Define $$ \begin{cases} X_1\sim Lognormal(ln(\mu_1), \sigma^2) \\ X_2\sim Lognormal(ln(\mu_2), \sigma^2) \end{cases} $$ where $\mu_2>\mu_1>0$ and that there is a definite proportion, $\eta\in(0,1)$, between $X_1$ and $X_2$ such that $$ \begin{cases} f_1(x)=\frac{\eta}{x\sigma\sqrt{2\pi}}e^{-{\frac{(ln(x)-ln(\mu_1))^2 \,\,\,\,\,\,}{2\sigma^2}}} \\ f_2(x)=\frac{1-\eta}{x\sigma\sqrt{2\pi}}e^{-\frac{(ln(x)-ln(\mu_2))^2 \,\,\,\,\,\,}{2\sigma^2}} \end{cases} $$ where $f_1$ and $f_2$ represent the $\eta$-scaled PDF's of $X_1$ and $X_2$, respectively.

Based on the above definitions, note that $\int_{x=0}^\infty f_1(x)\,dx\,+\int_{x=0}^\infty f_2(x)\,dx=1$.

Given $\mu_1$, $\mu_2$, $\sigma$, and $\eta$, how is the overlapping area of the two probability distribution curves, $OVL=f(\mu_1,\mu_2,\sigma,\eta)$, defined?

Please see a illustrative plot below, where $OVL=f(\mu_1=5,\mu_2=10,\sigma=20\%,\eta=50\%)$ is highlighted in yellow:


I am able to perform numerical approximation for $OVL$ using the trapezoidal rule, but I need to express $OVL$ explicitly and I am not sure how to do so.

  • 3
    $\begingroup$ Related Percentage of overlapping regions of two normal distributions $\endgroup$ – user2974951 Oct 15 '20 at 6:03
  • 3
  • 1
    $\begingroup$ Thanks. I have seen these, but they refer to normal distributions and also do not have the proportion parameter defined in my question. $\endgroup$ – Matthew Hui Oct 15 '20 at 7:01
  • 1
    $\begingroup$ What statistical meaning do you suppose the yellow area to have? (I cannot recognize any -- and it is not equivalent to any function of your $f_1$ and $f_2.$) What is the function $f$ you want to integrate?? $\endgroup$ – whuber Oct 15 '20 at 16:12
  • 1
    $\begingroup$ Thank you for the references. The problem is that OVL does not transform meaningfully when you analyze the distribution of $\exp(X)$ (the lognormal distribution) rather than $X$ itself (the normal distribution), so we're stuck right at the beginning: what do you hope this "overlapping area" represents? $\endgroup$ – whuber Oct 16 '20 at 14:00

With help from a friend we solved the problem ourselves:

Let $\{\tau\in\Bbb{R}^+|f_1(\tau,\mu_1,\sigma,\eta)=f_2(\tau,\mu_2,\sigma,\eta)\}$, then $$ \begin{align} \frac{\eta}{\tau\sigma\sqrt{2\pi}}e^{-\frac{(\ln\tau-\ln{\mu_1})^2}{2\sigma^2}}&=\frac{1-\eta}{\tau\sigma\sqrt{2\pi}}e^{-\frac{(\ln\tau-\ln{\mu_2})^2}{2\sigma^2}} \\ e^\frac{(\ln\tau-\ln{\mu_1})^2-(\ln\tau-\ln{\mu_2})^2}{2\sigma^2}&=\frac{\eta}{1-\eta}\quad (\because\tau>0) \\ \frac{2\ln\tau\ln{\tau_2}-2\ln\tau\ln{\tau_1}+(\ln{\mu_1})^2-(\ln{\mu_2})^2}{2\sigma^2}&=\ln\frac{\eta}{1-\eta} \\ \ln\tau&=\frac{\sigma^2[\ln\eta-\ln(1-\eta)]}{\ln{\mu_2}-\ln{\mu_1}}+\frac{\ln{\mu_2}+\ln{\mu_1}}{2} \\ \tau&=e^{\frac{\sigma^2[\ln\eta-\ln(1-\eta)]}{\ln{\mu_2}-\ln{\mu_1}}+\frac{\ln{\mu_2}+\ln{\mu_1}}{2}} \end{align} $$ Since $\mu_2>\mu_1$ and $\eta\in(0,1)$, $\exists!\tau\in\Bbb{R}^+|f_1=f_2$

$\forall x\in\Bbb{R}^+$,



$=\ln\frac{1-\eta}{\eta}+\frac{(\ln{\mu_1})^2-(\ln{\mu_2})^2}{2\sigma^2}+\frac{\ln{\mu_2}-\ln{\mu_1}}{\sigma^2}\ln{x}$, which is strictly increasing on $x\quad(\because\mu_2>\mu_1)$

$\therefore\frac{f_2(x,\mu_2,\sigma,\eta)}{f_1(x,\mu_1,\sigma,\eta)}$ is strictly increasing on $x$

It follows that:

$\begin{cases} f_2(x,\mu_2,\sigma,\eta)>f_1(x,\mu_1,\sigma,\eta)\quad\forall x>\tau \\ f_2(x,\mu_2,\sigma,\eta)<f_1(x,\mu_1,\sigma,\eta)\quad\forall x<\tau \end{cases}\quad\quad\ldots\ldots(*)$

The $\eta$-scaled CDF's of $X_1$ and $X_2$, denoted as $F_1$ and $F_2$, are defined as follows: $\begin{cases} F_1(x,\mu_1,\sigma,\eta)=\frac{\eta}{2}[1+erf(\frac{\ln{x}-\ln{\mu_1}}{\sqrt{2}\sigma})] \\ F_2(x,\mu_2,\sigma,\eta)=\frac{1-\eta}{2}[1+erf(\frac{\ln{x}-\ln{\mu_2}}{\sqrt{2}\sigma})] \end{cases}$, where $erf(z)=\frac{2}{\sqrt\pi}\int_0^z{e^{-t^2}dt}$

By (*),

$\begin{align} OVL=f(\mu_1,\mu_2,\sigma,\eta)&=F_2(\tau,\mu_2,\sigma,\eta)+[1-F_1(\tau,\mu_1,\sigma,\eta)] \\ &=1+F_2(\tau,\mu_2,\sigma,\eta)-F_1(\tau,\mu_1,\sigma,\eta) \end{align}$


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.