Question 1

Exercise 11 page 363

The price of one share of stock in the Pilsdorff Beer Company (see Exercise 8.2.12) is given by \(Y_n\) on the nth 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 (a) \(\ge 100\). (b) \(\ge 110\). (c) \(\ge 120\).

if \(X_n = Y_{n+1} - Y_n\) , then \(Y_{365} = Y_{364} + X_{364}\) \[Therefore, \\ Y_{n+1} = Y_1 + \Sigma^{n=364}_{n=1} X_n\]

so If \(Y_1 = 100\) and \(\mu_{X_i} =\), then: \[ E[Y_{n+1}] = E[Y_1] + \Sigma^{n=364}_{n=1} E[X_n] =E[Y_1] = 100 \]

Also,

\[ {\sigma}_{Y_{365}}^2 = 365 \cdot \frac{1}{4} = 91.25 \]

Ans. (a) \(\ge 100\).

Prob_a <- pnorm(q = 100, mean = 100,sd = sqrt(91.25), lower.tail = FALSE)

Prob_a
## [1] 0.5

The probability that \(Y_{365}\) when \(\ge 100\) is 0.5

  1. \(\ge 110\).
Prob_b <- round(pnorm(q = 110, mean = 100, sd = sqrt(91.25), lower.tail = FALSE),4)

Prob_b
## [1] 0.1476

The probability that \(Y_{365}\) when \(\ge 110\) is 0.1476

  1. \(\ge 120\).
Prob_c <- round(pnorm(q = 120, mean = 100, sqrt(91.25), lower.tail = FALSE),4)

Prob_c
## [1] 0.0181

The probability that \(Y_{365}\) when \(\ge 120\) is 0.0181

Question 2

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

The probability density function for the exponential distribution is \(p_X(X=k) = {n \choose k} p^{k}q^{n-k}\), where \(q = 1-p\).

The moment generating function is: \[ g(t) = \sum^{n}_{j=0} e^{tj}p_X(j) \] \[ = \sum^{n}_{k=0} e^{tk}p_X(X=k) \] \[ = \sum^{n}_{k=0} e^{tk}{n \choose k} p^{k}q^{n-k} \] \[ = \sum^{n}_{k=0} {n \choose k}(pe^t)^{k}q^{n-k} \]

In order to find the expected value and variance we will need to calculate \(g'(0)\) and \(g''(0)\).

\[ g'(t) = npe^t(pe^t +q)^{n-1} \] \[E(X) = \mu_1 = g'(0) = np(p+q) = np \]

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

\[\mu_2 = g''(0) = n(n-1)p^2 +np = np(np -p +1)\]

\[\sigma^2 = \mu_2-\mu_1^2 = np(np -p +1) - (np)^2\] \[\sigma^2 = np(1-p)=npq\]

Question 3

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

The probability density function for the exponential distribution is \(f_X(x) = \lambda e^{- \lambda x}\), for \(x > 0\).