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


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.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.