Download Review, Study Q`s for Expected Utility I realize that I`ve made some

Review, Study Q's for Expected Utility
I realize that I've made some assumptions which may not be reasonable given the prerequisites about
your level of understanding of probability. Here are some general rules for probability:
The probability of an event, expressed as fraction or decimal form, represents the number of chances in
100 that the event will occur. i.e. If probability is p = 0.45, that means that there 45/100 times the event
will occur. This is generally the format we will use but its possible that you may see a ¼ probability
somewhere. That's just that 1 out of every 4 times or 25 out of 100 the event will come about.
If all possible outcomes are represented, their probabilities must add up to 1 or 100%. So, for example,
flipping a coin, the probability of heads is 0.50. That means the probability of tails is 1-0.50 = 0.50 as
well. More complex random events, such as rolling a 6 sided die, have the same property. So you have
a 1/6 or 0.1667 probability of rolling a '6' (or any other single number).
OR Events
If we're interested in a range of possible outcomes, so long as their probabilities are independent
(meaning that the likelihood of one outcome doesn't in any way depend on the other occurring), then
the probability of one or the other being the outcome is determined by adding up their probabilities.
Thus, the probability of either a 5 OR a 6 coming up on the six sided die is 1/6 + 1/6 = 2/6 or 1/3.
AND Events
If instead we are interested in knowing how likely multiple sequences of independent events will come
about, we determine that by multiplying the individual probabilities together. So the probability of
rolling two sixes is the product of the two probabilities: 1/6 x 1/6 = 1/36 .
Now recall that the expected value of an event is the sum of the payoffs from each possible outcome,
weighted by (multiplied by) the probability of that outcome. So if I offer you $100 if a coin flip comes
up heads and -$10 if it comes up tails, the expected value is 0.5x100 + 0.5x(-10) = $45.
What about expected utility?
Well, if an individual is risk neutral, that means their optimal choice is directly determined by the
expected value of the payoffs. So E(U) = U(E(π)) . In words, the expected utility is just the utility of
the expected value. However, if the individual has any other risk attitude, the expected utility is
determined by the weighted sum of the utilities from each of the possible outcomes, where the weights
are determined by the probability of that outcome.
Take for example the square root utility function, which describes a risk averse individual,
U (x) =
x
Let's look at the expected utility for this individual
from the coin flip gamble mentioned above.
E(U) = 0.5 x U(100) + 0.5 x (U(-10)) = 0.5 x 10 + 0.5 x (-3.16) = 6.84. The more risk averse the
individual, the more curved the utility function (smaller utility difference as x gets higher).
For a risk loving or seeking individual, the utility function will be something like U(x) = x2
the effect would be the opposite, making riskier propositions with lower expected values have higher
expected utility (try it).
Problems:
1. Solve for both the expected value and expected utility with each of these three utility functions:
U(x) =  x , U(x) = x, U(x) = x2 for the following pairs of lotteries.
Solutions
(a note on notation, x^(1/2) is equivalent to the square root of x and x^2 is x squared.
Left
Right
0.
