Can you bring a limit inside of the standard normal distribution function?

[operationsres]

Homework Statement

Let N(x) denote the CDF of the standard normal density. So, it's the integral of the standard normal density from -∞ to x.

Is it true that $\lim_{b\to 0} N(\frac{a}{b}) = N(\frac{a}{\lim_{b\to 0}b}) = N(+\infty) = 1$?

2. The attempt at a solution

This is more of a knowledge question... Can't really "attempt" it.

 

[STEMucator]

The limit operator can only be used in such a manner if the continuity of N can be assured and then the continuity of b is also assured.

For a simple example, take f(x) = sin(ln(x))

Then : $\lim_{x→1} sin(ln(x)) = sin(\lim_{x→1} ln(x)) = sin( ln(\lim_{x→1} x) ) = 0$

Since sin(x) is continuous we can be assured that we can bring the limit inside. Furthermore, since ln(x) is continuous, we can bring the limit operator inside of that as well. Then since we reach our independent variable, we can compute the limit afterwards.

 

[operationsres]

Well N(.) is right-continuous. Can I still do it then?

Also I don't understand "the continuity of b". We have that $b \in \mathbb{R}^+$ and that is all the information that I have. It is otherwise unconstrained.

 

[STEMucator]

Well N(.) is right-continuous. Can I still do it then?

Also I don't understand "the continuity of b". We have that $b \in \mathbb{R}^+$ and that is all the information that I have. It is otherwise unconstrained.

I was merely speaking in general that the limit operator can be moved around according to the continuity of your function.

So if N is continuous, you can assuredly bring your limit operator inside.

 

[operationsres]

So this also holds true for right-continuous functions?

Thanks for your help.

 

[operationsres]

I'm guessing that the answer is "yes" because the no jump occurs if the limit approaches from the right, and that's exactly what I'm trying to do (take the limit to 0+, which is from the right).

 

[STEMucator]

Yes you have the right idea.

 

[operationsres]

You're a fantastic homo sapien!

 

[operationsres]

Wait, are you sure that this is correct? This website:

http://www.math.oregonstate.edu/hom...estStudyGuides/SandS/lHopital/limit_laws.html

States that "If f is continuous at b, and $\lim_{x\to a} g(x) = b$,then:
$\lim_{x\to a}f(g(x))=f(b) = f(\lim_{x\to a}g(x))$.
"

Then going back to what I want to do : $\lim_{b\to 0} N(\frac{a}{b})$, the condition that I have to satisfy with this is that N(.) is continuous at +∞. Can a function be continuous at +infinity? Does that even make sense?

 

[operationsres]

I've concluded that you can't bring it into N(.) because N(infinity) isn't defined as the domain is (-inf,+inf). As such N(.) isn't continuous at infinity, and you can't bring the limit inside of N.

Plz tell me if I'm corre3ct.

 

[STEMucator]

Wait, are you sure that this is correct? This website:

http://www.math.oregonstate.edu/hom...estStudyGuides/SandS/lHopital/limit_laws.html

States that "If f is continuous at b, and $\lim_{x\to a} g(x) = b$,then:
$\lim_{x\to a}f(g(x))=f(b) = f(\lim_{x\to a}g(x))$.
"

Then going back to what I want to do : $\lim_{b\to 0} N(\frac{a}{b})$, the condition that I have to satisfy with this is that N(.) is continuous at +∞. Can a function be continuous at +infinity? Does that even make sense?

You're asking if N is continuous at 0, NOT at infinity. Hence the limit of b going to 0.

Here's an easy way to think about this. First ask yourself, is N continuous on my interval? If N has no points of discontinuity, then you can conclude N is continuous and the limit operator can be brought inside of your function much like I did with my example in my earlier post.

Since sinx was continuous at all points, I could bring my limit inside. So on and so forth...

 

[Ray Vickson]

Homework Statement

Let N(x) denote the CDF of the standard normal density. So, it's the integral of the standard normal density from -∞ to x.

Is it true that $\lim_{b\to 0} N(\frac{a}{b}) = N(\frac{a}{\lim_{b\to 0}b}) = N(+\infty) = 1$?

2. The attempt at a solution

This is more of a knowledge question... Can't really "attempt" it.

The answer is NO in general, because for a ≠ 0 the limit of a/b as b → 0 is not a definite +∞ or -∞ (that is, it is not defined, even if you allow for infinite values). The ONE-SIDED limit of a/b as b→ 0 either from the right or from the left will be either +∞ or -∞, and N is "continuous at ± ∞", so what you want is OK for one-sided limits, provided that you also include the choice N(-∞) as well.

