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v primeru neodivisnosti spremenljivk velja enakost
Dokaz
pride direktno iz lastnosti korelacijskega koeficienta primer $E(Y)=E(X)=\mu = 0$$$|r(X,Y)|=\left| \frac{E((X-E(X))(Y-E(Y)))}{\sqrt{E(X^2)-E^2(X)}\cdot \sqrt{E(Y^2)-E^2(Y)}} \right| =\left| \frac{E(XY)}{\sqrt{E(X^2)E(Y^2)}} \right| \leq 1$$
2. Jensenova neenakost
Ce je $g$ konveksna ($g''(x)>0$), potem
$$E(g(X))\geq g(E(X))$$
Ce je $g$ konkavna potem:
$$E(g(X))\leq g(E(X))$$
Dokaz
$$g(x) \geq a+bx, \text{ (Vsaka tangenta je pod krivuljo (konveksnost))}$$$$g(X) \geq a+bX$$$$E(g(X))\geq E(a+bX)=a+bE(X)=a+b\mu=g(\mu)=g(E(X))$$