1. 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 \(σ^2 = 1/4\). If \(Y_1 = 100\), estimate the probability that \(Y_365\) is

\(F(x) = p(X <= x)\)

n <- 365
y1 <- 100
var <- n * .25
sd <- (sqrt(var))
mean <- 0
  1. \(\geq\) 100
a = pnorm(100-100, mean, sd)
a
## [1] 0.5
  1. \(\geq\) 110
b = 1- pnorm(110-100, mean, sd)
b
## [1] 0.1475849
  1. \(\geq\) 120
c = 1-pnorm(120-100, mean, sd)
c
## [1] 0.01814355
  1. Calculate the expected value and variance of the binomial distribution using the moment generating function.

reference= https://onlinecourses.science.psu.edu/stat414/node/72

Moment Generating Function: \[g(t) = E(e^{tX}) = \sum\limits_{k=0}^\infty\frac{\mu_k*t^k}{k!} = E (\sum\limits_{k=0}^\infty\frac{X^ktk}{k!}) = \sum\limits_{j=1}^\infty e^{tx_j}p(x_j)\]

expected value: The rth moment about the origin can be found by evaluating the rth derivative of the moment-generating function at t = 0. That is:

\(M^{(r)}(0)=E(X^r)\)

variance: The variance of X can be found by evaluating the first and second derivatives of the moment-generating function at t = 0. That is:

\(σ^2=E(X^2)−[E(X)]^2=M″(0)−[M′(0)]^2\)

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

probability density function: \(\lambda e ^{-\lambda x}\)

moment generating function for the probability density function: \(g(t) = \frac{\lambda}{\lambda - 1}\) for \(t < \lambda\)

I’m sort of lost from here!