See how to score a lottery. Give each outcome a happiness score, weight it by probability, and add up.
The key: π° money does not pile up happiness at a steady rate. The richer you are, the less one more coin delights you. So the average prize, paid for sure, feels better than the gamble itself. That difference is the Jensen gap.
For example, flip a coin: heads pays 100 won, tails pays nothing. The average prize is 50 won. But offer a sure 40 won instead, and many friends pick the 40. The sure thing is easier on the heart.
A consumer with Bernoulli utility u(x)=xβ evaluates the lottery L paying 150 with probability 1/2 and 50 with probability 1/2.
Compute the expected value, the expected utility, and the certainty equivalent directly, and verify that the Jensen inequality below holds strictly.
E[L]=21ββ 150+21ββ 50=100
U(L)=21β150β+21β50ββ9.659
U(L)β€u(E[L])
A 50-50 lottery over 150 and 50, u(x)=xβ.Show proof
The expected value is E[L]=100. Expected utility is U(L)=21β150β+21β50ββ21β(12.247+7.071)β9.659.
The certainty equivalent c solves u(c)=U(L): cββ9.659, so cβ93.30. The utility of the expected value is u(E[L])=100β=10>9.659=U(L) β strict Jensen for concave u.
JensenΒ gap=u(E[L])βU(L)β10β9.659=0.341>0
The risk premium is E[L]βcβ100β93.30=6.70. In the figure this reads off as the vertical segment above the expected value (the Jensen gap) and the certainty-equivalent marker sitting left of the expected value.
The same lottery L is evaluated by a risk-neutral consumer with Bernoulli utility u(x)=x.
Find the expected utility, the certainty equivalent, and the Jensen gap, and show the risk premium is zero.
U(L)=21ββ 150+21ββ 50=100=E[L]
The same lottery with linear utility u(x)=x β risk neutrality.Show proof
With linear utility, U(L)=21ββ 150+21ββ 50=100=E[L]. The expectation operator is linear, so a linear u passes it straight through.
The certainty equivalent is c=U(L)=100=E[L]. The Jensen gap is u(E[L])βU(L)=100β100=0, and the risk premium E[L]βc=0 β the risk-neutral consumer is indifferent between the lottery and its expected value.
In the figure the chord lies exactly on the curve.