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Suppose X is a realvalued random variable and let P X denote the distribution of X Then E ( X − c ) = ∫ R x − c d P X ( x) The medians of X are defined as any number m ∈ R such that P ( X ≤ m) ≥ 1 2 and P ( X ≥ m) ≥ 1 2 Why do the medians solveProof lnexy = xy = lnex lney = ln(ex ·ey) Since lnx is onetoone, then exy = ex ·ey 1 = e0 = ex(−x) = ex ·e−x ⇒ e−x = 1 ex ex−y = ex(−y) = ex ·e−y = ex 1 ey ex ey • For r = m ∈ N, emx = e z }m { x···x = z }m { ex ···ex = (ex)m • For r = 1 n, n ∈ N and n 6= 0, ex = e n n x = e 1 nx n ⇒ e n x = (ex) 1 • For r rational, let r = m n, m, n ∈ NE g(X) ≥ g( EX ) Source Wolfram MathWorld a) Proof Let g(X) be −√$, for all positive real numbers g(x) is a convex function as seen by the graph to the right Since g(X) satisfies the requirements for Jensen's inequality, we can use it to directly prove the inequality

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