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I found a MGF with 2 variables, n and t, to do with a transformation to an average of a sum of independent random variables following the same distribution. The limit of this MGF as n approaches infinity is equal to 0 and I am wondering what distribution this follows? Edit: I'm wondering about the limiting distribution of Zn

$f(x)=0.75*0.25^{x-1} , x=1, 2, 3, ...$

$Y_n=\sum_{i=1}^{n}{X_i}$

$Z_n=\frac{3}{2}\sqrt{n}\bar{Y_n}-2\sqrt{n}$

$M_\bar{Y_n}(t)=[M_{Y_n}(\frac{t}{n})]^n=[M_{X_n}(\frac{t}{n})]^{n^2}=[\frac{0.75e^\frac{t}{n}}{1-0.25e^\frac{t}{n}}]^{n^2}$

$M_{Z_n}(t)=E(e^{tz})=E(e^{t(\frac{3}{2}\sqrt{n}\bar{Y_n}-2\sqrt{n})})=E(e^{\frac{3t}{2}\sqrt{n}\bar{Y_n}}e^{-2t\sqrt{n}})=e^{-2t\sqrt{n}}E(e^{\frac{3t}{2}\sqrt{n}\bar{Y_n}})=e^{-2t\sqrt{n}}M_{\bar{Y_n}}(\frac{3t\sqrt{n}}{2})$

$\lim_{n->\infty}(M_{Z_n}(t))=\frac{1}{\infty}=0$

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  • $\begingroup$ Welcome to CV Ricardo. Your question would be much improved by providing the MGF and the distribution of the variables. You can get lots of LaTeX markup working by delimiting it with $, so for example, $\sum_{i=1}^{n}{x_{i}}$ produces $\sum_{i=1}^{n}{x_{i}}$. You can edit your question using the "edit" link in the lower left. $\endgroup$
    – Alexis
    May 24, 2018 at 17:36
  • $\begingroup$ One (trivial) option would be the "delta distribution". That is the normal distribution with zero mean and zero variance. $\endgroup$ May 24, 2018 at 19:17
  • $\begingroup$ I think the "delta distribution" and other degenerate RV's have MGF equal to $1$, not $0$. $\endgroup$
    – knrumsey
    May 24, 2018 at 19:41
  • $\begingroup$ Huh. I always thought of the delta distribution as $\lim_{\sigma\to0}\phi(x;0,\sigma)$, which would lead the MGF as going to 0. Might be wrong of the definition though. $\endgroup$ May 24, 2018 at 19:46
  • $\begingroup$ @AndreasStorvikStrauman, The MGF of a normal distribution is $e^{\mu t + \sigma^2 t^2/2}$. So even in the limit, we get $\phi(t) \rightarrow e^0 = 1$. $\endgroup$
    – knrumsey
    Apr 13, 2022 at 18:20

2 Answers 2

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Your limit is wrong. $X_i$ has geometric distribution with mean $1/(1-p)=4/3$, variance $(1-p)/p^2$ which by CLT implies that $\sqrt{n}\bar{Y}_n-\sqrt{n}/(1-p)$ approaches a normal random variable with mean 0 and variance $(1-p)/p^2$

$Z_i$ has mean $(3/2)\sqrt{n}(4/3)-2\sqrt{n}=0$, and should also correspondingly also approach a normal random variable via CLT.

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  • $\begingroup$ Thank you so much! Is my notation for $M_{\bar{Y_n}}(t)$ correct? $\endgroup$
    – Ricardo723
    May 24, 2018 at 20:55
  • $\begingroup$ Also, Is it correct to notate $\bar{Y_n}=\frac{4}{3}$? $\endgroup$
    – Ricardo723
    May 24, 2018 at 21:27
  • $\begingroup$ That's ambiguous. I'm assuming you defined $\bar{Y}_n = \frac{1}{n}\sum_{i=1}^n X_i$, in which case it's a sample mean, not the true mean. $\endgroup$
    – Alex R.
    May 24, 2018 at 22:13
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There does not exist an MGF that is identically equal to zero:

Let $f:D\to\mathbb{R}$, then for $f$ to be a valid pdf then $\int_Df(x)\mathrm{d}x=1$ and $f(x)\geq0~\forall~x\in D$.

For the MGF to be identically zero then $\int_{D}e^{tx}f(x)\mathrm{d}x$ would have to be 0 for all values of $t$. However, for $t=0$ we get $$\int_De^{0 x}f(x)\mathrm{d}x=1\neq0$$.

This also holds in the discrete case:

$$\sum_{x\in D}e^{tx}p(x)\leadsto\sum_{x\in D}e^{0 x}p(x)=1$$

Edit: I now realize that I'm answering your title and not your actual question. Sorry about that, heh.

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  • $\begingroup$ haha that's ok, I guess I should have been a bit more specific in the title. $\endgroup$
    – Ricardo723
    May 24, 2018 at 20:02
  • $\begingroup$ But it could definitely be some shape of $e^x$ if you look at the solutions to the equations above in the answer. $\endgroup$ May 24, 2018 at 20:06
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    $\begingroup$ This is completely wrong. There is in fact no MGF that's identically zero. For starters at $t=0$, $\int_D f(x)dx=1$ is a must, whereas your initial assumption is that its 0, but later you say it's 1. This has nothing to do with derivatives either, since if $g\geq 0$, $\int_D g(x)dx=0$ iff g(x)=0 almost everywhere. $\endgroup$
    – Alex R.
    May 24, 2018 at 20:13
  • $\begingroup$ Yup. You're right. Brain fart. $\endgroup$ May 24, 2018 at 20:18
  • $\begingroup$ @AlexR. Brain unfarted? $\endgroup$ May 24, 2018 at 20:30

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