N Selina Assignment 9 - Data 605

Noori Selina

Excercise 11 on page 363

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 \(n\)-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

1. ≥ 100. - There is a 50% chance

1-pnorm(100-100,0,9.552487)

## [1] 0.5

2. ≥ 110. - Theres a 14.8% chance

1-pnorm(110-100,0,9.552487)

## [1] 0.1475849

3. ≥ 120 - There is a 1.81% chance

1-pnorm(120-100,0,9.552487)

## [1] 0.01814355

2. Calculate the expected value and variance of the binomial distribution using the moment generating function Given the moment generating function (MGF) for a binomial distribution, as expressed by \(g(t) = (pe^t + q)^n\), where \(p\) represents the probability of success, \(q\) represents the probability of failure, and \(n\) denotes the number of trials.

To calculate the expected value, we differentiate the MGF once with respect to \(t\), yielding \(g'(t) = np e^t (pe^t + q)^{n-1}\).

We then evaluate the first derivative at \(t = 0\) to find the expected value: \(g'(0) = np(p + q)^{n-1}\).

For the variance, we proceed by taking the second derivative of the MGF with respect to \(t\), which gives \(g''(t) = np e^t (pe^t + q)^{n-1} + n(n-1)p^2 e^{2t} (pe^t + q)^{n-2}\).

Then, we evaluate the second derivative at \(t = 0\) to determine the variance: \(g''(0) = np(p + q)^{n-1} + n(n-1)p^2 (p + q)^{n-2}\).

These expressions provide a convenient method to compute the expected value and variance of a binomial distribution using its moment generating function.

3. Calculate the expected value and variance of the exponential distribution using the moment generating function. Calculate the expected value and variance of the exponential distribution using the moment generating function.

Given: \(g(t) = \frac{1}{1 - \beta t}\) for \(t < \frac{1}{\beta}\)

To find the expected value, differentiate the moment generating function \(g(t)\) once with respect to \(t\), yielding \(g'(t) = \beta \left( 1 - \beta t \right)^{-2}\). Evaluate \(g'(0)\) to get the expected value: \[ \text{Expected value} = \beta \]

To find the variance, differentiate the moment generating function \(g(t)\) twice with respect to \(t\), yielding \(g''(t) = 2 \beta^2 \left( 1 - \beta t \right)^{-3}\). Evaluate \(g''(0)\) to get the variance: \[ \text{Variance} = 2\beta^2 \]