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I would like to derive the Cumulative Distribution Function of a random variable ($Z$) which is a linear combination of two log normal random variables with different parameters. i.e.

$Z = aX +bY$ ; $X \sim \text{Lognormal}({\mu_1},{\sigma_1^2})$ and $Y \sim \text{Lognormal}({\mu_2},{\sigma_2^2})$ ; where, $a$ and $b$ are positive constants

$f_X (x) = \frac{1}{x \sigma \sqrt{2 \pi}} e^{\frac{-(\ln(x)-\mu)^2}{2\sigma^2}}$

$f_Y (y) = \frac{1}{y \sigma \sqrt{2 \pi}} e^{\frac{-(\ln(y)-\mu)^2}{2\sigma^2}}$

Please note that $X$ and $Y$ are independent.

In wikipedia, I only found the solutions for $X*Y$ or $aX$. How do I approach these kind of problems in general ? Any references would be appreciated.

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2 Answers 2

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I'll solve a similar one, using convolution: $$ Z=X+Y \Rightarrow X=Z-Y,\\ X\sim LogNormal(m_1, s_1),\\ Y \sim LogNormal(m_2, s_2) $$

Keep in mind if $x \leq 0$, $f_X(x) = 0$, so we want $x,y: f(x)=f(z-y)>0, f(y)>0 \Rightarrow z-y>0, y<z$. Therefore you get
$$ f_Z(z) = \int_{0}^{z}f_X(z-y)f_Y(y)dy = \int_{0}^{z}\frac{1}{z(z-y)\sqrt{2 \pi s_1s_2}} e^{-\frac{(\log (z-y)-m_1)^2}{2s_1}}e^{-\frac{(\log y-m_2)^2}{2s_2}}dy $$ You will have to rewrite $\frac{1}{z(z-y)}$ in separate fractions, and solve for $y$. Keep in mind you ca do the substitution $$ \int\frac{1}{z-y}e^{-(\log^2(z-y))}dy = -\int e^{-(\log^2(z-y))}d(\log(z-y)) $$

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  • $\begingroup$ Thank you. I am new to stats and I am not able to comprehend how these work in general. So, could you please give me some references to understand these concepts. $\endgroup$ Jul 29, 2020 at 17:26
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    $\begingroup$ Look up convolution of random variables, Fubini's/Tonelli's theorem $\endgroup$
    – Alex
    Jul 29, 2020 at 17:27
  • $\begingroup$ After doing partial fractions, do I need to go with integration by parts, if so, it seems very complex. I tried and didn't lead anywhere $\endgroup$ Aug 3, 2020 at 22:40
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The distribution of $Z$ has no closed-form expression. Thus it is not possible to get the explicit expression for the cumulative distribution function (CDF) of $Z$. But it is possible to find CDF numerically or use an approximation for $Z$ (if approximation is used, then one must be carefull since behaviour of CDF of $Z$ for small values is very different from the behaviour for large values).

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