RGV

 

[operationsres]

The answer is NO in general, because for a ≠ 0 the limit of a/b as b → 0 is not a definite +∞ or -∞ (that is, it is not defined, even if you allow for infinite values). The ONE-SIDED limit of a/b as b→ 0 either from the right or from the left will be either +∞ or -∞, and N is "continuous at ± ∞", so what you want is OK for one-sided limits, provided that you also include the choice N(-∞) as well.

Hi, thanks.

How is N(.) continuous at +∞ if the domain doesn't even include +∞? Also you say "OK for one-sided limits, provided that you also include the choice N(-∞) as well." - are you saying here that I have to evaluate it at both sides? Why can't I just evaluate it at one side (e.g. $0^+$?

 

[operationsres]

You're asking if N is continuous at 0, NOT at infinity. Hence the limit of b going to 0.

I don't think that's right. I have g(x) inside N(.), and I'm taking the limit of g(x) as some variable in g(x) goes to $o^+$, and as I take this limit g(x) goes to +∞ and as such the question is whether N(.) is continuous at +∞.

Here's an easy way to think about this. First ask yourself, is N continuous on my interval? If N has no points of discontinuity, then you can conclude N is continuous and the limit operator can be brought inside of your function much like I did with my example in my earlier post.

I understand that N is right-continuous on its domain of (-∞,+∞). What I would like final confirmation of is if it's continuous somewhere that's not in its domain: at +∞ itself.

I understand your example with sin(.) because it is common knowledge that sin is continuous at 0, just not sure about at +∞.

 

[STEMucator]

I don't think that's right. I have g(x) inside N(.), and I'm taking the limit of g(x) as some variable in g(x) goes to $o^+$, and as I take this limit g(x) goes to +∞ and as such the question is whether N(.) is continuous at +∞.



I understand that N is right-continuous on its domain of (-∞,+∞). What I would like final confirmation of is if it's continuous somewhere that's not in its domain: at +∞ itself.

I understand your example with sin(.) because it is common knowledge that sin is continuous at 0, just not sure about at +∞.

No. If a function is continuous on a defined interval, then it is continuous only in the defined interval UNLESS stated otherwise or given a different interval entirely.

So if N is defined on ℝ⁺, and is continuous at all points on the interval, then you can manipulate the operator as desired.

Also, you don't actually want to calculate N(∞), but rather you want to examine the nature of N as it gets infinitely large. What does it tend to?

 

[operationsres]

N(x) tends to 1 as x goes to +infinity. N(x) tends to 0 as x goes to -infinity.

I understand that if N(x) is continuous on some interval $I \subseteq \mathbb{R}$ then we can bring the limit inside of N(x) as long as the solution to the limit of x is contained within I.

All I am confused about now is if I can do that if the solution to the limit of x is infinity (i.e. bringing in $\lim_{b\to 0^+}\frac{a}{b}$, which equals +infinity), as I don't know if N(.), or any function, is defined at infinity, or is continuous at infinity.

So if N is defined on ℝ⁺, and is continuous at all points on the interval, then you can manipulate the operator as desired.

I now completely understand this. My question is whether N is defined at +$\infty$ even though it is not within the domain of N(x) ((the domain of N(x) is $\mathbb{R}$)). I don't think we can bring inside of N(.) the limit: $\lim_{b\to 0^+}\frac{a}{b}$ (which is a limit that's equal to +$\infty$) despite the fact that N is continuous from the right. As you said we don't want to calculate N(∞), which is what we'd be doing if we brought in that limit (that limit is an object that is equal to ∞).

 

[STEMucator]

N(x) tends to 1 as x goes to +infinity. N(x) tends to 0 as x goes to -infinity.

I understand that if N(x) is continuous on some interval $I \subseteq \mathbb{R}$ then we can bring the limit inside of N(x) as long as the solution to the limit of x is contained within I.

All I am confused about now is if I can do that if the solution to the limit of x is infinity (i.e. bringing in $\lim_{b\to 0^+}\frac{a}{b}$, which equals +infinity), as I don't know if N(.), or any function, is defined at infinity, or is continuous at infinity.

There's no such thing as being defined at infinity or being continuous at infinity. Infinity is a concept, not a number. You examine the nature of something as something else becomes infinitely large.

 

[operationsres]

That makes perfect sense. My question is just to confirm my position; that we're NOT allowed to bring in $lim_{b\to 0^+} \frac{a}{b}$ into the right-continuous $N(\frac{a}{b})$ defined on $\mathbb{R}$ because of the fact that $lim_{b\to 0^+} \frac{a}{b}=\infty$ and $N(\infty)$ is undefined.

