ are well defined to equal 0 and 1 respectively, so that's not what I'm asking :).

Please focus on whether I can push the limits inside of N(.) under both a > 0 and a < 0 under the condition that N(.) is right-continuous.

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Refresher: right continuous at $c$ means that $\lim_{x \to c^+}f(x) = f(c)$.

Last edited: Oct 17, 2012
2. Oct 17, 2012

### Ray Vickson

The answer is YES for ANY legitimate cdf, not just for the normal cdf N(.). I have already stated this about 3 times, but for some reason you seem not to believe the answer. All you need to is fall back on _defintions_ involving limits of +∞ or -∞ and use standard properties of a cdf F(x).

Things would be a bit different if you were talking about finite limits; then you really would need to distinguish between limits from the left or from the right, at least for a cdf having jump discontinuities (but not for continuous ones like N(.)).

RGV

3. Oct 17, 2012

### operationsres

Consider $\lim_{b\to 0^-} N(\frac{a}{b})$ with a < 0. we have that $\frac{a}{b}\to +\infty$ from the left, and you say in this case it is perfectly fine to push the limit inside N(.) even though it is right continuous. Okay, I get this.

But I would like an explanation of why we (i) can't automatically put the limit inside N(.) if we're doing a finite limit from the left , (ii) we are allowed to put the limit inside N(.) if we're doing an infinite limit from the left.

Also N(.) is right-continuous according to wikipedia (which is distinct from "continuous" like you say?).

Also my friend is a maths post-doc and he said that I can only push the limit inside when it approaches from the left if the function is left-continuous or continuous, not right-continuous (as is the case with N(.)), which adds to my confusion. I guess he's wrong.

Last edited: Oct 17, 2012
4. Oct 17, 2012

### Ray Vickson

For any F(x) obtained by integrating a density, F is both right and left continuous---just plain continuous. It is different if you have a mixed distribution (partly discrete and partly continuous) or discrete. In those cases there will be points at which F is continuous from the right but not from the left (at least if you use the more-or-less standard convention that F(x) = Pr{X x}. On the other hand, the complementary cumulative G(x) = 1-F(x) = Pr{X > x} would be continuous from the left but not from the right in those cases.

RGV

5. Oct 17, 2012

### operationsres

So when Wiki says "Every cumulative distribution function F is (not necessarily strictly) monotone non-decreasing (see monotone increasing) and right-continuous.", this doesn't exclude that it could also be left continuous AS WELL AS right continuous, making it plain old continuous?

Thanks for your help!!! I just thought that $\text{Right Continuous} \Rightarrow \neg \text{Left Continuous}$ which is where I was getting confused.

6. Oct 17, 2012

### Ray Vickson

Nope. It is like saying "all men are human". That does not imply that all humans are men.

RGV