The price of one share of stock in the Pilsdoff 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:
By definition, we know that \(S_n = X_1 + X_2 + X_3 + ... + X_n\). We are told that \(X_n = Y_{n+1} - Y_n\), so subbing this in, we get: \[S_n = (Y_2 - Y_1) + (Y_3 - Y_2) + (Y_4 - Y_3)+ ... + (Y_{n+1} - Y_n)\]
Simplifying by canceling out terms, we are left with: \[S_n = Y_{n+1} - Y_1\] Since we are asked to find \(Y_{365}\), we will use \(n=364\) to calculate \(S_{364}=Y_{365}-Y_1\).
Finally, we can substitute this into our equation: \[S_n^* = \frac{S_n - n\mu}{\sqrt{n} \sigma}\]
We’ll define our common variables, \(n\), \(\mu\), \(\sigma^2\), \(\sigma\), and \(Y_1\):
n <- 364
mu <- 0
var <- 1/4
stdev <- sqrt(var)
y1 <- 100y365 <- 100
s364 <- y365-y1
sn1 <- (s364 - (n*mu)) / (sqrt(n) * stdev)
pnorm(sn1, lower.tail=FALSE)## [1] 0.5y365 <- 110
s364 <- y365-y1
sn2 <- (s364 - (n*mu)) / (sqrt(n) * stdev)
pnorm(sn2, lower.tail=FALSE)## [1] 0.1472537y365 <- 120
s364 <- y365-y1
sn3 <- (s364 - (n*mu)) / (sqrt(n) * stdev)
pnorm(sn3, lower.tail=FALSE)## [1] 0.01801584Calculate the expected value and variance of the binomial distribution using the moment generating function.
The moment generating function is:
\[M_x(t) = \Sigma e^{tx}p(x)\] The binomial distribution is:
\[{ n \choose j}p^jq^{n-j}\]
Substituting this in, we get: \[M_x(t) = \Sigma e^{tx} \cdot { n \choose j}p^jq^{n-j}\] This further evaluates to:
\[M_x(t) = (pe^{t}+q)^n \]
We also know that:
\[E(x) = \mu_1 = M_x'(0)\] and \[\sigma^2 = \mu_2 - \mu_1^2 = M_x''(0) - M_x'(0)^2\]
We know that the first derivative of the moment generating function evaluated at \(t=0\) is the expected value, so we calculate this as: \[M'_x(t) = n(pe^t+q)^{n-1}\cdot pe^t\] Evaluated at \(t=0\): \[E(x)= \mu_1 = npe^0(pe^0+q)^{n-1}\] \[= np(p+q)^{n-1}\] We know that \(p+q=1\), so: \[ E(x)= \mu_1 = np\]
By definition, we know that the variance is:
\[\sigma^2 = \mu_2 − \mu_1^2\]
Let’s first calculate \(M''_x(t)\):
\[M''_x(t) = \mu_2 = [npe^t(pe^t+q)^{n-1}]+[npe^t \cdot (n-1)(pe^t+q)^{n-2} \cdot pe^t]\]
Evaluated at \(t=0\):
\[= [npe^0(pe^0+q)^{n-1}]+[npe^0 \cdot (n-1)(pe^0+q)^{n-2} \cdot pe^0]\] \[= [np(p+q)^{n-1}]+[np \cdot (n-1)(p+q)^{n-2} \cdot p]\] Once again, we know that \(p+q=1\), so this simplifies to: \[\mu_2 = np+ np^2(n-1)\] Finally, subbing in for our variance equation \(\sigma^2 = \mu_2 − \mu_1^2\):
\[\sigma^2 = np+ np^2(n-1) - (np)^2\] \[= np+ n^2p^2 - np^2 - n^2p^2\]
\[\sigma^2= np- np^2\]
Calculate the expected value and variance of the exponential distribution using the moment generating function.
We will start out by creating the moment function:
\[M_x(t) = \int_0^\infty e^{tx}(\lambda e^{-\lambda x})dx\]
\[= \lambda \int_0^\infty e^{x(t -\lambda)}dx\] When \(\lambda < t\): \[= \lambda [\frac{1}{t-\lambda}e^{-\infty}-\frac{1}{t-\lambda}e^{0}]\]
\[M_x(t)= \frac{-\lambda}{t-\lambda} = \frac{\lambda}{\lambda - t} \]
\[M_x'(t) = \frac{0( \lambda - t) - \lambda(-1)}{(\lambda - t)(\lambda - t)}\] \[= \frac{\lambda}{\lambda^2 - 2 \lambda t + t^2}\]
Evaluated at \(t=0\): \[E(x) = \mu_1 = \frac{\lambda}{\lambda^2} = \frac{1}{\lambda}\]
\[M_x''(t) =\mu_2 = \frac{0(\lambda^2 - 2 \lambda t + t^2) - \lambda (-2 \lambda + 2t)}{(\lambda^2 - 2 \lambda t + t^2)(\lambda^2 - 2 \lambda t + t^2)}\]
\[ = \frac{ 2\lambda^2 - 2 \lambda t}{\lambda^4 - 2 \lambda^3 t + \lambda^2t^2 - 2 \lambda^3 t + 4\lambda^2 t^2 - 2\lambda t^3 + t^2 \lambda^2 - 2 \lambda t^3 + t^4}\]
\[ = \frac{ 2\lambda(\lambda - t)}{\lambda^4 + t^4 - 4 \lambda^3 t + 6\lambda^2t^2 - 4 \lambda t^3}\]
Evaluated at \(t=0\): \[ = \frac{ 2\lambda^2}{\lambda^4} = \frac{2}{\lambda^2}\]
Finally, subbing in for our variance equation \(\sigma^2 = \mu_2 − \mu_1^2\): \[\sigma^2 = \frac{2}{\lambda^2} - (\frac{1}{\lambda})^2\]
\[= \frac{2}{\lambda^2} - \frac{1}{\lambda^2} \]
\[\sigma^2 = \frac{1}{\lambda^2}\]