35 by lotus 1 2v 1 e y 2 2v 2 dudv v 1 0 1 0 1 2v

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: D (1) = 0.4, pD (0.99) = 0.1. The CLT pertains to sums of random variables, so we will apply the Gaussian approximation to ln(Y ), which is a sum of a large number of random variables: 365 ln Y = ln Dk . k=1 By LOTUS, the mean and variance of ln(Dk ) are given by µ = E [ln(Dk )] = 0.5 ln(1.01) + 0.4 ln(1) + 0.1 ln(0.99) = 0.00397. 2 σ 2 = Var(Dk ) = E [Dk ] − µ2 = 0.5(ln(1.01))2 + 0.4(ln(1))2 + 0.1(ln(0.99))2 − µ2 = 0.00004384. √ Thus, ln Y is approximately Gaussian with mean 365µ = 1.450 and standard deviation 365σ 2 = 0.127. Therefore, P {Y ≥ c} = P {ln(Y ) ≥ ln(c)} ln(c) − 1.450 ln(Y ) − 1.450 ≥ =P 0.127 0.127 ln(c) − 1.450 ≈Q . 0.127 In particular: (a) P {Y ≥ 3} ≈ Q(−2.77) ≈ 0.997. (b) P {Y ≥ 4} ≈ Q(−0.4965) = 0.69. (c) The median is the value c such that P {Y ≥ c} = 0.5, which by the Gaussian approximation is e1.450 = 4.26. (This is the same result one would get by the following argument, based on the law of large numbers. We expect, during the year, the stock to increase by one...
View Full Document

Ask a homework question - tutors are online