Let L t(, 2 ), > 1. a) In the lectures, we computed the value-at-risk and expected shortfall for normal random variables. Now compute VaR(L) and ES(L) where L t(, 2 ) for some > 1. Where did one need the assumption > 1 (as opposed to > 0)? Hint: Use that ES(L) = E(L | L > VaR); the resulting integral can be computed directly (i.e., you can easily find an antiderivative). b) Show that lim1 ES(L) VaR(L) = 1 and interpret the result from a risk management perspective. Hint: First show that lim1 ES(L) VaR(L) = 1 limx ft (x)(1+x 2/) x 1t(x) . Then, after applying l'Hopital, you should be able to cancel all ft (x) terms