Homework Week 9
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 variance \(\sigma^2 = \frac{1}{4}\). If \(Y_1 = 100\), estimate the probability that \(Y_365\) is
- ≥ 100. given the values above, find \(P(\frac{Y_365-Y_1}{\sqrt{365-1}} \geq \frac{0}{\sqrt{364}})\)
y_365_gt <- 100
P_gt_100 <- 1 - pnorm( q = (y_365_gt - y_1)/sqrt(364),
mean = mu, sd = sigma,
lower.tail = TRUE )
paste( 'P =', P_gt_100 )## [1] "P = 0.5"- ≥ 110. given the values above, find \(P(\frac{Y_365-Y_1}{\sqrt{365-1}} \geq \frac{10}{\sqrt{364}})\)
y_365_gt <- 110
P_gt_110 <- 1 - pnorm( q = (y_365_gt - y_1)/sqrt(364),
mean = mu, sd = sigma,
lower.tail = TRUE )
paste( 'P =', P_gt_110 )## [1] "P = 0.147253696840055"- ≥ 120. given the values above, find \(P(\frac{Y_365-Y_1}{\sqrt{365-1}} \geq \frac{20}{\sqrt{364}})\)
y_365_gt <- 120
P_gt_120 <- 1 - pnorm( q = (y_365_gt - y_1)/sqrt(364),
mean = mu, sd = sigma,
lower.tail = TRUE )
paste( 'P =', P_gt_120 )## [1] "P = 0.0180158431091167"2. Calculate the expected value and variance of the binomial distribution using the moment
generating function.
This video describes using the MGF to find the expected value and variance for a binomial distribution which I summarize below:
The Moment Generating function for an binomial random variable is given as: \[M(t) = ( (1-p) + pe^t)^n\] finding the first derivative (moment): \[M'(t) = n\mathfrak{L}^{n-1}(t)pe^t, \mbox{ where } \mathfrak{L} = (1-p + pe^t) \] the expected value is found as: \[M'(0) = npe^0\mathfrak{L}^{n-1}(0) = np\] \[\therefore E(X) = M'(0) = np\] to calculate the variance, need to find the second derivative (moment): \[M''(t) = np(e^t(n-1)\mathfrak{L}^{n-2}\cdot pe^t + \mathfrak{L}^{n-1}e^t)\] now to evaluate… \[M''(0) = np( 1\cdot(n-1)\cdot1\cdot p + 1\cdot1) = n(n-1)p^2 + np\] the variance is calculated as follows: \[var( X ) = M''_X(0) - M'_X(0)^2 = n(n-1)p^2 + np - n^2p^2\] \[=n^2p^2 - np^2 +np - n^2p^2 = np(1-p)\]
3. Calculate the expected value and variance of the exponential distribution using the moment
generating function. This video describes the calculation of the expected value which I summarize below:
The Moment Generating function for an exponential random variable is given as: \[M(t) = \frac{\lambda}{\lambda - t} \dashrightarrow \lambda(\lambda-t)^{-1}, \mbox{ given the condition: } t \lt \lambda\] where \(\lambda =\) the rate parameter
finding the first derivative: \[M'(t) = -1 \cdot \lambda(\lambda-t)^{-2} \cdot -1 = \frac{ \lambda}{(\lambda-t)^2}\] The expected value is then found as: \[E(x) = M'(0) = \frac{\lambda}{(\lambda-0)^2} = \frac{\lambda}{\lambda^2} = \frac{1}{\lambda}\] To calculate the variance, find the second derivative: \[M''(t) = \frac{2\lambda}{(\lambda - t)^3} \vert_{t=0} = \frac{2}{\lambda^2}\] The variance is calculated as follows: \[var( X ) = M''_X(0) - M'_X(0)^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}\]