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 \(\mu\) = 0 and variance \(\sigma ^ 2\) = 1/4. If Y1 = 100, estimate the probability that Y365 is
(a) >= 100
(b) >= 110
(c) >= 120

Solution

\((a)\) >= 100 

n <- 365
mu <- 0
y <- 100
mu_y <- mu + y
var <- 1/4 * (n)
sd <- sqrt(var)
x1 <- 100

pnorm(x1, mu_y, sd, lower.tail = F)
## [1] 0.5

\((b)\) >= 110

x2 <- 110
pnorm(x2, mu_y, sd, lower.tail = F)
## [1] 0.1475849

\((c)\) >= 120

x3 <- 120
pnorm(x3, mu_y, sd, lower.tail = F)
## [1] 0.01814355

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

Solution
The binomial function: \(b(x, n, p) = \frac{n!}{x!(n-x)!}p^xq^{n-x}\) ,where \(q = 1-p\). Moment generating function:
\[Mx(t) = \sum_{x=0}^{n} e^{xt} \frac{n!}{x!(n-x)!}p^xq^{n-x} = \sum_{x=0}^{n} (pe^t)^x \frac{n!}{x!(n-x)!}q^{n-x} = (q+pe^t)^n\] Take derivative with respect to t \[\frac{dM(t)}{dt} = pe^t n(q+pe^t)^{n-1}\] Expected value E(X) at t= 0:
\[\frac{dM(t)}{dt} = pe^0 n(q+pe^0)^{n-1} = pn(q + p) ^ {n-1} = pn(1-p+p) ^ {n-1} = np\] \[E(x^2) = np(q + p)^{n−2}(q + np) = np(q + np)\] The variance is V(X)=E(X2)−E(X)2: \[V(X) = E(x^2) - (E(x))^2 = np(q + np) - (np)^2 = np(q+np) - n^2p^2 = npq\] Above equation found, E(X) = np and V(X) = npq.

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

Solution
Exponential distribution:

\[f(x) = \lambda e ^ {-\lambda x}\] Moment generating function:
\[Mx(t) = \int_{-\infty}^\infty e^{tx}fx(x)dx = \int_0^\infty e^{tx}\lambda e^{-\lambda x}dx = \lambda \int_0^\infty e^{(t - \lambda)x}dx = \frac{\lambda}{t- \lambda}\] Expected value E(X), substitute t = 0:
\[E(x) = \frac{\lambda}{(t- \lambda)^2} = \frac{1}{\lambda}\]

The variance is V(X)=E(X2)−E(X)2, substitute t = 0:
\[E(x^2) = \frac{2\lambda}{(t- \lambda)^3} = \frac{2}{\lambda^2}\] \[E(X) ^ 2 = \frac {2\lambda} {(\lambda - 0) ^ 3} - \frac {1} {\lambda ^ 2}\]

\[V(X) = \frac {2\lambda} {\lambda ^ 3} - \frac {1} {\lambda ^ 2} \]

\[V(X) = \frac {2} {\lambda ^ 2} - \frac {1} {\lambda ^ 2} \] \[V(X) = \frac {1} {\lambda ^ 2}\]

Above equation found, E(X) = \(\frac {1} {\lambda}\) and V(X) = \(\frac {1} {\lambda ^ 2}\).