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Suppose $Z \sim \mathcal{N}(0,1)$.

Suppose $X$ is a lognormally distributed random variable, defined as $X:=X_0exp^{(-0.5\sigma^2+\sigma Z)}$, in other words, $X$ is log-normal with $\mathbb{E}[X]=X_0$.

Suppose we are interested in the variable of the type $Y:=\frac{1}{1+X}$

Question: Does the distribution of $Y$ have any name? Does it have a well defined PDF and CDF?

Distributions such as $Y$ arise often in finance, because interest rates might be modelled as exponential martingales (i.e. their distribution at a specific point in time would correspond to the variable $X$ defined above). Then, Bond prices would actually have a distribution corresponding to the variable $Y$ (that is a zero-coupon bond maturing in one year. If the bond matures in "$n$" years, then the denominator is of power $n$: $(1+X)^n$)

I ran a simple simulation in Python to plot $X$ and $Y$, with $X_0=0.01$, $\sigma=0.2$. Then I get a log-normal distribution for $X$ (as, of course, expected):

enter image description here

For $Y$, the shape of the graph resembles a log-normal random variable, but rotated around its mean axis (i.e. longer left-tail instead of longer right-tail): just by eyeballing the graph, I would think that perhaps the PDF and CDF are well defined, but before diving into attempting the algebra, I wanted to check here whether this problem has a standard solution?

enter image description here

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I will answer a simplified version, so leave the generalization as an exercise. Let $Z$ be a standard normal random variable so $X=e^Z$ is standard lognormal. Since $X>0 $ we have $Y=\frac1{1+X}$ is in the unit interval. Let $\phi, \Phi$ be the density and cdf (cumulative distribution ) functions of the standard normal, then we find $$ \DeclareMathOperator{\P}{\mathbb{P}} F_Y(y)=\P(Y \le y)= 1-\Phi\left( \ln(\frac{1-y}{y})\right) $$ and by differentiation the density is $$f_Y(y)=\frac{\phi\left( \ln(\frac{1-y}{y}) \right)}{y(1-y)} $$ The factor in the denominator leads the thoughts to something logistic, and,in fact, this is a Logit-normal distribution.


That relationship seems important and need a simpler derivation, just from the definitions. Since $Z$ is standard normal, so symmetric about zero, $-Z$ have the same distribution, so to represent (the distribution of ) $X$ we can as well use $X=e^{-Z}$. Then $$ Y=\frac1{1+X}=\frac1{1+e^{-Z}}=\frac{e^Z}{1+e^Z} $$ and it follows directly that $\operatorname{logit}(Y)$ is a standard normal distribution, without any need of deriving the density function.

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    $\begingroup$ That's a great answer, thank you very much. $\endgroup$ – Jan Stuller Nov 16 '20 at 12:42
  • $\begingroup$ The yield "$X$" on the bond is an annualized quantity (it expresses the yield per year). The formula in my original question for the bond price is valid for a bond that matures in 1 year. A bond that matures in "$n$" years would look like this: $$Y=\frac{1}{(1+X)^n}$$ Expanding the denominator gives a Sum of Lognormal variables. Looking at this Cross-Validated post here, the Sum of Lognormals is approximately log-normal. So the Logit-Normal would also hold for $n\neq1$? $\endgroup$ – Jan Stuller Nov 17 '20 at 8:04

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