I have the following probability density function
$$ {f_X}\left( x \right) = \frac{{{a_2} - {a_1}}}{2}erf\left( {\frac{x}{{b }}} \right) + \frac{{{a_1}}}{2}erf\left( {\frac{{x + 1}}{{ b }}} \right) - \frac{{{a_2}}}{2}erf\left( {\frac{{x - 1}}{{ b }}} \right). $$
Given specific values of $a_1$, $a_2$, and $b$, by using Matlab, I observed that $f_X (x)$ can be well approximated by a Gaussian PDF
$$ {f_G}\left( x \right) = \frac{1}{{\sqrt {2\pi \delta _G^2} }}{{\mathop{\rm e}\nolimits} ^{ - \frac{{{{\left( {x - {\mu _G}} \right)}^2}}}{{2\delta _G^2}}}}. $$
What I'm trying to do is to find $\mu_G$ and $\delta_G^2$ as functions of $a_1$, $a_2$, and $b$.
=================
My efforts: I used the series representation of the error function $$ erf\left( x \right) = \frac{2}{{\sqrt \pi }}\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{n!\left( {2n + 1} \right)}}} $$ to obtain $$ {f_X}\left( x \right) = \frac{1}{{\sqrt \pi }}\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{n!\left( {2n + 1} \right){b^{2n + 1}}}}\left[ {\left( {{a_2} - {a_1}} \right){x^{2n + 1}} + {a_1}{{\left( {x + 1} \right)}^{2n + 1}} - {a_2}{{\left( {x - 1} \right)}^{2n + 1}}} \right]}. $$ Then by using the binomial expansion, I further obtained $$ {f_X}\left( x \right) = \frac{1}{{\sqrt \pi }}\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{n!\left( {2n + 1} \right){b^{2n + 1}}}}\sum\limits_{k = 0}^{2n} {\left( {_k^{2n + 1}} \right){x^k}\left\{ {{a_1} + {{\left( { - 1} \right)}^{2n - k}}{a_2}} \right\}} } \\ \;\;\; = \frac{1}{{\sqrt \pi }}\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{n!{b^{2n + 1}}}}\sum\limits_{k = 0}^{2n} {\left( {_k^{2n}} \right)\frac{{{x^k}\left( {{a_1} + {{\left( { - 1} \right)}^k}{a_2}} \right)}}{{\left( {2n + 1 - k} \right)}}} }. \;\;\;\;\;\;\;\;\;\; (1) $$ On the other hand, the Gaussian pdf can be rewritten as $$ \begin{array}{l} {f_G}\left( x \right) = \frac{1}{{\sqrt {2\pi \delta _G^2} }}\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{n!{{\left( {2\delta _G^2} \right)}^n}}}{{\left( {x - {\mu _G}} \right)}^2}} \\ \;\;\;\;\;\;\;\;\;\; = \frac{1}{{\sqrt \pi }}\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{n!{2^{n + 1/2}}\delta _G^{2n + 1}}}\sum\limits_{k = 0}^{2n} {\left( {_k^{2n}} \right){x^k}{{\left( { - {\mu _G}} \right)}^{2n - k}}} } \\ \;\;\;\;\;\;\;\;\;\; = \frac{1}{{\sqrt \pi }}\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{n!{{\left( {\sqrt 2 {\delta _G}} \right)}^{2n + 1}}}}\sum\limits_{k = 0}^{2n} {\left( {_k^{2n}} \right){x^k}{{\left( { - 1} \right)}^k}\mu _G^{2n - k}} }. \;\;\;\;\;\;\;\;\;\; (2) \end{array} $$ Although (1) and (2) have similar forms, I failed to find relations between $\delta_G^2$, $\mu_G$ and $a_1$, $a_2$, and $b$.
Any suggestions are appreciated! Thank you and best regards.
After playing around with the above summations without any results, I realize that we can simply derive the mean and variance of X via $f_X(x)$.
By using Mathematica, we obtain the mean and variance as follows:
where $a = a_2$, $b = a_1$, and $c = b$ in the original post.