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:
# Define the mean and standard deviation for the sum of differences
mean_sum <- 0 # The mean of the sum is 0 since each X_i has a mean of 0
# The variance of the sum of 364 independent variables, each with a variance of 1/4, is 364 * (1/4) = 91
# Therefore, the standard deviation is the square root of the variance
std_dev_sum <- sqrt(91) # Standard deviation of the sum
# (a) P(Y365 >= 100)
# This is equivalent to P(sum of X_i >= 0)
p_a <- 1 - pnorm(0, mean = mean_sum, sd = std_dev_sum)
# (b) P(Y365 >= 110)
# This is equivalent to P(sum of X_i >= 10)
p_b <- 1 - pnorm(10, mean = mean_sum, sd = std_dev_sum)
# (c) P(Y365 >= 120)
# This is equivalent to P(sum of X_i >= 20)
p_c <- 1 - pnorm(20, mean = mean_sum, sd = std_dev_sum)
# Output the results
print(paste("P(Y365 >= 100):", p_a))## [1] "P(Y365 >= 100): 0.5"print(paste("P(Y365 >= 110):", p_b))## [1] "P(Y365 >= 110): 0.147253696840055"print(paste("P(Y365 >= 120):", p_c))## [1] "P(Y365 >= 120): 0.0180158431091167"Below is an explanation of the above solution
The explanations for the probabilities \(P(Y_{365} \geq 100)\), \(P(Y_{365} \geq 110)\), and \(P(Y_{365} \geq 120)\) are expanded below:
The probability \(P(Y_{365} \geq 100)\) is equivalent to \(P(X_1 + X_2 + \ldots + X_{364} \geq 0)\). Here, the \(X_i\) variables represent the daily stock price changes, which are assumed to be independent and identically distributed with a mean of 0. According to the Central Limit Theorem, the sum of a large number of such variables will approximate a normal distribution. Since the mean of the sum is 0, the distribution is symmetric around 0, and thus the probability of the sum being greater than or equal to 0 is 0.5.
For \(P(Y_{365} \geq 110)\), we calculate the probability \(P(X_1 + X_2 + \ldots + X_{364} \geq 10)\). The sum of the \(X_i\) variables is approximated by a normal distribution with a mean of 0 and a standard deviation of \(\sqrt{91}\) (since the variance of each \(X_i\) is \(1/4\), and there are 364 variables). The normal approximation is used to find the probability that the sum is at least 10, which is computed as \(1 - \text{norm.cdf}\left(\frac{10}{\sqrt{91}}\right)\).
Similarly, \(P(Y_{365} \geq 120)\) is calculated as \(P(X_1 + X_2 + \ldots + X_{364} \geq 20)\). Using the normal approximation for the sum of the \(X_i\) variables, we determine the probability that this sum exceeds 20, which translates to the stock price being at least 120 on the 365th day. This is computed as \(1 - \text{norm.cdf}\left(\frac{20}{\sqrt{91}}\right)\).
These detailed explanations emphasize the application of the Central Limit Theorem and the rationale behind using the normal distribution to approximate the distribution of the sum of a large number of independent, identically distributed random variables.
Calculate the expected value and variance of the binomial distribution using the moment generating function.
From Example 10.2 page 367, the moment-generating function (MGF) for a binomial distribution with parameters \(n\) (number of trials) and \(p\) (success probability) is:
\[M(t) = (pe^t + q)^n\]
where \(q = 1 - p\).
The first moment (expected value) is found by differentiating \(M(t)\) with respect to \(t\) and evaluating at \(t = 0\):
\[E(X) = M'(0) = npe^0 = np\]
The second moment about the origin is \(E(X^2) = M''(0)\). The variance is \(\text{Var}(X) = E(X^2) - [E(X)]^2\):
\[E(X^2) = M''(0) = n(n-1)p^2 + np\]
\[\text{Var}(X) = E(X^2) - [E(X)]^2 = np(1-p) = npq\]
Calculate the expected value and variance of the exponential distribution using the moment generating function
The MGF for an exponential distribution with rate \(\lambda\) is:
\[M(t) = \frac{\lambda}{\lambda - t}\]
for \(t < \lambda\).
The first derivative of \(M(t)\) gives the first moment:
\[M'(t) = \frac{\lambda}{(\lambda - t)^2}\]
Evaluating the first derivative at \(t = 0\) gives:
\[E(X) = M'(0) = \frac{1}{\lambda}\]
The second moment about the origin is found by taking the second derivative of \(M(t)\) and evaluating at \(t = 0\):
\[M''(t) = \frac{2\lambda}{(\lambda - t)^3}\]
\[E(X^2) = M''(0) = \frac{2}{\lambda^2}\]
The variance is \(\text{Var}(X) = E(X^2) - [E(X)]^2\):
\[\text{Var}(X) = \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2 = \frac{1}{\lambda^2}\]