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Suppose $Y\sim N(\mu,\sigma)$

I would like to investigate the distribution of:

$$\frac{1}{1+Y}$$

Does the distribution exist and is it well defined? Does it have analytically computable moments?

Googling hasn't lead to any concrete results, so I have attempted to derive its PDF using the method of derived distributions (I believe I have succeeded), and I have tried to compute it's mean (I haven't succeeded).

My work is as follows: let $X:=\frac{1}{1+Y}$, then:

$$F_X(x)=\mathbb{P}\left(X\leq x\right)=\mathbb{P}\left(\frac{1}{1+Y}\leq x\right)=\mathbb{P}\left(\frac{1}{x}-1\leq Y\right)=1-\mathbb{P}\left(Y\leq \frac{1}{x}-1\right)=1-F_Y\left(\frac{1-x}{x}\right)$$

Now (setting $d:=\frac{1-x}{x}$)

$$f_X(x):=\frac{\partial}{\partial x}\left(F_X\left(x\right)\right)=\frac{\partial}{\partial x}\left(1-F_Y\left(d\right)\right)=-\frac{\partial}{\partial d}F_Y(d)\frac{\partial d}{\partial x}=-f_Y(d)\left(\frac{-1}{x^2}\right)$$

So we finally have the PDF of $X$ as:

$$\frac{1}{x^2}\frac{1}{\sigma \sqrt{t2 \pi}}\exp\left\{-\frac{(d-\mu)^2}{2t\sigma^2}\right\}=\frac{1}{x^2}\frac{1}{\sigma \sqrt{t2 \pi}}\exp\left\{-\frac{(\frac{1-x}{x}-\mu)^2}{2t\sigma^2}\right\}$$

To simplify the above, I will do the following:

$$\left(\frac{1-x}{x}-\mu\right)^2=\left(\frac{1}{x}-1-\mu\right)^2=\frac{1}{x^2}-2(1-\mu)\frac{1}{x}+(1-\mu)^2+4\mu=\left(\frac{1}{x}-(1-\mu)\right)^2+4\mu$$

So that:

$$f_X(x)=\frac{1}{x^2}\frac{1}{\sigma \sqrt{t2 \pi}}\exp\left\{-\frac{\left(\frac{1}{x}-(1-\mu)\right)^2}{2t\sigma^2}\right\}e^{-\left(\frac{4\mu}{2t\sigma^2}\right)}$$

Now attempting to compute the expectation:

$$\mathbb{E}\left[X\right]=\int_{h=-\infty}^{h=\infty}hf_X(h)dh=e^{-\left(\frac{4\mu}{2t\sigma^2}\right)}\int_{h=-\infty}^{h=\infty}\frac{1}{h}\frac{1}{\sigma \sqrt{t2 \pi}}\exp\left\{-\frac{\left(\frac{1}{h}-(1-\mu)\right)^2}{2t\sigma^2}\right\}dh$$

I will use the substitution $v=\frac{1}{h}$, with $dh=\frac{-1}{v^2}dv$:

$$\mathbb{E}\left[X\right]=e^{-\left(\frac{4\mu}{2t\sigma^2}\right)}\int_{v=0}^{v=0}\frac{-1}{v}\frac{1}{\sigma \sqrt{t2 \pi}}\exp\left\{-\frac{\left(v-(1-\mu)\right)^2}{2t\sigma^2}\right\}dv$$

The above is where I get stuck, and the fact that the integral limit is from zero to zero makes me think the expectation might no be well defined, in the sense that it is not analytically computable?

Any further tips and hints would be greatly appreciated.

Jan Stuller
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    I am not sure the last substitution is fine. Note that in the definition of integral with $\infty$, we must use limit. In addition, if $h\rightarrow +\infty$, then $1/h \rightarrow 0^{+}$. – TrungDung Dec 05 '20 at 19:06
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    Google `reciprocal normal distribution`; in particular, [see](https://math.stackexchange.com/questions/646428/mean-and-variance-of-reciprocal-normal-distribution). // Perhaps worst case: If $\mu = -1,$ then the denominator has a mode at $0.$ – BruceET Dec 05 '20 at 19:34
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    The expectation is zero: this follows directly from my analysis at https://stats.stackexchange.com/a/299765/919. Consequently, all higher moments are zero, too: https://stats.stackexchange.com/questions/244202. As far as existence goes, that's settled by observing there is zero chance that $1/(1+Y)$ will be undefined. – whuber Dec 05 '20 at 19:44
  • @BruceET, thank you so much, that is really helpful. So I suppose I could just say that $$\hat{Y}:=1+Y\sim N(\mu+1,\sigma)$$ and in fact the above is then the "inverse normal" (or reciprocal normal): right? – Jan Stuller Dec 05 '20 at 19:45
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    @whuber: many thanks for posting that resource. I've had a look and it's quite a bit to take in. I will have a proper read tomorrow: it seems to exactly address what I am interested in, so many thanks indeed. – Jan Stuller Dec 05 '20 at 19:50
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    You may translate that analysis to any particular instance. The one insight is to analyze the behavior of the integral in a neighborhood of $1+Y\approx 0,$ which corresponds to $|h|\ge T$ in your integral for arbitrarily large numbers $T.$ In that case you can obtain a positive lower bound (say $M$) for the exponential, leaving you to evaluate $$M\left(\left\{\int_{-\infty}^{-T} + \int_{T}^\infty\right\} \frac {1}{|h|}\mathrm{d}h\right),$$ which diverges. – whuber Dec 05 '20 at 19:57

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