# Conditional expectation with a given property

Let $$f$$ and $$F$$ denote the probability density and distribution of a random variable $$X$$. Assume that $$f$$ is twice continuously differentiable. Letting $$g(k) \equiv \mathbb{E}(X|_{X>k})$$, I'm interested in finding distributions (or a class of distributions) for which $$g^{\prime \prime}(k)>0$$.

I've started by assuming that $$X$$ follows a distribution from the Pearson family (sufficiently general for my particular application). Letting $$\lambda(k)\equiv \frac{f(k)}{\overline{F}(k)}$$ denote the inverse Mills ratio,

$$\mathbb{E}(X|_{X>k})=\frac{ \left(-b_0-b_1 k-b_2 k^2\right)}{2 b_2-a_1}\lambda(k)+\mathbb{E}(X)$$ (see Inverse Mills ratio for non normal distributions.)

Thus: $$g^{\prime \prime}(k)=\frac{ -2b_2}{2 b_2-a_1}\lambda(k) + \frac{ -2b_1 -4b_2 k}{2 b_2-a_1}\lambda^{\prime}(k) +\frac{ \left(-b_0-b_1 k-b_2 k^2\right)}{2 b_2-a_1}\lambda^{\prime\prime}(k)$$

I haven't been able to draw any conclusion. Restricting attention to the Pearson class, can you find a general characterization of the distributions with the property $$g^{\prime \prime}(k)>0$$?

• An approach. For an exponential distribution with parameter 1, Jul 14, 2019 at 2:40

An approach. For an exponential distribution with parameter 1, $$g(k)=k+1$$ (for $$k\ge 0$$), so $$g''(k)=0$$. To get $$g''(k)\gt 0$$, you need a density function decaying slower than exponential.
• $g^{\prime\prime}(k)$ has the same sign as $[1-F(k)][f(k)+kf^{\prime}(k)]+f^{\prime}\int_k^{+\infty}x f(x) \mathrm{d} x + 2f(k)g(k)$. But I still could not see sufficiency in terms of the pdf decay. What am I missing? Jul 15, 2019 at 13:31
• For large $x,\ f'(x) \lt 0$. My answer seems to require something like $f'(x) \gt -e^{-x}$. But I haven't looked at in detail. In any case, the negative term in $g''(k)$ would be smaller? Jul 15, 2019 at 21:17