Problem 11 on page 363.

11. The price of one share of stock in the Pilsdorff Beer Company (see Exercise8.2.12) 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 \(\sigma^2 = 1/4\). If \(Y_1 = 100\) estimate the probability that \(Y_{365}\) is:-

\(Y_{365}−Y_1=X_1+X_2..X_{364}\)

\(\mu = 0\)

\(\sigma^2 = 1/4\)

(a) \(\geq\) 100.

## [1] 0.5

(b) \(\geq\) 110.

## [1] 0.1472537

(c) \(\geq\) 120.

## [1] 0.01801584

2. Calculate the expected value and variance of the binomial distribution using the moment generating function.

The moment generating function for a binomial random variable X:

\(M(t)=[(1-p) +pe^t]^n\).

Calculating the first Derivative of M(t) with respect to t, we get

\(M'(t) = n(pe^t)[(1-p)+pe^t]^{n-1}\)

Setting t = 0

\(M'(0) = n [ 1 - p + p ]^{n-1} p(1) = np\)

Expected value:- \[ E(X) = \mu = M'(0) = np \]

To find the variance, we first need to take the second derivative of M(t) with respect to t. Doing so, we get:

\(M''(t) = n[1 - p + pe^t]^{n-1}(pe^t)+ (pe^t) n(n-1)[1 - p + pe^t]^{n-2}(pe^t)\)

Setting t = 0

\(M''(0) = np + n^2p^2 -np^2\)

Variance: = \(M''(0) = n(n-1)p^2 + np\) = \(\sigma^2 = M''(0) - [M'(0)]^2 = np(1-p)\)

or,

\(\sigma^2 = np(1-p)\)

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

The exponential distribution function is given as: \(f(x)= \lambda e^{−\lambda x}\)

Moment generating function for the distribution is:

\(M_X(t)=E(e^{tx})=\int ^∞_0e^{tx}f(x)dx\)

\(=E(e^{tx})=\int ^∞_0e^{tx}λe^{−λx}dx\)

\(=\lambda \int^∞_0e^{tx}e^{−λx}dx\)

\(=\lambda \int^∞_0e^{(t−λ)x}dx\)

\(=\lambda \int ^∞_0e^{(t−λ)x}dx\)

So, \(M_X(t)=\frac {\lambda}{\lambda−t}\)

For t=0,

The first derivative

\(M_X'(0) = E(x)= \frac {\lambda}{(\lambda -0)^2} = \frac{1}{\lambda}\) = \(\mu_1\) = expected value E(x)

The second derivative For t = 0,

\(M_X''(0) = E(X^2)= \frac {2\lambda}{(\lambda-t)^3}\)

\(M_X''(0) = E(X^2)= \frac{2}{\lambda^2}\)

\(V(X)=E(X^2)−E(X)^2=\frac { 2 }{ \lambda ^{ 2 } } -\frac { 1 }{ \lambda ^{ 2 } }=\frac{1}{\lambda^2}\) = V(X) = Variance = \(\sigma^2\)