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$Y$ is a random variable with $$M(t) = \frac{1}{(2-\exp(t))^s}.$$

Does $$\frac{Y-E(Y)}{\sqrt{\operatorname{Var}(Y)}}$$ converge in distribution as $s$ tends to infinity?

I let $Z = \frac{Y-E(Y)}{\sqrt{\operatorname{Var}(Y)}}$. Differentiating the MGF of $Y$ and let $t = 0$ we have $$E(Y) = s,$$$$E(Y^2) = s^2 +2s,$$$$\operatorname{Var}(Y) = 2s.$$ Thus Mgf of Z: \begin{align*}E(\exp(tz)) &= E\left(\exp\left(\frac{t(y-s)}{\sqrt{2s}}\right)\right)\\ &=E\left(\exp\left(\frac{ty}{\sqrt{2s}}\right)\exp\left(\frac{-ts}{\sqrt{2s}}\right)\right)\\ &=E\left(\exp\left(\frac{ty}{\sqrt{2s}}\right)\exp\left(\frac{-ts}{\sqrt{2s}}\right)\right)\\ &=\exp\left(\frac{-ts}{\sqrt{2s}}\right)\frac{1}{\left(2-\exp\left(\frac{t}{\sqrt{2s}}\right)\right)^s}. \end{align*} But I'm not sure how to proceed - seems to me it tends to $0$?

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up vote 1 down vote accepted

As you have determined: $$ M_Z(t) = \mathbb{E}(\exp(t Z ) ) = \frac{\exp\left(-t \frac{s}{\sqrt{2s}}\right)}{\left( 2 - \exp\left(\frac{t}{\sqrt{2s}} \right) \right)^s} = \left( \frac{\exp\left(-t \frac{1}{\sqrt{2s}}\right)}{ 2 - \exp\left(\frac{t}{\sqrt{2s}} \right) } \right)^s $$ In order to determine large $s$ behavior, use Taylor series: $$ \frac{\exp\left(-t \frac{1}{\sqrt{2s}}\right)}{ 2 - \exp\left(\frac{t}{\sqrt{2s}} \right) } = \frac{1- \frac{t}{\sqrt{2s}} + \frac{t^2}{4 s} + o\left(\frac{t^2}{s}\right)}{ 1- \frac{t}{\sqrt{2s}} - \frac{t^2}{4 s} + o\left(\frac{t^2}{s}\right) } = 1 + \frac{t^2}{2s} + o\left(\frac{t^2}{s}\right) $$ Thus $$ M_Z(t) = \left( 1 + \frac{t^2}{2s} + o\left(\frac{t^2}{s}\right) \right)^s $$ The large $s$ limit is now easy to find: $$ \lim_{s\to \infty} M_Z(t) = \exp\left( \frac{t^2}{2} \right) $$ This is the mgf of the standard normal variable, as expected by CLT.

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