is it possible to show that for a constant $0<k<1$ that $\mathbb{E}[X | X>k] \geq k$? Apologises if this seems obvious, probability theory is not my thing.
2026-03-27 01:43:28.1774575808
Conditional Expectation Greater than Constant
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$$\mathbb{E}[X| X > k] = \int_{-\infty}^{\infty} x f_{X|X>k}(x) dx = \int_{k}^{\infty} x f_{X|X>k}(x) dx \geq k \int_{k}^{\infty} f_{X|X>k}(x) dx = k$$ because the pdf for all values of $x$ when $x \leq k$ under the conditional distribution must be $0$.