Central Limit Theorem & Generating Functions:

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 _n_th 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 \(\mu = 0\) and variance \(\sigma = 1 / 4\). If \(Y_1\) = 100, estimate the probability that \(Y_{365}\) is:

  1. \(\ge 100\)

  2. \(\ge 110\)

  3. \(\ge 120\)

Solutions:

# greater than or equal to 100
1 - pnorm((100 - 100) / (0.5 * sqrt(365 - 1)))
## [1] 0.5
# greater than or equal to 110
1 - pnorm((110 - 100) / (0.5 * sqrt(365 - 1)))
## [1] 0.1472537
# greater than or equal to 120
1 - pnorm((120 - 100) / (0.5 * sqrt(365 - 1)))
## [1] 0.01801584

Probability that \(Y_365\) for \(\ge 100\) is 0.5, for \(\ge 110\) it’s 0.147 and for \(\ge 120\) it’s 0.0180.

2. Calculate the expected value and variance of the binomial distribution using the moment generating function.

The moment generating function (MGF) is: \(M_z (t) = Expected(e^{tx})\)

The binomial probability mass function (PMF) is: \(P_x = \binom{n}{x}p^x (1 - p)^{n - x}\)

The combination of MGF and PMF is: \(M_z (t) = \sum_{x = 0}^{n} e^{tx} * \binom{n}{x}p^x (1 - p)^{n - x}\)

Simplified when t is a real number: M(t) = \((pe^t + 1 - p)^n\)

First instant of expected value: M’(t) = \(npe^t(pe^t - p + 1)^{n-1}\)

At 0 we get this expected value: M’(0) = \(npe^0(pe^0 - p + 1)^{n-1} = np(p - p + 1)^{n - 1} = np(1)^{n - 1} = np\)

For the variance:

\(M^n(t) = npe^t(pe^t - p + 1)^{n-2}(npe^t -p +1)\)

\(M^n(0) = npe^0(pe^0 - p + 1)^{n-2}(npe^0 -p +1) = np(np - p + 1)\)

Therefore the expected binomial distribution is np and the variance is np(np - p + 1).

3. Calculate the expected value and variance of the exponential distribution using the moment generating function.

The exponential distribution: \(\lambda e^{-\lambda x}\) when x \(\ge 0\)

\(M(t)=\int^\inf_0e^{tx}*\lambda e^{-\lambda x}=-\frac{\lambda}{t-\lambda}\)

The expected value evaluated at 0: \(M'(t) = \dfrac{\lambda}{(t-\lambda)^2}\)

\(M'(0) = 1\)

\(M^n(t) = -\dfrac{2\lambda}{(t-\lambda)^3}\)

\(M^n(0) = -\dfrac{2\lambda}{(-\lambda)^3} = 2 / \lambda ^2\)

The expected value is \(1 / \lambda\) and the variance is \(2 / \lambda ^2\)