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
Since \(X_n = Y_{n+1} − Y_n\), we can also say \(Y_{n+1} = X_n + Y_n\), or in the case of year 1 and year 2, \(Y_2 = X_1 + Y_1\).
Accordingly, \(Y_{365} = Y_1 + X_1 + X_2 + X_3...X_{364}\).
Because we know \(Y_1 = 100\) and \(X_1, X_2, X_3...X_{365}\) have \(\mu = 0\), we know that the expected value of \(Y_1\) is \(100 + 0 + 0 + 0...0 = 100\).
Likewise, the variance of \(Y_1, X_1, X_2, X_3...X_{364}\) is \(\frac{1}{4}\) for each item. Therefore, the variance for \(Y_{365} = 365 \times \frac{1}{4} = 91.25\). Therefore, the standard deviation \(= \sqrt{91.25} = 9.552\). Using \(\mu = 0\) and \(\sigma = \sqrt{91.25}\)
(a) ≥ 100.
pnorm(100 - 100, mean = 0, sd = sqrt(91.25), lower.tail = FALSE)## [1] 0.5(b) ≥ 110.
pnorm(110 - 100, mean = 0, sd = sqrt(91.25), lower.tail = FALSE)## [1] 0.1475849(c) ≥ 120.
pnorm(120 - 100, mean = 0, sd = sqrt(91.25), lower.tail = FALSE)## [1] 0.01814355The probability mass function for the binomial distribution is:
\(\binom{n}{k}p^k(1 - p)^{n-k}\)
Therefore, the moment generating function for the binomial distribution is:
\(g(t) = \sum\limits_{j=1}^{n}\binom{n}{k}p^k(1 - p)^{n-k}e^{tj}\) \(= (1 - p + pe^t)^n\)
Calculating the first derivative:
\(g'(t) = npe^t(pe^t - p + 1)^{n-1}\)
Evaluating for t = 0 to determine the mean \(\mu_1\):
\(g'(0) = n \times p \times e^0(p \times e^0 - p + 1)^{n-1} = np\)
Calculating the second derivative:
\(g''(t) = npe^t(pe^t - p + 1)^{n-2}(npe^t - p + 1)\)
Evaluating for t = 0 to determine \(\mu_2\):
\(g''(0) = npe^0(pe^0 - p + 1)^{n-2}(npe^0 - p + 1) = np(np + 1 - p)\)
The variance is \(\mu_2 - \mu_1^2 = np(np + 1 - p)np = n^2p^2(np + 1 - p)\)
The probability density function for the exponential distribution is:
\(\lambda e^{-\lambda x}\)
Therefore, the moment generating function for the binomial distribution is:
\(g(t) = \frac{\lambda}{\lambda - t}\) for \(t < \lambda\)
Calculating the first derivative:
\(g'(t) = \frac{\lambda}{(\lambda - t)^2}\)
Evaluating for t = 0 to determine the mean \(\mu_1\):
\(g'(0) = \frac{1}{\lambda}\)
Calculating the second derivative:
\(g''(t) = \frac{2 \lambda}{(\lambda-t)^3}\)
Evaluating for t = 0 to determine \(\mu_2\):
\(g''(0) = \frac{2}{\lambda^2}\)
The variance is \(\mu_2 - \mu_1^2 = \frac{2}{\lambda^2} - (\frac{1}{\lambda})^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}\)