## Bringing limit inside a right-continuous function.

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.

-------------------------

Refresher: right continuous at $c$ means that $\lim_{x \to c^+}f(x) = f(c)$.
 PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study

Recognitions:
Homework Help
 Quote by 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. ------------------------- Refresher: right continuous at $c$ means that $\lim_{x \to c^+}f(x) = f(c)$.
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

 Quote by 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

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.

Recognitions:
Homework Help

## Bringing limit inside a right-continuous function.

 Quote by operationsres In case 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, which adds to my confusion.
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

 Quote by Ray Vickson For any F(x) obtained by integrating a density, F is both right and left continuous---just plain continuous.
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.

Recognitions:
Homework Help
 Quote by 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.
Nope. It is like saying "all men are human". That does not imply that all humans are men.

RGV