The price of one share of stock in the Pilsdorff Beer Company (see Exer- cise 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 \(\sigma^2 = 1/4\). If \(Y_1 = 100\), estimate the probability that \(Y_{365}\) is
\(\geq 100\).
\(\geq 110\).
\(\geq 120\).
First, lets’ calculate the probability that \(Y_{365} \geq 100\):
Given: \(X_n = Y_{n+1} − Y_n\) we can solve for \(Y_{n+1}\):
\(\Rightarrow Y_{n+1} = Y_n + X_n\) which means that if \(n+1=365\) then \(n=364\).
SO: The mean of \(Y_{365}\) is: \(\mu_{(Y_{365})} = Y_1 + n \mu = 100 +364 \times 0 = 100\)
The variance of \(Y_{365}\) is: \(\sigma^2_{(Y_{365})} = n \sigma^2 = 364 \times \frac{1}{4} = 91\)
The standard deviation of \(Y_{365}\) is: \(\sigma_{(Y_{365})} = \sqrt{91} \approx 9.54\)
Now, we are going to use these values to convert each of the given values (100, 110, and 120) to z-scores \(\left(z = \frac{x-\mu}{\sigma}\right)\) and we can use the Normal Distribution Table in Appendix A page 500 from the Probability text.
The value of \(0\) is right in the middle of the curve in page 500, that indicates that there is 50% probability of \(Y_{365} \geq 100\)
\(x=110\) Then, \(\left(z = \frac{110-100}{9.54}\right) = 1.05\)
Based on the table in page 500 (which shows the area under the curve only between 0 and d), the corresponding value of \(1.05\) is \(0.3531\) so the probability that \(Y_{365} \geq 110\) is: \(0.5-0.3531 = 0.1469\) which is about 15%.
\(x=120\) Then, \(\left(z = \frac{120-100}{9.54}\right) = 2.10\)
Based on the table in page 500, the corresponding value of \(2.10\) is \(0.4821\) so the probability that \(Y_{365} \geq 120\) is: \(0.5-0.4821 = 0.0179\) which is about 2%.
Let’s check our answers with this R-Code:
# Loading the necessary library
library(pracma)
# Given data
Y_1 <- 100
n <- 364 # As we're computing the difference until the 365th day
# Compute mean and standard deviation of Y_365
mean_Y_365 <- Y_1
std_dev_Y_365 <- sqrt(n / 4)
# Function to compute cumulative probability using z-score
cumulative_probability <- function(mean, std_dev, x) {
z_score <- (x - mean) / std_dev
return(1 - 0.5 * (1 + erf(z_score / sqrt(2))))
} #erf() function is to calculate the error function value of the given input; the formula is below this code
# (a) Probability that Y_365 is >= 100
prob_a <- cumulative_probability(mean_Y_365, std_dev_Y_365, Y_1)
cat("Probability that Y_365 is >= 100:", prob_a, "\n")## Probability that Y_365 is >= 100: 0.5# (b) Probability that Y_365 is >= 110
prob_b <- cumulative_probability(mean_Y_365, std_dev_Y_365, 110)
cat("Probability that Y_365 is >= 110:", prob_b, "\n")## Probability that Y_365 is >= 110: 0.1472537# (c) Probability that Y_365 is >= 120
prob_c <- cumulative_probability(mean_Y_365, std_dev_Y_365, 120)
cat("Probability that Y_365 is >= 120:", prob_c, "\n")## Probability that Y_365 is >= 120: 0.01801584\(erf(x) =\frac{2}{\sqrt{\pi}} \int _{0}^{x} e^{-t^2} dt\)
Calculate the expected value and variance of the binomial distribution using the moment generating function.
\[ \begin {align} M_x(t) = E(e^{tx}) &= \sum_{x=0}^{n} e^{tx} p(x) \\ &=\sum_{x=0}^{n} e^{tx} \binom{n}{x} p^x(1-p)^{n-x} \end{align} \]
so to find the expected value and the variance, we can take the derivatives of MGF where \(t=0\):
Since we are calculating for the binomial distribution, The MGF is:
\(M_x(t) = (pe^t + q)^n = (pe^t + 1-p)^n\) (Example 10.2 page 367)
Expected Value: \(E(x) = M'_x(0)\) which is the first moment; we use the chain rule:
\(E(x) = M'_x(0) = n(1-p+pe^t)^{n-1} (pe^t) = n \cdot p\)
Variance: \(Var(X) = E(X^2) - (E(X))^2\) is the square of the first moment subtracted from the second moment.
Now, let’s calculate the second moment; we use the product rule \((uv)'=u'v+uv'\) and the chain rule \((u^n)' = n(u^{n-1}) \cdot u'\): where:
\(u=npe^t\) and \(v=(1-p+pe^t)^{n-1}\)
\(u'=npe^t\) and \(v'=(n-1)(1-p+pe^t)^{n-2} \cdot (pe^t)\)
\(\Rightarrow E(X^2) = M''_x(0) = npe^t \cdot \left((1-p+pe^t)^{n-1}\right) +npe^t \cdot \left( (n-1)(1-p+pe^t)^{n-2} \cdot (pe^t)\right)\)
Plug in the \(t=0\), we get: \(M''_x(0)= n \cdot p (1-p+np)\)
\[ \begin{align} \Rightarrow Var(X) &= E(X^2) - (E(X))^2 \\ &= n \cdot p (1-p+np) - (n \cdot p)^2\\ &=np(1-p+np-np)\\ &=np(1-p) \end{align} \]
Calculate the expected value and variance of the exponential distribution using the moment generating function.
\[ \begin {align} M_x(t) = E(e^{tx}) &=\int_0^\infty e^{tx} \lambda e^{- \lambda x} dx \\ &= \int_0^\infty \lambda e^{(t-\lambda)x} dx \\ &=\frac{\lambda}{\lambda - t} \end{align} \]
Expected Value: \(E(x) = M'_x(0)\) which is the first moment; we use the quotient rule \((\frac{u}{v})'=\frac{u'v-uv'}{v^2}\):
\[ \begin{align} M'_x(0) &= \frac{(\lambda-t)0-\lambda(-1)}{(\lambda - t)^2}\\ &=\frac{\lambda}{(\lambda -t)^2}\\ &=\frac{\lambda}{\lambda^2}\\ &=\frac{1}{\lambda} \end{align} \]
Variance: \(Var(X) = E(X^2) - (E(X))^2\) is the square of the first moment subtracted from the second moment.
Now, let’s calculate the second moment; we use the quotient rule and the chain rule:
\[ \begin{align} \Rightarrow E(X^2) = M''_x(0)&= \left(\frac{\lambda}{(\lambda-t)^2}\right)' \\ &= \frac{-2 \lambda (\lambda-t)(-1)}{(\lambda -t)^4} \\ &=\frac{2 \lambda (\lambda -t)}{(\lambda -t)^4}\\ &= \frac{2 \lambda}{(\lambda-t)^3} \end{align} \]
When \(t=0\):
\[ \begin{align} E(X^2) = \frac{2\lambda}{\lambda^3}=\frac{2}{\lambda^2} \end{align} \]
\[ \begin{align} \Rightarrow Var(X) &= E(X^2) - (E(X))^2\\ & =\frac{2}{\lambda^2}-\frac{1}{\lambda^2}\\ &=\frac{1}{\lambda^2} \end{align} \]