Chapter 9: Page 363

    1. 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 \(\sigma^{2}\) = 1/4. If \(Y_{1}\) = 100, estimate the probability that \(Y_{365}\) is:
\[ X_{n} = (Y_{n+1} - Y_{n}); X_{1} = (Y_{2} - 100) \]
\[ S_{n} = X_{1} + X_{1} + X_{2} +... + X_{n} \]
\[ S_{365} = {(Y_{2} - 100) + (Y_{3} - Y_{2}) + ... + (Y_{365} - Y_{364})}\]
\[ \mu_{365} = 0 \]
\[ \sigma_{365} = \sqrt (1/4 * 365) = 9.5525 \]
  1. \(\geq\) 100.
pnorm((100-100), mean = 0, sd = 9.55, lower.tail = FALSE)
## [1] 0.5
  1. \(\geq\) 110.
pnorm((110-100), mean = 0, sd = 9.55, lower.tail = FALSE)
## [1] 0.147522
  1. \(\geq\) 120.
pnorm((120-100), mean = 0, sd = 9.55, lower.tail = FALSE)
## [1] 0.01811927
  1. Calculate the expected value and variance of the binomial distribution using the moment generating function
Binomial Distribution -> Probability Mass Function

\[ f(x) = C(n,x)p^{x}(1-p)^{n-x}\]

Moment Generating Function

\[ M(t) = \Sigma_{x=o^{n}} e^{tx}C(n,x)p^{x}(1-p)^{n-x}\] \[ M(t) = \Sigma_{x=o^{n}} (pe^{t})^{x}C(n,x)(1-p)^{n-x}\] \[ M(t) = [(1-p)+pe^{t}]^{n} \]

Expected Value calculation
First derivative of the MGF

\[ M'(t) = n(pe^{t})[(1-p) + pe^{t}]^{n-1} \]

Evaluate at t = o

\[ M(o) = n(pe^{o})[(1-p) + pe^{o}]^{n-1} = np \]

Variance calculation
Differentiate the MGF, then evaluate this derivative at t = 0, then find M’’(t)

\[ M''(t) = n(n-1)(pe^{t})^{2}[(1-p) + pe^{t}]^{n-2} + n(pe^{t})[(1-p) + pe^{t}]^{n-1} \] \[ \sigma^{2} = M''(o) - [M'(o)]^{2} = n(n-1)p^{2} + np - (np)^{2} = np(1-p) \]

  1. Calculate the expected value and variance of the exponential distribution using the moment generating function
Exponential Distribution -> Probability Density Function

\[ f(x,\lambda) = \lambda e^{-\lambda x} \]

Moment Generating Function

\[ M(t) = E(e^{tX}) \] \[ M(t) = \int_{0}^{\infty} e^{tx}\lambda e^{-\lambda x}dx \] \[ M(t) = \lambda \int_{0}^{\infty} e^{(t-\lambda)x} dx \] \[ M(t) = \lambda/(\lambda - t) \]

Expected Value Calculation

\[ E(X) \int M(X) dx \]

\[ E(X) = \int \lambda/(\lambda - t) dt \]

\[ E(X) = 1/\lambda \]

Variance Calculation

\[ \sigma^{2} = M''(o) - [M'(o)]^{2} \]

\[ \sigma^{2} = E(X^{2}) - E(X)^{2}) \]

\[ \sigma^{2} = 2/\lambda^{2} - (1/\lambda)^{2} = 1/\lambda^{2} \]