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Suppose $F_X(x)$ is CDF of random variable $X$. I am going to find the value of $x$ such that the following equation is satisfied: $$ ax+bxF_X(x)+ c \int_{x}^{\infty} \left[ 1-F_X(y) \right] \text{d}y = 0 $$

where $a,b,c$ are known parameters.

Is there any way to solve this equation?

Amin
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  • Newton-Raphson. – Kurt G. Feb 15 '24 at 20:16
  • Is it possible to have a simpler expression analytically? – Amin Feb 16 '24 at 00:20
  • Unlikely. When you set $c=0$ and take as $F_X$ the CDF of the exponential distribution we already have a problem that calls in the Lambert $W$-function. – Kurt G. Feb 16 '24 at 08:58
  • What is $c$ is not equal to 0 and parameters are set such that we are sure that there exists a solution to this equation? – Amin Feb 16 '24 at 15:48
  • See my first comment. That method is dead simple to implement, say, in python. Existence of a solution and finding it in closed form are two different things. You should also work a bit yourself and not assume that others know all the answers to every question. – Kurt G. Feb 16 '24 at 17:43

1 Answers1

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If you want it to be true for all $x$, you'll have to solve for $F_X(x)$.

We'll start by differentiating both sides wrt $x$: $$a+bF_X(x)+bx\frac{\text d F_X(x)}{\text dx}-c(1-F_X(x))=0.$$ Rearranging, $$bx\frac{\text dF_X(x)}{\text d x}+(b+c)F_X(x)+a-c=0.$$ So we got ourselves a differential equation in $F_X(x)$ which can be solved via power series and Frobenius method.

Supposing $F_X(x)=\sum_{n\geq 0}a_nx^n$, we'd get $$\sum_{n\geq 0}[b(n+1)+c]a_nx^n=c-a$$ Therefore, for $n=0$ you can find $a_0\neq 0$ and for $n\geq 1$ you'll find that $a_n=0$ so we'll use Frobenius: $$n=0: a_0=\frac{c-a}{b+c}$$ $$n\geq 1: F_X(x)\sim x^r\implies [b(r+1)+c]x^r=0\implies r=-\frac{b+c}{b}$$ Finally, combining both we'd get that the equation holds true $\forall x$ if $$F_X(x)=\frac{c-a}{b+c}+Cx^{-\frac{b+c}{b}}$$

Joan S. Guillamet F.
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