Question

Consider lottery X∼ distributed as (−10,1/3;0,1/3;+10,1/3) and 'white noise' ε∼ distributed as (−5,1/2;+5,1/2). First generate lottery Y∼ by attaching white noise to the worst outcome of X∼ (i.e., -10) and then generate lottery Z∼ by attaching white noise instead to the best outcome of X∼( i.e., −10). (a). Compute E(Y∼),E(Z∼),var(Y∼) and var(Z∼) (b). Draw the cumulative distributions of Y∼ and Z∼. Can you say that one is riskier than the other? Why or why not? (c). If a decision maker has a quadratic utility such as u(w)=w−0.01w2 compute E[u(Y∼)] and E[u(Z∼)]. Are you surprised by the fact that E[u(Y∼)]=E[u(Z∼)] ? (d). Choose a utility function such that u′′′>0 then show that E[u(Y∼)]

Answer 1

(a) E(Y∼) = -3.33, E(Z∼) = 3.33, var(Y∼) = 8.89, var(Z∼) = 8.89. (b) Cumulative distributions of Y∼ and Z∼ would be similar as their means and variances are identical. It's not possible to determine which one is riskier based solely on mean and variance. (c) E[u(Y∼)] = 33.26, E[u(Z∼)] = 33.26. It’s not surprising that E[u(Y∼)] = E[u(Z∼)] due to identical means. (d) For u''' > 0, E[u(Y∼)] would be increasing due to the convexity of the utility function.

Step-by-step explanation:

(a) The expected value of Y∼ is calculated by adding the mean of the 'white noise' (-3.33) to the worst outcome of X∼ (-10), resulting in -3.33. Similarly, the expected value of Z∼ is determined by adding the mean of the 'white noise' (3.33) to the best outcome of X∼ (+10), resulting in 3.33. The variance for both Y∼ and Z∼ is 8.89, as the variance remains the same when adding a constant to a random variable.

(b) The cumulative distributions of Y∼ and Z∼ might not provide conclusive evidence on which is riskier since their means and variances are identical. Although one attaches noise to the worst outcome and the other to the best, these metrics alone don't determine risk. Further analysis beyond mean and variance is needed to ascertain riskiness.

(c) With a quadratic utility function u(w) = w - 0.01w^2, both E[u(Y∼)] and E[u(Z∼)] are calculated to be 33.26. This similarity is anticipated due to the identical means of Y∼ and Z∼, leading to the same expected utility.

(d) For a utility function with u''' > 0, the expected utility, E[u(Y∼)], would be increasing because of the function's convexity. Convexity implies a higher risk aversion, influencing decision-making by favoring more certainty.

Answer 2

To compute E(Y) and var(Y), add the mean of the white noise distribution to the worst outcome of X and compute variances of the distributions. For E(Z) and var(Z), attach white noise to the best outcome of X and compute variances of the distributions.

Step-by-step explanation:

In this question, we are asked to compute the expected value (E) and variance (var) of the random variables Y and Z. The random variables Y and Z are generated by attaching white noise to the worst and best outcomes of the lottery X, respectively.

To compute E(Y), we add the mean of the white noise distribution to the worst outcome of X and multiply it by the probability of that outcome occurring. In this case, we have E(Y) = (-10 + (-5)) * 1/3 = -15/3 = -5.

Similarly, to compute var(Y), we first find the variance of the white noise distribution, which is (5 - (-5))^2 * 1/2 = 100/2 = 50. Then we add the variance of the white noise distribution to the variance of the worst outcome of X, which is (10 - (-10))^2 * 1/3 = 400/3. So var(Y) = 400/3 + 50 = 450/3 = 150/1 = 150.

To compute E(Z) and var(Z), we follow the same steps as above, except we attach the white noise to the best outcome of X. So E(Z) = (10 + (-5)) * 1/3 = 5/3 and var(Z) = 400/3 + 50 = 150.