# Distribution of $\frac{1}{1+X}$ if $X$ is Lognormal

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): 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? 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.
• 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 $n\neq1$? Nov 17, 2020 at 8:04