2
$\begingroup$

Question: If we have the moment generating function of S as $g_s(t) = (\frac {1+e^t}2)^n$

Define $Y = \frac {S- \frac n2}{\sqrt n}$.

Show that the moment generating function of Y is $\lim\limits_{n \to \infty} g_Y(t) → e^{t^2/8}$

Does anyone mind explaining how I could transform the moment generating function of S into Y?

1 Answers 1

1

$$g_{Y}(t)=E(e^{tY})=E(e^{t\frac {S- \frac n2}{\sqrt n}})=E(e^{\frac{t}{\sqrt{n}}S})e^{-t\frac{\frac{n}{2}}{\sqrt{n}}}=g_{S}(\frac{t}{\sqrt{n}})e^{\frac{-t\sqrt{n}}{2}}$$

Using Taylor-Exansion on $e^\frac{t}{2\sqrt{n}}$ and $e^\frac{-t}{2\sqrt{n}}$, we have

$$g_{Y}(\frac{t}{\sqrt{n}})=(\frac {1+e^{\frac{t}{\sqrt{n}}}}2)^n = (e^\frac{t}{2\sqrt{n}}\frac{e^\frac{-t}{2\sqrt{n}}+e^\frac{t}{2\sqrt{n}}}{2})^n=e^\frac{t\sqrt{n}}{2}(\frac{2+\frac{t^2}{4n}+\mathcal{o(\frac{1}{n})}}{2})^n=e^\frac{t\sqrt{n}}{2}(1+\frac{t^2}{8n}+\mathcal{o(\frac{1}{n})})^n=e^\frac{t\sqrt{n}}{2}e^{nlog({(1+\frac{t^2}{8n}+\mathcal{o(\frac{1}{n})})}}=e^\frac{t\sqrt{n}}{2}e^{{\frac{t^2}{8}+\mathcal{o(1)}}}$$

Finally, we have

$$\lim\limits_{n \to \infty} g_Y(t) → e^{t^2/8}$$

  • 0
    Thank you so much! That made a lot of sense. I had one quick question about the notation: What does the o(1/n) mean? My professor also had it in his notes but provided no explanation. – 2017-01-13
  • 0
    it is called "Little-o notation, please read section :Related asymptotic notations of the following link https://en.wikipedia.org/wiki/Big_O_notation – 2017-01-13