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Thread: Limit of standard normal CDF(x) as x goes to ∞?

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    Limit of standard normal CDF(x) as x goes to ∞?




    Let N:\mathbb{R} \to [0,1] be the CDF of the standard normal density function. I want to evaluate:

    \lim_{b\to 0^+} N(\frac{a}{b}). Clearly this is equal to 1 (we have that a > 0).

    What I would like to know is whether I can bring the limit inside. The CDF is right-continuous and so I would usually have no issues with bringing it inside. However is N(\infty) defined? If it is NOT defined then since lim_{b\to 0^+}\frac{a}{b} is itself equal to \infty, it seems that I'm not allowed to bring the limit inside of N(.) in this instance? But if it is defined at +infinity then I can bring the limit inside.
    Last edited by derksheng; 10-15-2012 at 10:37 PM.

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    Re: Limit of standard normal CDF(x) as x goes to ∞?

    Quote Originally Posted by derksheng View Post
    \lim_{b\to 0^+} N(\frac{a}{b}). Clearly this is equal to +\infty.
    Seems to me the limit would be 1?
    I don't have emotions and sometimes that makes me very sad.

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    Re: Limit of standard normal CDF(x) as x goes to ∞?

    Lol. Sorry. Of course it's 1. been typing too much latex in last 5 hours.

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    Re: Limit of standard normal CDF(x) as x goes to ∞?

    Actually I might have been too quick in asserting that the domain is \mathbb{R}. I was advised that we can put in \infty into the argument of erf(x), so maybe we can do it for N(x) if this person is correct.
    Last edited by derksheng; 10-15-2012 at 10:54 PM.

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    Re: Limit of standard normal CDF(x) as x goes to ∞?

    I see no reason why we couldn't extend the domain to the extended Reals. Then for any CDF we should have F(-\infty) = 0 and F(\infty) = 1
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    Re: Limit of standard normal CDF(x) as x goes to ∞?

    And as a side note you need to have more than just "right continuous" to be able to bring a limit inside the integral.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Limit of standard normal CDF(x) as x goes to ∞?

    I am confident that we can extend it to +\infty because we would just be taking the integral of the PDF from -\infty to +\infty which is well defined as equal to 1.

    However I'm less confident about fixing x = -\infty in N(x), because I haven't done the math to show that we can integrate from -\infty to -\infty and get 0 without any problems.

    It doesn't help that these notes make the claim that the domain is \mathbb{R}:
    http://ocw.mit.edu/courses/electrica...JF08_lec05.pdf

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    Re: Limit of standard normal CDF(x) as x goes to ∞?

    But then again those notes must be wrong here because the domain is obviously AT LEAST \mathbb{R} \cup \{\infty\}.

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    Re: Limit of standard normal CDF(x) as x goes to ∞?

    Quote Originally Posted by derksheng View Post

    However I'm less confident about fixing x = -\infty in N(x), because I haven't done the math to show that we can integrate from -\infty to -\infty and get 0 without any problems.
    The integral of the standard normal pdf for x from -Infinity to -Infinity will be zero.

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    Re: Limit of standard normal CDF(x) as x goes to ∞?

    Okay good so the domain is the extended reals! Great!!!

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    Re: Limit of standard normal CDF(x) as x goes to ∞?

    Quote Originally Posted by Dason View Post
    And as a side note you need to have more than just "right continuous" to be able to bring a limit inside the integral.
    However I'm not bringing it "inside" the integral. For N(x), x simply means the integral of the N(0,1) PDF from -\infty \text{ to } x. (Which you already knew of course, just making sure I'm right!). So the limit won't be going in the integral but in \int_{-\infty}^{lim_{x\to +\infty} x} = \int_{-\infty}^{+\infty}.
    Last edited by derksheng; 10-16-2012 at 02:40 AM.

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    Re: Limit of standard normal CDF(x) as x goes to ∞?

    To finalize this thread, N(.) is both right and left continuous so it is always fine to bring in the limit.

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    Re: Limit of standard normal CDF(x) as x goes to ∞?


    But in a measure theoretic sense you are bringing it inside the limit because what is happening is

    \int_{-\infty}^x f(t) dt = \int_{-\infty}^\infty f(t) I(t \leq x) dt

    Further note that continuity isn't all you need to bring limits inside!

    Let f_n(x) = 0 \mbox{ if } x \leq n, 1 \mbox{ if } x > n+1, x-n \mbox{ if } n < x \leq n+1

    Now f_n is continuous for all n. Also for any value of x we have f_n(x) \rightarrow 0 as n \rightarrow \infty. However

    \int_{\infty}^{\infty}f_n(x) dx = \infty for all n.

    So in this case 0 = \int_{-\infty}^\infty 0 dx
    = \int_{-\infty}^\infty \lim_{n \rightarrow \infty} f_n(x) dx
    \neq \lim_{n \rightarrow \infty}  \int_{-\infty}^\infty f_n(x) dx = \infty

    If there is one thing I learned in measure theory it's that you need to be careful when moving limits around.
    I don't have emotions and sometimes that makes me very sad.

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