# Find CDF of $aX+bY$; Given that $X \sim \text{Lognormal}(\mu_1,\sigma_1^2)$ and $Y \sim\text{Lognormal}(\mu_2,\sigma_2^2)$

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.

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 $$x,y: f(x)=f(z-y)>0, f(y)>0 \Rightarrow z-y>0, y. Therefore you get
$$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))$$
The distribution of $$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).