Mean Excess Loss Function I have some problems in this question.
Let $X$ be a random variable with mean excess loss function $e(x)=x+1$. Find survival function $S(x)$, pdf $f(x)$, and hazard rate function $h(x)$.
The formula for mean excess loss function is $e(d) = \frac{\int_d^{\infty} S(x) dx}{S(d)}$. But, how can we solve for $S(x)$? I'm stuck here.
 A: Let's begin with a mindless calculation, then consider when it might be applicable.
Multiplying both sides by the denominator $S(d)$ to clear the fraction simplifies the equation.  Convert it to a differential equation by differentiating both sides:
$$\begin{aligned}
S(d) + (1+d)S^\prime(d) &= e^\prime(d)S(d) + e(d) S^\prime(d) \\
&= \frac{\mathrm{d}}{\mathrm{d}d} \left(e(d)S(d)\right)= \frac{\mathrm{d}}{\mathrm{d}d} \int_d S(x)\mathrm{d}x \\
& = - S(d)
\end{aligned}$$
Multiply both sides by the differential element $\mathrm{d}d$ and use basic algebra to rearrange this into a differential
$$\frac{\mathrm{d}S}{S} = \frac{-2\,\mathrm{d}d}{1+d}.$$
Integrate both sides (starting, say, at $d=1$) to obtain
$$\log(S) = -2\log(1+d) + C$$
for some constant of integration $C$.  Exponentiating both sides yields
$$S(d) = \frac{e^C}{(1+d)^2}.$$

Now for the thinking part. This formula is not a survival function, because it does not decrease monotonically from $1$ to $0$ for all $d.$  Information is missing from the question: the given formula for $e$ is not a valid excess loss.
I will guess the formula for $e$ is supposed to hold only on some interval $[x_0, \infty).$  If (for simplicity, and because in many applications variables are non-negative) we assume the support of the random variable is $[0,\infty),$ then the relationship $1=S(0)=e^C$ shows $C=0.$  In this case, necessarily $S(d)=0$ for all $d\lt 0.$
As an example of why this consideration is important, suppose $e(x)=1+x$ only for $x\ge 2.$  Choosing $C$ so that $e^C=9$ gives the solution
$$S(d) = \left\{\matrix{\frac{9}{(1+d)^2}&d\ge 2 \\ 1&\text{otherwise,}}\right.$$
which is perfectly valid: it's the survival function of a Type II Pareto distribution.
