Q 9.3.11: The price of one share of stock in the Pilsdorff Beer Company (see Exercise 8.2.12) is given by \(Y_n\) on the 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 \(\sigma^2 = \frac{1}{4}\). If \(Y_1 = 100\), estimate the probability that \(Y_{365}\) is
\[ S_n ^* = \frac{S_n - n \mu}{\sigma \sqrt{n}} \]
\[ \begin{align} P(X_{364} \geq 100) &= 1 - P(X_{364} < 100) \\ &= 1 - P(Z < \frac{100 - 100 - 0(1/4)}{(1/2) \sqrt{364}}) \\ &= 1 - \phi(0) & \text{using z-table} \\ &= 1- 0.5 \\ &= 0.5 \end{align} \]
\[ \begin{align} P(X_{364} \geq 110) &= 1 - P(X_{364} < 110) \\ &= 1 - P(Z < \frac{110 - 100 - 0(1/4)}{(1/2) \sqrt{364}}) \\ &= 1 - \phi(1.05) & \text{using z-table} \\ &= 1- 0.8531 \\ &= 0.1469 \end{align} \]
\[ \begin{align} P(X_{364} \geq 120) &= 1 - P(X_{364} < 120) \\ &= 1 - P(Z < \frac{120 - 100 - 0(1/4)}{(1/2) \sqrt{364}}) \\ &= 1 - \phi(2.09) & \text{using z-table} \\ &= 1- 0.9817 \\ &= 0.0183 \end{align} \]
Calculate the expected value and variance of the binomial distribution using the moment generating function.
We know that the binomial distribution is discrete, so we will be using the discrete case from the moment generation piecewise functions.
\[ \begin{align} M_X(t) &= E[e^{tx}] \\ &= \sum_{k=0}^{n} e^{tx} \binom{n}{k} p^k (1-p)^{n-k} \\ &= \sum_{k=0}^{n} \binom{n}{k} (pe^t)^k (1-p)^{n-k} \\ &= (pe^t + 1 -p)^n & \text{due to binomial expansion} \end{align} \]
Now we have our mgf for binomial. We then calculate the first moment by taking the prime of \(M_X(t)\) and the second moment by taking the second derivative of the mgf.
\[ M_X^{\prime}(t) = \frac{d}{dt} M_X(t) = n(pe^t +1-p)^{n-1} (pe^t) \\ M_X^{\prime \prime}(t) = \frac{d}{dt} M_X^{\prime}(t) = n(n-1)(pe^t + 1-p)^{n-2}(pe^t)^2 + n(pe^t + 1-p)^{n-1}(pe^t) \] So, our first and second moment when \(t=0\),
$$ \[\begin{align} E[X] = M_X ^{\prime}(0) &= n(pe^0 +1-p)^{n-1}(pe^0) \\ &= np \\ \\ E[X^2] = M_X^{\prime \prime}(0) &= n(n-1)(pe^0 + 1-p)^{n-2}(pe^0)^2 + n(pe^0 + 1-p)^{n-1}(pe^0) \\ &= n(n-1)p^2 +np \\ \\ Var(X) &= E[X^2] - (E[X])^2 \\ &= n(n-1)p^2 +np -n^2p^2 \\ &= np(1-p) \end{align}\] $$
Calculate the expected value and variance of the exponential distribution using the moment generating function.
For the continuous distribution,we will the use continuous case from the moment generating piecewise function.
\[ \begin{align} M_X(t) &= E[e^{tx}] \\ &= \int_0 ^\infty e^{tx} \lambda e^{-\lambda x} \ dx \\ &= \lambda \int_0 ^\infty e^{-(\lambda-t)x} \ dx & \text{apply u-sub} \\ &= \frac{\lambda}{\lambda - t}, \ \text{for}\ t < \lambda \\ \end{align} \]
Applying the quotient rule,
\[ \begin{align} M_X^{\prime}(t) &= \frac{\lambda}{(\lambda -t)^2} \\ M_X^{\prime \prime}(t) &= \frac{2 \lambda}{(\lambda -t)^3} \end{align} \]
Then, let \(t = 0\),
\[ \begin{align} E[X] &= M_X^{\prime}(0) = \frac{\lambda}{\lambda^2} = \lambda \\ Var(X) &= E[X^2]-(E[X])^2 \\ &= M_x^{\prime \prime}(0) - (M_X^{\prime}(0))^2 \\ &= \frac{2 \lambda}{\lambda^3} - \frac{1}{\lambda^2} \\ &= \frac{2 \lambda^3 - \lambda^3}{\lambda^5} \\ &= \frac{1}{\lambda^2} \end{align} \] # Resources:
Used the Z-Table for the phi functions.