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Assume that $d$ is a random variable with pdf $f_d(x)$, is the following true?

$$E\left[\frac{1}{d^2}\bigg|d>1\right]\leq E\left[\frac{1}{d^2}\bigg|d>1 \cap d<2\right]$$

My answer is yes, since by additional condition on the right side of above equation we are deleting large $d$ area therefore, removing small $\frac{1}{d^2}$ area. Thus, the average should increase. Is that right? How can I see a mathematical solution?

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    Calculate the conditional pdf's and compare the two (ugly) integrals. – 2017-01-09
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    @SoumikGhosh *Certainly not!* (Shivers at the idea...) – 2017-01-10
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    @Susan Can you show that $$E(d^{-2}\mid d\geqslant2)\leqslant2^{-2} – 2017-01-10

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Begin with: the definition of conditional expectation when given an event, and the linearity of expectation.

$$\begin{align}\mathsf E(d^{-2}\mid 11)\mathsf E(d^{-2}\mid d>1)-\mathsf P(d\geqslant 2)\mathsf E(d^{-2}\mid d\geqslant 2)}{\mathsf P(1

Take it from there.

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    Much more complicated than required. – 2017-01-10