Hazal Gunduz
1. Exercise 11 (page 363) The price of one share of stock in the Pilsdorff Beer Company (see Exercise 8.2.12) is given by \(Y_{n}\) on the nth day of the year. Finn observes that the differences \(X_{n} = Y_{n+1} - Y_{n}\) appear to be independent random variables with a common distribution having mean µ = 0 and variance \(σ^2 = 1/4\). If \(Y_{1} = 100\), estimate the probability that \(Y_{365}\) is
(a) ≥ 100.
# a.Greater than or equal to 100
1 - pnorm((100 - 100) / (0.5 * sqrt(365 - 1)))## [1] 0.5P(\(Y_{365}\) ≥ 100) = 0.5
(b) ≥ 110.
# b.Greater than or equal to 110
1 - pnorm((110 - 100) / (0.5 * sqrt(365 - 1)))## [1] 0.1472537P(\(Y_{365}\) ≥ 110) = 0.147
(c) ≥ 120.
# c.Greater than or equal to 120
1 - pnorm((120 - 100) / (0.5 * sqrt(365 - 1)))## [1] 0.01801584P(\(Y_{365}\) ≥ 120) = 0.018
2. Calculate the expected value and variance of the binomial distribution using the moment generating function.
Moment generating function is:
\(M_z(t) = Expected(e^{tx})\)
Binomial distribution function is:
\(P_{x} = {n \choose x} p^x (1 - p)^{n-x}\)
Moment generating function and binomial distribution is:
\(M_z(t) = \sum_{x=0}^{n} e^{tx} * {n\choose x} p^x (1 - p)^{n-x}\)
M(t) = \((pe^{t} + 1 - p)^n\)
M’(t) = \(npe^{t}(pe^{t} - p + 1)^{n-1}\)
M’(0) = \(npe^0(pe^0 - p + 1)^{n-1} = np(p - p + 1)^{n-1} = np(1)^{n-1} = np\)
\(M^n(t) = npe^t(pe^t - p + 1)^{n-2}(npe^t - p + 1)\)
\(M^n(0) = npe^0(pe^0 - p + 1)^{n-2}(npe^0 - p + 1) = np(np - p + 1)\)
Expected binomial distribution is \(np\) and the variance is: \(np(− p + 1)\)
3. Calculate the expected value and variance of the exponential distribution using the moment generating function.
Exponential distribution is:
\(f(x) = λe^{-λx}\)
Moment generating function is:
\(M(t) = \int_{0}^{\inf} e^{tx} * λe^{-λx} = - λ / t - λ\)
Expected value is:
M′(t) = \(λ / (t - λ)^2\)
M′(0) = 1
\(M^n(t) = - 2λ / (t - λ)^3\)
\(M^n(0) = - 2λ / (- λ)^3 = 2 / λ^2\)
The expected value is \(1 / λ\) and the variance is \(2 / λ^2\)