restart eq := erf(x)=erf(Pi); solve(eq, x) assuming x::real Ly1JJGVyZkc2JCUqcHJvdGVjdGVkR0koX3N5c2xpYkc2IjYjSSJ4R0YoLUYkNiNJI1BpR0Ym LUknUm9vdE9mRzYkJSpwcm90ZWN0ZWRHSShfc3lzbGliRzYiNiMsJi1JJGVyZkdGJDYjSSNfWkdGJCIiIi1GKzYjSSNQaUdGJSEiIg== # Where does this equality come from? # # I results from a mapping of continuous random variable X # (Exponential(1) distributed in the illustration above) to a # standard gaussian random variable G. # # Illustration with(Statistics): G := RandomVariable(Normal(0, 1)): X := RandomVariable(Exponential(1)): g := Quantile(G, Probability(X <= x)) assuming x::positive KiYtSSdSb290T2ZHNiQlKnByb3RlY3RlZEdJKF9zeXNsaWJHNiI2IywoKiYtSSRlcmZHRiU2I0kjX1pHRiUiIiItSSRleHBHRiU2I0kieEdGKEYwRjBGMSEiIiIiI0YwRjBGNiNGMEY2 # Let us assume now that X is already a standard gaussian # random variable with(Statistics): X := RandomVariable(Normal(0, 1)): g := Quantile(G, Probability(X <= x)) KiYtSSdSb290T2ZHNiQlKnByb3RlY3RlZEdJKF9zeXNsaWJHNiI2IywmLUkkZXJmR0YlNiNJI19aR0YlIiIiLUYsNiMsJComSSJ4R0YoRi8iIiMjRi9GNUY2ISIiRi9GNUY2 # Obviously 'g' = x L0kiZ0c2IkkieEdGJA== # How can I "simplify" to get ?