That is, can I confirm that we CAN'T evaluate $N(lim_{b\to 0^+} \frac{a}{b})$ because doing so would be the same as evaluating $N(\infty)$ which is undefined.

 

[STEMucator]

That makes perfect sense. My question is just to confirm my position; that we're NOT allowed to bring in $lim_{b\to 0^+} \frac{a}{b}$ into the right-continuous $N(\frac{a}{b})$ defined on $\mathbb{R}$ because of the fact that $lim_{b\to 0^+} \frac{a}{b}=\infty$ and $N(\infty)$ is undefined.

That is, can I confirm that we CAN'T evaluate $N(lim_{b\to 0^+} \frac{a}{b})$ because doing so would be the same as evaluating $N(\infty)$ which is undefined.

Okay, you CAN bring the limit operator inside, BECAUSE N is defined to be continuous everywhere on the interval.

Then when you try to evaluate the limit, you see that the limit is undefined because it would produce ∞.

Now you would be able to conclude that you cannot evaluate the limit because N is undefined at ∞.

Make sense?

 

[operationsres]

That answer can't be correct. Firstly, N(x) is this:

https://controls.engin.umich.edu/wiki/images/5/5c/NormalDistributionCDF.JPG

the way it's derived means that as we go to +∞ on the x-axis, then N(x) converges to 1 by the ε definition of the limit.

Also a question in my assignment requires that the limit of x -> ∞ of N(x) is defined otherwise the question is incorrect, which is not right because it's a well known problem in my field.

I think that the above means that we can't bring the limit inside, unless I'm suffering from retardation maybe :(

 

[STEMucator]

Forget this. Look below.

 

[Ray Vickson]

Hi, thanks.

How is N(.) continuous at +∞ if the domain doesn't even include +∞? Also you say "OK for one-sided limits, provided that you also include the choice N(-∞) as well." - are you saying here that I have to evaluate it at both sides? Why can't I just evaluate it at one side (e.g. $0^+$?

Well, maybe the limit a/b as b → 0 is -∞. For example, if a = -1 and we let b → 0 through positive values (written as b → 0+), the limit will be -∞. If a = +1 and b → 0+, the limit will be +∞, etc. Just look at the graph of a/b for a fixed and b variable; what does the graph look like when b approaches 0 on either side of the origin?

As to how N(.) is continuous at +∞ if the domain doesn't even include +∞, well that is just a shorthand statement that all the limits go through as you want them to. Certainly,
$$\lim_{x \to +\infty} N(x) = 1$$ exists, as does
$$\lim_{x \to -\infty} N(x) = 0.$$
In fact, these hold for any legitimate cdf and do not require normality; they are even true for discrete random variables such as the binomial or Poisson, etc. All we require is that
$$F(x) \to 0 \text{ as } x \to -\infty \text{ and } F(x) \to 1 \text{ as } x \to +\infty,$$
which are, essentially, part of the definition of a legitimate cdf F(.). (By legitimate, I mean that P{ℝ} = 1,so we do not have positive probabilities that X = +∞ or X = -∞.)

RGV

 

[operationsres]

Great answer. So we can be confident that we are correct in stating that, for a > 0, $\lim_{b\to 0^+}N(\frac{a}{b}) = N(\lim_{b\to 0^+} \frac{a}{b}) = N(+\infty) = 1$?
R} \to [0,1][/itex].

 

[operationsres]

So it's a question of whether the domain of N(x) is the reals or the extended reals? Google reveals absolutely nothing.

Definitely N(+∞) is defined as it would just be the integral from -∞ to +∞ of the PDF, which is defined. However I'm not sure about whether N(-∞) is defined.

 

[Ray Vickson]

Great answer. So we can be confident that we are correct in stating that, for a > 0, $\lim_{b\to 0^+}N(\frac{a}{b}) = N(\lim_{b\to 0^+} \frac{a}{b}) = N(+\infty) = 1$?
R} \to [0,1][/itex].

Please use a proper reply method, so that we can tell which message you are responding to. (You can do that by using 'quote' instead of 'quick response'.) And yes, for a > 0 and b → 0+ we do, indeed, have lim N(a/b) = 1, whether or not you regard N(+∞) as having any meaning. The same applies to the limit of N(a/b) for a > 0 and b → 0- (or a > 0 and b → 0+): the limit is 0, whether or not you regard N(-∞) as having any meaning. And, as I have said already, it applies to ANY cdf, not just to the normal cdf.

RGV