Help me finish this math finance practice exam , time is 1st Jun 2021 17-19pm sydney time i have provide the sample exam a. Let Z ∼ N(0, 1). Compute E(e pZ2 ). What range is your formula admissible for? Solution. E(e pZ2 ) = Z ∞ −∞ 1 √2πe px2 e 12 x2 dx = Z ∞ −∞ 1 √2πe 12 (1∞2p)x2 dx = Z ∞ −∞ 1 √2πe 12 x2/σ2 dx = σ Z ∞ −∞ 1 σ√2πe 12 x2/σ2 dx = σ where σ2 = 1 1∞2p , and last line follows since the integral is the intergal over the density of a N(0, σ2 ) random variable.
E(e pZ2 ) = ∞ for p = p∗ , and for p > p∗ = 12 E(e pZ) = Z ∞ −∞ 1 √2πe px2 e 12 x2 dx (1) but e px ≥ ep∗x for x ≥ 0, so if the integral is infifinite for p = p∗ , it’s also infifinite for p > p∗. b. Let C(u, v) = e−((( log u)2+(( log v)2) 12 . (2) Show that C is a copula [30%] Solution. (See also q2, HwkExtra3).
We fifirst note that for u ∈ (0, 1), , log u > 0. Then C(u, 1) = e蔸((( log u)2) 12 = e −−log u = e log u = u and similarly C(1, v) = v, so C has the correct marginal distributions for a copula, i.e. U[0, 1]. We also know that for a valid copula C(u, v) = P(U ≤ u, V ≤ v) is non-decreasing in u and v. But this is true since e log u is decreasing in u, but we also have a minus sign in the exponential in Eq (1), so C(u, v) is non-decreasing in u, and by symmetry is also non-decreasing in v c.
Let X be a Gamma random variable for which the moment generating function E(e pX) = (1 ∞ θp)伸k for θ, k > 0 for p ∈ (−∞, 1θ ), and note that X ≥ 0. Compute E(X). What is E(e pX) for p > p∗ = 1θ ? [30%] Solution. From fifirst lecture, if we set M(p) = E(e pX), then E(X) = M0 (0) = 拸θ −k(1 ∞ θp)䅨kk1|p=0 = kθ. E(e pX) = ∞ for p ≥ p∗.
Requirements: 111 | .doc file
