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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:

Plotting

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.

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    $\begingroup$ Related Percentage of overlapping regions of two normal distributions $\endgroup$ – user2974951 Oct 15 at 6:03
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    $\begingroup$ 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 at 7:01
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    $\begingroup$ 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 at 16:12
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    $\begingroup$ 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 at 14:00
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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 $\eta\in(0,1)$, $\exists!\tau\in\Bbb{R}^+|f_1=f_2$


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

$\ln{f_2(x,\mu_2,\sigma,\eta)}-\ln{f_1(x,\mu_1,\sigma,\eta)}$

$=\ln\frac{1-\eta}{x\sigma\sqrt{2\pi}}-\ln\frac{\eta}{x\sigma\sqrt{2\pi}}+\frac{(\ln{x}-\ln{\mu_1})^2}{2\sigma^2}-\frac{(\ln{x}-\ln{\mu_2})^2}{2\sigma^2}$

$=\ln\frac{1-\eta}{\eta}+\frac{(\ln{\mu_1})^2-(\ln{\mu_2})^2}{2\sigma^2}+\frac{\ln{\mu_2}-\ln{\mu_1}}{\sigma^2}\ln{x}$, 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 (*),

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

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