Mathematics Asked by Diego Andres Gomez Polo on February 10, 2021
I’ve been trying to prove that for a given random variable $X$, with pdf := $f_X(x) = e^{-x}1_{[0, infty)}$, the moment generating function of $Y:= sqrt{n}(overline{X}-1)$, where $overline{X} $ is the sample mean of $X$ for $n$ samples, is equal to:
$$
M_n(t) = bigg[e^{frac{t}{sqrt{n}}} – bigg(frac{t}{sqrt{n}} bigg)e^{frac{t}{sqrt{n}}} bigg]^{-n}
$$
For $t < sqrt{n}$
First i calculated the expected value $mu$ and the variance $sigma ^2$ of $X$.
$$
mu = int_{mathbb{R}} x e^{-x}mathbb{1}_{[0,infty)}dx = int_{0}^{infty} x e^{-x}dx = 1
$$
$$
sigma ^2 = int_{mathbb{R}} x^2 e^{-x}mathbb{1}_{[0,infty)}dx – mu ^2 = int_{0}^{infty} x^2 e^{-x}dx – 1 = 2 – 1 = 1
$$
Then i used the fact that $overline{X} thicksim mathcal{N}(mu, frac{sigma ^2}{n})$ to get $overline{X} thicksim mathcal{N}(1, frac{1}{n})$, my goal was to use this together with the random variable transformation theorem to get the density function of $Y$ and compute its mgf.
In this case, the theorem states that:
$$
begin{align*}
pdf(Y) = f_Y(y) &= f_{overline{X}}(frac{y}{sqrt{n}}+1)frac{d}{dy}bigg(sqrt{n}(y-1)
bigg)\
\
&= frac{1}{frac{1}{sqrt{n}}sqrt{2pi}}expBigg({-frac{1}{2}}bigg(frac{frac{y}{sqrt{n}}+1-1}{frac{1}{sqrt{n}}} bigg)^2 Bigg )sqrt{n}\
\
&= frac{n}{sqrt{2pi}}e^{-frac{1}{2}y^2}
end{align*}
$$
Which is not integrable in a usefull way, so i am basically stuck here and don’t know what else to do
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