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If u is strictly concave and 0 < p < 1, then u(E(x)) > E(u(x)). Jensen’s Inequality Derivation for the two-point support case

Jensen's inequality

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Page 1: Jensen's inequality

If u is strictly concave and 0 < p < 1, then u(E(x)) > E(u(x)).

Jensen’s Inequality

Derivation for the two-point support case

Page 2: Jensen's inequality

x is a random variable with realizations x1 and x2, x1 < x2.u is the utility function. u is increasing and strictly concave.p = prob{x = x1}, 1 – p = prob{x = x2}

Model Setup

Page 3: Jensen's inequality

E(x) = px1 + (1-p)x2

E(u(x)) = pu(x1) + (1-p)u(x2)

Expectation of x and

Expectation of u(x)

Page 4: Jensen's inequality

Graph of Model Setup

Page 5: Jensen's inequality

Similar Triangles

Page 6: Jensen's inequality

Larger : Base = x2 - x1 Height = u(x2) – u(x1).

Smaller : Base = E(x) - x1 Height = ?

Base and Height

Page 7: Jensen's inequality

Larger : Base = x2 - x1 Height = u(x2) – u(x1).

Smaller : Base = E(x) - x1 Height = ?

E(x) - x1 = px1 + (1 – p)x2 – x1 = (1 – p)(x2 –x1).

A little bit of algebra

Page 8: Jensen's inequality

Larger : Base = x2 - x1 Height = u(x2) – u(x1).

Smaller : Base = (1 – p)(x2 –x1).

Height = (1 – p)[u(x2) – u(x1)].

Conclusion

Page 9: Jensen's inequality

u(x1) + (1 – p)[u(x2) – u(x1)] =

u(x1) - (1 – p)u(x1) + (1 – p)u(x2) =

pu(x1) + (1 – p)u(x2) = E(u(x))

More Algebra

Page 10: Jensen's inequality

Justification of Labeling