E(x) =
$6
0.8 * $8 = $6.40
U(x) = x^(1/2)
U(x) = x^2
1. E(x) =
U(x) = x^(1/2)
E(U) = 6^(1/2) = 2.45
E(U) = 6^2 = 36
$2
E(U) = 2^(1/2) = 1.41
E(U) = 0.8 * (8^(1/2)) + 0 * 0^(1/2) = 2.26
E(U) = 0.8 * (8^2) + 0* 0^2 = 38.4
0.8 * $8 = $6.40
E(U) = 0.8 * (4^(1/2)) + (0.2) * -(4^(1/2)) = 1.2
U(x) = x^2
2. E(x) =
U(x) = x^(1/2)
U(x) = x^2
3. E(x) =
U(x) = x^(1/2)
U(x) = x^2
4. E(x) =
U(x) = x^(1/2)
U(x) = x^2
5. E(x) =
U(x) = x^(1/2)
U(x) = x^2
E(U) = 2^2 = 4
E(U) = 0.8 * (4^2) + 0.2* -(4^2) = 12.8 – 3.2 = 9.6
.25*6 + .75*0= $1.5
0.2 * $8 + 0.8 * 0= $1.6
E(U) = 0.25*(6^(1/2)) = 0.6125
E(U) = 0.25*6^2 = 9
$30
E(U) = 30^(1/2) = 5.48
E(U) = 30^2 = 900
E(U) = 0.2 * (8^(1/2)) + (0.8) * 0^(1/2) = 0.566
E(U) = 0.2 * (8^2) + 0.8* 0^2= 9.6
0.8 * $40 = $32
E(U) = 0.8 * (40^(1/2)) + (0.2) * 0^(1/2) = 5.06
E(U) = 0.8 * (40^2) + 0.2* (0^2) = 1280
0.5*4 + 0.5*3.2 = $3.6
0.5 * 7.7 + 0.5 * $0.2 = $3.95 = 1.41
E(U) = 0.5*(4^(1/2)) + 0.5*(3.2^(1/2))= 1.89 E(U)=0.5 * (7.7^(1/2)) +
(0.5) * (0.2^(1/2)) = 1.61
E(U) = 0.5*(4^(2)) + 0.5*(3.2^(2))= 10.24
-$6
E(U) = -(6^(1/2)) = -2.45
E(U) =- (6^2) = -36
E(U)=0.5*(7.7^(2)) +
(0.5) * (0.2^(2)) =59.31
0.8 * -$8 = -$6.40
E(U) = 0.8 * -(8^(1/2)) + 0 * 0^(1/2) = -2.26
E(U) = 0.8 * -(8^2) + 0* 0^2 = -38.4
6.E(x) =
0.5*2 + 0.5*1.2 = $1.6
U(x) = x^(1/2)
E(U) = 0.5*(2^(1/2)) + 0.5*(1.2^(1/2) = 1.25
U(x) = x^2
7. E(x) =
0.5*5.7 + 0.5*-1.8 = $1.95
E(U) = 0.5* (2^2) + 0.5*(1.2^2) = 2.72
.25*-6 + .75*0= -$1.5
E(U) = 0.5 * (5.7^(1/2)) +
0.5*-(1.8^(1/2)) = 0.522
E(U) = 0.5 * (5.7^2) + 0.5* -(1.8^2)
=14.625
0.2 * -$8 + 0.8 * 0= -$1.6
U(x) = x^(1/2)
E(U) = 0.25*-(6^(1/2)) = -0.6125 E(U) = 0.2*-(8^(1/2)) + (0.8) * 0^(1/2) = -0.566
U(x) = x^2
E(U) = 0.25*-(6^2) = - 9
E(U) = 0.2 * -(8^2) + 0.8* 0^2= -9.6
8. E(x) =
0.5*-4 + 0.5*-3.2 = -$3.6
0.5 * -7.7 + 0.5 * -$0.2 = -$3.95
U(x) = x^(1/2)
E(U) = 0.5*-(4^(1/2)) + 0.5*-(3.2^(1/2))= -1.89 E(U)=0.5 * -(7.7^(1/2))
0.5* -(0.2^(1/2)) = -1.61
U(x) = x^2
E(U) = 0.5*-(4^(2)) + 0.5*-(3.2^(2))= -10.24
E(U)=0.5*-(7.7^(2)) +
0.5* -(0.2^(2)) = -59.31
So, now try to find pairs of choices which would demonstrate the certainty effect. How about the
reflection effect?
Which choice would a risk neutral individual make in each case? Risk averse (with U(x) = x^1/2).
A note on expected utility calculations. Clearly any loss needs to be calculated as potential loss of
utility. This can be confusing if you see that U(x) = x^1/2 but then try calculating the square root of -4
(which doesn't exist in real numbers). The key here is that the utility loss(or disutility) is determined by
the magnitude of the lost payment. So -4 would lead to -(U(4)) or -(4^(1/2))= -2, right?