Problem11 page 363

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) \(\geq\) 100.

ANS:

\(Y_{365} - Y_1 = X_1 + X_2 .. X_{364}\)

E[X] = \(\mu\) = 0

\(P(Y_{365} - Y_1 \geq 0)\)

a <- 0/sqrt(364)
sigma <- sqrt(1/4)
1 - pnorm(q = a, mean = 0, sd = sigma, lower.tail = TRUE)
## [1] 0.5
  1. \(\geq\) 110.

\(P(Y_{365} - Y_1 \geq 10)\)

a <- 10/sqrt(364)
sigma <- sqrt(1/4)
1 - pnorm(q = a, mean = 0, sd = sigma, lower.tail = TRUE)
## [1] 0.1472537
  1. \(\geq\) 120.

\(P(Y_{365} - Y_1 \geq 20)\)

a <- 20/sqrt(364)
sigma <- sqrt(1/4)
1 - pnorm(q = a, mean = 0, sd = sigma, lower.tail = TRUE)
## [1] 0.01801584
  1. Calculate the expected value and variance of the binomial distribution using the moment generating function.

MGF

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

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

mean =E(X) = \[\sum x.p(x) = np \]

Variance

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

\[ E(X^2) - [E(X)]^2 \]

#Lets take the values of last set from exercise above.
n=120
p=0.018

m =n*p
m
## [1] 2.16
v= (n^2 * p ) - m^2
v
## [1] 254.5344
  1. Calculate the expected value and variance of the exponential distribution using the moment generating function.

\[f_X(x) = \lambda e^ {-\lambda x} I_{(0,\infty}) (x)\]

\[ Mx(t) = \lambda / ( \lambda - t) \] \[E(X) = 1 / \lambda\]

\[E(X^2) = 2/ \lambda ^2 \]

\[E(X) - [E(X)]^2 = 1/\lambda ^2 \]