pg. 363 Ex. 11

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 \(\sigma^2= \frac{1}{4}\). If \(Y_1 = 100\), estimate the probability that \(Y_{365}\) is

Note that \(E(x) = \mu = 0\), \(\sigma = \frac{1}{2}\), \(n = 364\).

This can be modeled as a Gaussian random walk of 364 steps. The probability distribution is given by the normal curve: \(N(100, n\sigma^2) = N(100,91)\), since the starting point of the price is 100 USD.

We can use pnorm to calculate the probabilities.

(a) \(\le\) 100.

1 - pnorm(100,100,91)
## [1] 0.5
normalPlot(100,91,c(100,Inf))

There is a 50% chance the stock price will equal or exceed it’s starting value of $100 after a year.

(b) \(\le\) 110.

1 - pnorm(110,100,91)
## [1] 0.4562483
normalPlot(100,91,c(110,Inf))

There is a 45.6% chance that the stock price will increase to $110 or more after a year..

(c) \(\le\) 120.

1 - pnorm(120,100,91)
## [1] 0.4130212
normalPlot(100,91,c(120,Inf))

There is a 41.3% chance that the stock price will increase to $120 or more after a year.

Binomial Distribution - Moment Generating Function

Following example 10.2 in the text:

\[ g(t) = \sum_{j=0}^n e^{tj} {{n}\choose{j}}p^jq^{n-j} \\ g(t) = \sum_{j=0}^n {{n}\choose{j}}(pe^t)^jq^{n-j} \\ g(t) = (pe^t+q)^n \\ g'(t) = n(pe^t+q)^{n-1}pe^t \\ g''(t) = n(n-1)(pe^t+q)(pe^t)^2 + n(pe^t+q)^npe^t \\ g'(0) = n(p+q)^{n-1}p =np \\ g''(0) = n(n-1)(p+q)^{n-1}p^2 + n(p+q)^np \\ g''(0) = n(n-1)p^2 + np \\ \boxed{\mu = \mu_1 = g'(0) = np} \\ \sigma^2 = \mu_2-\mu_1^2 = g''(0) - g'(0)^2 \\ \sigma^2 = n(n-1)p^2 + np - (np)^2 \\ \sigma^2 = np[(n-1)p +1 -np] \\ \sigma^2 = np[(np-p) +1 -np] \\ \boxed{\sigma^2 = np[1-p]} \] Note that since \(q = 1 - p\), \(p + q = 1\).

Exponential Distribution - Moment Generating Function

Starting from the text Example 10.16:

\[ g(t) = \int_0^\infty e^{tx}\lambda e^{-\lambda x} dx \\ g(t) = \frac{\lambda e^{(t-\lambda )x}}{t -\lambda}|_0^{\infty} \\ g(t) = \frac{\lambda}{\lambda-t} \\ g'(t) = \frac{\lambda}{(\lambda-t)^2} \\ g'(0) = \frac{\lambda}{\lambda^2} = \frac{1}{\lambda} \\ g''(t) = \frac{2\lambda}{(\lambda-t)^3} \\ g''(0) = \frac{2\lambda}{\lambda^3} = \frac{2}{\lambda^2} \]

As above: \(\mu = g'(0) = \lambda^{-1}\) as expected

\(\sigma^2 = g''(0) - g'(0)^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \lambda^{-2}\), as expected