How do I solve the following equation for $x$
$$\newcommand{\erf}{\operatorname{erf}}\frac{1}{2}\left((f-1)\cdot\erf\left(\dfrac{c-x}{\sqrt2\,b}\right)-f\erf\left(\dfrac{r-x}{\sqrt2\,d}\right)\right)=y$$
I need this to solve this problem :
I have a distribution, built by the sum of two normal distributions, one multiplied by $(1-f)$ and the other multiplied by $f$. The sum of both is multiplied by $g$. like this (where $0 \leq f < 1$) :
$$\left(\frac{(1-f) e^{-\frac{1}{2} \left(\frac{x-c}{b}\right)^2}}{\sqrt{2 \pi } b}+\frac{f e^{-\frac{1}{2} \left(\frac{x-r}{d}\right)^2}}{\sqrt{2
\pi } d}\right)
$$
I need to know the $x$ where $P(X < x) = y$.
First equation is the cumulative distribution or I did a mistake somewhere?
Answer
If you must do this analytically then you need a convolution. I would not.
If your two normal distributions $X_1\sim \mathcal N(a,b^2)$ and $X_2\sim \mathcal N(c,d^2)$ are independent then an easier approach is to say $fX_1\sim \mathcal N\left(fa,f^2 b^2\right)$ and $(1-f)X_2\sim \mathcal N\left((1-f)c,(1-f)^2d^2\right)$ so $$X=fX_1+(1-f)X_2 \sim \mathcal N\left(fa+(1-f)c,f^2 b^2+(1-f)^2d^2\right)$$ and given the mean and variance of $X$ you can solve $P(X\le x)=y$ in the usual way for a normal distribution.
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