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I've got a equation as follows:

$\mu(\tau)-\mathbb{E}Y=\frac{2\tau-1}{1-\tau}\cdot\int_{\mu(\tau)}^{\infty}(y-\mu(\tau))dF(y)$,

where F is a cdf of random variable Y.

What I need to understand a proof of existence and uniqueness of solution $\mu(\tau)$ to this equation. I quote a proof below:

Let $T_F(\mu)=\int_{\mu(\tau)}^{\infty}(y-\mu(\tau))dF(y)$ and $\alpha(\tau)=(2\tau-1)/(1-\tau)$. As discussed by DeGroot (1970, p.246), $T_F(\mu)$ is a convex function of $\mu$ (and is therefore continuous in $\mu$) and satisfies

(A. 1) $T_F(\mu)\geq \mathbb{E}Y-\mu$, $\quad \lim_{\mu\rightarrow\infty}T_F(\mu)=0$, $\quad\lim_{\mu\rightarrow -\infty}[T_F(\mu)-(\mathbb{E}Y-\mu)]=0$.

Also, for $\tau$ in $(0,1)$, $\alpha(\tau)$ satisfies

$\alpha(\tau)>-1$, $\quad d\alpha(\tau)/d\tau=1/(1-\tau)^2>0$.

It follows that $\mu-m$ is greater (smaller) than $\alpha(\tau)T_F(\mu)$ for $\mu$ large (small) enough, so that a solution to equation (2.7) exists by the intermediate value theorem. Also, any such solution must be unique because the convexity of $T_F(\mu)$ and (A.1) imply that for $\mu'>\mu$, $0 \geq T_F(\mu')-T_F(\mu)\geq-(\mu'-\mu)$ (i.e. $T_F(\mu)$ is monotonic decreasing and has a "slope" of at least $-1$).

The existence of solution seems to be straighforward - because of convexity $T_F(\mu)$ has to be continuous, hence applying Bolzano's theorem to function $h(\mu)=\alpha(\tau)T_F(\mu)+\mathbb{E}Y-\mu(\tau)$ gives us the existence of solution.

But I've got some troubles with understanding that this solution is unique. Ok, I've got convexity and I can see the monotonicity, but what for I need a slope $-1$? How does their combination proof the uniqueness?

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1 Answer 1

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The uniqueness of solution is a consequence of monotonicity of $\mu-\alpha T_F(\mu)$ in $\mu$. For $\alpha\ge0$ the monotonicty is obvious. For $\alpha<0$, suppose $\mu_1>\mu_2$. Then \begin{align} (\mu_1-\mu_2)-\alpha(\tau)(T_F(\mu_1)-T_F(\mu_2))&\ge(\mu_1-\mu_2)-\alpha(\tau)(-(\mu_1-\mu_2))\\ &=(1+\alpha)(\mu_1-\mu_2)\\ &>0 \end{align}

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  • $\begingroup$ Thanks! But in this reasoning we don't need a convexity of $T_F(\mu)$, do we? $\endgroup$
    – Easy_
    Jul 5, 2017 at 11:34
  • $\begingroup$ @Easy_ No, we don't. The convexity of $T_F(\mu)$ was used to prove that the "slope" of $T_F(\mu)$ is at least $-1$, as you quoted. $\endgroup$ Jul 5, 2017 at 11:47
  • $\begingroup$ One last thing, I can see that because of convexity $T_F(\mu_1)-T_F(\mu_2)\geq T_F(\mu_1-\mu_2)$ which is because of (A. 1) greater or equal to $\mathbb{E}Y-(\mu_1-\mu_2)$, but it isn't exactly what I need. How precisely convexity gives us a "slope" of at least -1? $\endgroup$
    – Easy_
    Jul 5, 2017 at 14:59
  • $\begingroup$ @Easy_ By the property of convex functions, $$\frac{T_F(\mu_1)-T_F(\mu_2)}{\mu_1-\mu_2}\ge\lim_{\lambda\to-\infty}\frac{T_F(\mu_1)-T_F(\lambda)}{\mu_1-\lambda}$$ $\endgroup$ Jul 5, 2017 at 15:15
  • $\begingroup$ @Easy_ And now the third property in (A. 1) gives the result. $\endgroup$ Jul 5, 2017 at 15:17

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