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In a statistics problem, the following limit (on the left side) falls out from using various theorems on moment generating functions. The last step of the problem is to show that the limit on the left does indeed evaluate to the moment generating function of the standard distribution, as follows:

$$ \lim_{n \to \infty}\frac {e^{-\sqrt{\frac n 2}\cdot t}} {\left( 1-\sqrt{\frac 2 n}\cdot t\right)^{\frac n 2}} \ {\overset ? =}\ {\large e^{t^2/2}} \tag1 $$

I am beyond rusty with my limits, but it "almost" seems like it should be an elementary thing - the numerator and denominator individually look "tame". But I'd like to take a stab at evaluating it myself, with some help. I ask for comments about my reasoning below.

The limit is "simple" enough that Maxima computes it symbolically - if need be, I'll trace what the limit function does with this one at the LISP level to understand how one could go about it, but I'd rather avoid that for now.

(%i1) num : exp(-sqrt(n/2)*t)$
(%i2) denom : (1-sqrt(2/n)*t)^(n/2)$
(%i3) limit(num/denom, n, inf);
        t^2/2
(%o3) %e

De l'Hospital's rule seems to work as well, but the derivatives are more complicated than the numerator and denominator themselves, so it seems fruitless to pursue that. The integrals of the numerator and denominator don't look any simpler either (if one were to work de l'Hospital's rule backwards).

(%i4) limit(diff(num,n) / diff(denom,n), n, inf);
        t^2/2
(%o4) %e

Back to the limit: the limit of the denominator $\left(1-\sqrt{2/n}\cdot t\right)^{n/2}$ for $t>0$ is zero, so using the quotient rule is out.

Experimenting a bit, I've found that

$$\lim_{m\to \inf} {1 \over \left( 1-{t^2 \over m} \right)^{m\over 2}} = {\large e^{t^2 \over 2}},\tag2\label2$$

so I assume this is an "elementary" limit that one could look up in a table somewhere. But I don't need to evaluate it anyway - I just want to use it as a hint.

Comparing $\eqref{2}$ to the denominator of the original function, we get

$$\lim_{n\to\infty} {1\over{\left( 1-\sqrt{\frac 2 n}\cdot t\right)^{\frac n 2}}} ^{t\over\sqrt{2n}} = {\large e^{t^2/2}} \tag3$$

Now, I'm thinking of the following approach:

  1. Transform the original equation to a ratio that should be shown equal to 1:

    $$ \lim_{n \to \infty} {\frac {e^{-\sqrt{\frac n 2}\cdot t}} {\left( 1-\sqrt{\frac 2 n}\cdot t\right)^{\frac n 2}} \over {\large e^{t^2/2}}}\ {\overset ? =}\ 1 \tag4 $$

  2. Since we expect the ratio to be 1, we can raise it to $t/\sqrt{2n}$. If the ratio is indeed 1, this won't change its value. I imagine there should be a more detailed argument made why it may be valid to do so - I'm not sure whether it is...

    $$ \lim_{n \to \infty} \left( {e^{-\sqrt{\frac n 2}\cdot t}} \over {{\left( 1-\sqrt{\frac 2 n}\cdot t\right)^{\frac n 2}} \cdot {\large e^{t^2/2}}} \right)^{t\over\sqrt{2n}}=\ \ldots\ {\overset ? =}\ 1 \tag5\label5 $$

  3. Evaluate the limit, by decomposing it as follows:

    $$\lim_{n\to\infty} {A \over {B\cdot C}} = {{\lim_{n\to\infty}A} \over {\lim_{n\to\infty}B}\cdot{\lim_{n\to\infty}C}} \tag6$$

    where

    $$A=\exp\left(-\sqrt{n\over 2}\cdot t \cdot{t\over\sqrt{2n}}\right),\tag7$$ $$B=\left(1-\sqrt{2\over n}\cdot t \right)^\left({n\over 2}\cdot{t\over\sqrt{2n}} \right),{\rm and} \tag8$$ $$C=\exp\left( {t^2\over2}\cdot{t\over\sqrt{2n}} \right).\tag9$$

    Those are elementary (I think) and evaluate to $$\lim_{n\to\infty}A=\lim_{n\to\infty}B=\exp\left(-{t^2\over 2}\right),{\rm and} \lim_{n\to\infty}C=1. \tag{10}$$

  4. And indeed, we have

    $$ \lim_{n \to \infty} {\frac {e^{-\sqrt{\frac n 2}\cdot t}} {\left( 1-\sqrt{\frac 2 n}\cdot t\right)^{\frac n 2}} \over {\large e^{t^2/2}}} \ =\ {{\lim_{n\to\infty}A} \over {\lim_{n\to\infty}B}\cdot{\lim_{n\to\infty}C}} \ =\ {{e^{t\over 2}} \over {{e^{t\over 2}} \cdot 1}} \tag{11} \ =\ \large 1. $$

I imagine that $\eqref{5}$ would require some support, but I'm not sure how would I formulate it.

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  • $\begingroup$ Since $\lim_{n \to \infty} \frac{t}{\sqrt{2n}} = 0$, $(5)$ is much weaker than $(4)$. Are you exclusively asking for comments about your reasoning, or would you also welcome a "this is how you can easily do it" answer? $\endgroup$ Apr 26, 2020 at 18:02
  • $\begingroup$ I'd like both: comments about the reasoning, as well as "how you can easily do it". $\endgroup$ Apr 28, 2020 at 15:08
  • $\begingroup$ A good strategy is "logarithm + Taylor expansion". Do you want to give it a go yourself? $\endgroup$ Apr 28, 2020 at 15:14
  • $\begingroup$ Yes. I will give it a try. $\endgroup$ Apr 28, 2020 at 15:16

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