11 P. 363

11 The price of one share of stock in the Pilsdorff Beer Company (see Exercise 8.2.12) is given by Yn on the nth day of the year. Finn observes that the differences Xn = Yn+1 − Yn appear to be independent random variables with a common distribution having mean μ = 0 and variance \(\sigma^2 = 1/4\). If \(Y_1 = 100\), estimate the probability that \(Y_{365}\) is:

  1. \(\ge 100 = P\left(Z\ge \frac{100-100}{\sqrt{(365-1)\sigma^2}}\right)=0.5\).

  2. \(\ge 110= P\left(Z\ge \frac{110-100}{\sqrt{(365-1)\sigma^2}}\right)=0.01472\).

  3. \(\ge 120= P\left(Z\ge \frac{120-100}{\sqrt{(365-1)\sigma^2}}\right)=0.0180\).

We suppose that Xn is normal distrubuted

x <- c(100, 110, 120)
for (i in 0:length(x)){
  mu <- 0
  dx <- x[i]-100
  n <- 365-1
  q <- dx/sqrt(n)
  sd <- sqrt(1/4)
  p <- pnorm(q, mu, sd, lower.tail = FALSE)
  print(p)
}
## numeric(0)
## [1] 0.5
## [1] 0.1472537
## [1] 0.01801584
  1. Binomial distribution function X with parameter n and p.

\(M_X(t)=E[e^{Xt}]\\ =\sum_{k=0}^{n}e^{tk}\begin{pmatrix} n\\ k \end{pmatrix}p^k(1-p)^{}n-k\\ =\sum_{k=0}^{n}(pe^t)^k\begin{pmatrix} n\\ k \end{pmatrix}(1-p)^{}n-k\\ = (pe^t+1-p)^n\)

The mean \(\mu=E(X)=M_{X}^{'}(0)=\frac{d}{dt}M_X(t)|_{t=0}=n(pe^t+1-p)^{n-1}pe^t|_{t=0}=np\)

\(M_{X}^{''}(0)=\frac{d^2}{dt^2}M_X(t)|_{t=0}=\frac{d}{dt}M_X^{'}(t)|_{t=0}\\ = n(n-1)(pe^t+1-p)^{n-2}(pe^t)^2+n(pe^t+1-p)^{n-1}pe^t|_{t=0}\)

\(E(X^2)=M_{X}^{''}(0)=n(n-1)p^2+np\)

The variance is \(\sigma^2= E(X^2)-E^{2}(X)=n(n-1)p^2+np-n^2p^2=np(1-p)\)

  1. Exponential distribution with parameter \(\lambda\). \(M_X(t)=E[e^{Xt}]\\ =\int_{0}^{+\infty}e^{tx}\lambda e^{-\lambda x}dx\\ =\lambda\int_{0}^{+\infty}e^{(t-\lambda)x}dx\\ = \frac{\lambda}{\lambda-t} \quad for \quad t<\lambda\)

The mean \(\mu=E(X)=M_{X}^{'}(0)=\frac{d}{dt}M_X(t)|_{t=0}=\frac{\lambda}{(\lambda-t)^2}|_{t=0}=\frac{1}{\lambda}\)

\(E(X^2)=M_{X}^{''}(0)=\frac{d^2}{dt^2}M_X(t)|_{t=0}=\frac{d}{dt}M_X^{'}(t)|_{t=0}=\frac{2\lambda}{(\lambda-t)^3}|_{t=0}=\frac{2}{\lambda^2}\)

The variance is \(\sigma^2= E(X^2)-E^{2}(X)=\frac{1}{\lambda^2}\) Source: A First Course in Probability. Sheldom Ross.