Suggest edit — Jensen's Inequality

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title: "Jensen's Inequality"
source: https://www.jemoka.com/posts/kbhjensen_s_inequality/
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linear edition if \(f\) is convex, then for \(x,y \in \text{dom }f, 0 \leq \theta \leq 1\), then:
\begin{equation} f\qty(\theta x + \qty(1-\theta) y) \leq \theta f\qty(x) + \qty(1-\theta) f\qty(y) \end{equation}
probabilistic extension Let \(f\) be a convex function; that is, \(f&rsquo;&rsquo;\qty(x) \geq 0\); let \(x\) be a random variable. Then, \(f\qty(\mathbb{E}[x]) \leq \mathbb{E}\qty [f\qty(x)]\).
Further, if \(f\) is strictly convex, that is \(f&rsquo;&rsquo;\qty(x) &gt; 0\), then \(\mathbb{E}\qty [f\qty(x)] = f\qty(\mathbb{E}[x])\), that is, \(x\) is constant.
the basic case is thus \begin{equation} P\qty(z= x) = \theta , P\qty(z=y) = 1-\theta \end{equation}
concave edition Let \(f\) be a concave function; that is, \(f&rsquo;&rsquo;\qty(x) \leq 0\); let \(x\) be a random variable. Then, \(f\qty(\mathbb{E}[x]) \geq \mathbb{E}\qty [f\qty(x)]\).
Further, if \(f\) is strictly concave, that is \(f&rsquo;&rsquo;\qty(x) &lt; 0\), then \(\mathbb{E}\qty [f\qty(x)] = f\qty(\mathbb{E}[x])\), that is, \(x\) is constant.