# Inverse Gaussian, Limiting Distributions

I'm trying to understand the nature of limiting distributions and distributions, specifically

$1/Z_n \longrightarrow ~?$

where $Z_n\longrightarrow Z -Gaussian(0,1)$

I understand that the gamma distribution converges to the gaussian for a large enough $n$, so would it be inverted gamma until the $n$ is sufficiently large? But the wikipedia article for Inverted Gamma and Inverse Gaussian are completely different, even though they're both written as $IG( ~, ~)$

But I also have the same distribution as Inverted Gamma with $\alpha,\beta$ as parameters

Attempts: Let $Y=1/N$ where $N=Gaussian(0,1)$

$F(Y)=P(Y<y)=P(1/N<y)=P(N>1/y)=1-F(1/y)$

But this gets me nothing that looks like what Wikipedia has

http://en.wikipedia.org/wiki/Inverse_Gaussian_distribution

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Wikipedia says about an Inverse Gaussian

The name can be misleading: it is an "inverse" only in that, while the Gaussian describes a Brownian Motion's level at a fixed time, the inverse Gaussian describes the distribution of the time a Brownian Motion with positive drift takes to reach a fixed positive level.

So $Y=1/N(0,1)$ (which might be called a reciprocal Gaussian, or in one case a recinormal random variable) is not what Wikipedia calls an Inverse Gaussian random variable. In particular $Y=1/N(0,1)$ (which has a bimodal probability density and heavy tails) has no mean while the mean is one of the two parameters of a so-called Inverse Gaussian distribution.

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Ah that was really useful thanks, I haven't learned about Brownian Motion yet. Reciprocal makes a lot more sense –  myTotoro Nov 18 '12 at 16:44