Assignment 9

p363: #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 σ 2 = 1/4. If Y1 = 100, estimate the probability that Y365 is (a) ≥ 100. (b) ≥ 110. (c) ≥ 120.

n <- 365
PBC_mean <- 100
PBC_variance <- 1/4 * n
PBC_sd <-sqrt(PBC_variance)



1 - pnorm(100,PBC_mean,PBC_sd)
## [1] 0.5
1 - pnorm(110,PBC_mean,PBC_sd)
## [1] 0.1475849
1 - pnorm(120,PBC_mean,PBC_sd)
## [1] 0.01814355

Using this as guidance:

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

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

\(M(t) = \sum_{x=0^n}(pe^t)^x C(n,x)>p^x(1-p)^{n-x}\)

Combining terms:

\(M(t) = [(1-p)]+pe^t]^n\)

First derivative:

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

Expected value is at t=0

$ M(0) = n(pe0)[(1-p) + pe^0]{n-1} $

$ M(0) = np$

Variance is t=0 at second derivative

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

\(M''(0) = n(n-1)p^2 + np\)

\(\sigma^2 = np(1-p)\)

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

moment generating function for exponential distribution is

\(Mx(t) = \frac{1}{1-\beta t} for t < \frac{1}{\beta}\)

from here

I’m a bit lost in taking the derivatives and next steps here… Derivative would be:

$E(X) = M’x(0) = $