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I have this (differential) equation:

f(x)*x + F(x) = 1

where f(x) is a pdf and F(x) its corresponding cdf. Or similarly,

F'(x)*x + F(x) = 1

I can calculate x numerically, but I want an analytical solution. Perhaps there is a single analytical (exact and closed) solution in general (that would be great), although I only need it for the normal distribution.

* Added after first answer *

The equation comes from this one:

$$ argmin_x \{ x(F(x)-1) \} $$

I use the following code in R. It is not meant to be efficient. Once I clarify this problem, I will use a better numerical method (e.g. hill climbing if the function is convex or other methods).

optpred_bidloss_normcalc <- function(me, sd) {   RESO <- 100000   MIN <- 0 # It cannot be negative   MAX <- max(me, 0) + sd   # This is an approximation to the maximum    x <- rep(1,RESO)   y <- rep(1,RESO)   for(i in 1:RESO) {     loc <- MIN + (MAX-MIN)*(i-1)/(RESO-1)     x[i] <- loc       y[i] <- loc * (pnorm(loc, mean= me, sd= sd) - 1)   }    imin <- which.min(y)   sol <- x[imin]    sol } 

* Example *

> mymean <- 10 > mysd <- 0.5 > x <- optpred_bidloss_normcalc(mymean, mysd) > x [1] 9.00132 > x* dnorm(x,mean=mymean, sd=mysd) + pnorm(x, mean=mymean,sd=mysd) [1] 1.000015 

1 Answers 1