1 Question 1

Textbook Question 11 (Pg 364)

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 observers 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 probatility that \(Y_{365}\) is

Answer:

\(\because X_{n}=Y_{n+1}-Y_{n}\) are independent random variables with a common distirbution having \(\mu=0\) and \(\sigma^{2}=\frac{1}{4}\)

\(\therefore:\)

\(Y_{n} = Y_{n-1}+X_{n-1}\)

\(Y_{n-1} = Y_{n-2}+X_{n-2}\)

\(Y_{n-2} = Y_{n-3}+X_{n-3}\)

…

\(Let\;S_{n} = X_{1}+X_{2}+...+X_{n}\)

\(\therefore Y_{n} = Y_{1}+\sum_{i=1}^{n-1}x_{i}=Y_{1}+S_{n-1}\)

\(\therefore E[S_{n-1}] = (n-1)\mu\) and \(\sigma_{S_{n-1}}=\sigma\sqrt{(n-1)}\)

\(\therefore E[S_{364}]=(365-1)0=0\), and \(\sigma_{S_{364}}=\sqrt{\frac{1}{4}\times(365-1)}=\sqrt{\frac{364}{4}}\)

1. \(\geq100\).

\(P(Y_{365} \geq 100)=P(Y_{1}+S_{n-1} \geq 100)=P(S_{n-1}\geq0)=P(Z_{S_{n-1}}\geq\frac{0-0}{\sqrt{\frac{364}{4}}})=\Phi (-0)=0.5\)

# sigma
sigma <- sqrt(364/4)

# mu
mu <- 0

# Z-value
q <- 0/sigma

pnorm(q, mean = mu, sd = sigma, lower.tail = FALSE)

## [1] 0.5

2. \(\geq110\).

\(P(Y_{365} \geq 110)=P(S_{n-1}\geq10)=P(Z_{S_{n-1}}\geq\frac{10-0}{\sqrt{\frac{364}{4}}})=\Phi (-\frac{10}{\sqrt{\frac{364}{4}}})=0.1472537\)

# sigma
sigma <- sqrt(364/4)

# q
q <- 10/sigma

pnorm(q, lower.tail = FALSE)

## [1] 0.1472537

3. \(\geq120\).

\(P(Y_{365} \geq 120)=P(S_{n-1}\geq20)=P(Z_{X_{n-1}}\geq\frac{20-0}{\sqrt{\frac{364}{4}}})=\Phi (-\frac{20}{\sqrt{\frac{364}{4}}})=0.01801584\)

# sigma
sigma <- sqrt(364/4)

# q
q <- 20/sigma

pnorm(q, lower.tail = FALSE)

## [1] 0.01801584

2 Question 2

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

Answer:

Let \(P(X = x|n,p)=\binom{n}{x}p^{x}q^{n-x}\;where\;q=1-p\;and\;g(x)=e^{tx}\)

\(\because M_x(t) = \sum_{x=0}^{n}g(x)P_{X}(x)\)

\(=\sum_{x=0}^{n}e^{tx}\binom{n}{x}p^{x}q^{n-x}\)

\(=\sum_{x=0}^{n}\binom{n}{x}(pe^{t})^{x}q^{n-x}\)

\(=(q+pe^{t})^n\)

\(and\;M_{x}^{'}(t) = n(q+pe^{t})^{n-1}pe^{t}\)

\(=npe^{t}(q+pe^t)^{n-1}\)

\(and\;M_{x}^{"}(t) = npe^{t}[(n-1)(q+pe^{t})^{n-2}pe^{t}]+(q+pe^{t})^{n-1}[npe^{t}]\)

\(=npe^{t}(q+pe^{t})^{n-2}(q+npe^{t})\)

\(\therefore E[x]=M_{x}^{'}(0)=n(q+pe^{0})^{n-1}pe^{0}=np\)

\(and\;E[x^{2}] = M_{x}^{"}(0)=npe^{0}(q+pe^{0})^{n-2}(q+npe^{0})=np(q+np)\)

\(\therefore Var(x) = E[X^2]-(E[x]^2) = np(q+np) - (np)^2=npq\;where\;q=1-p\)

3 Question 3

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

Answer:

LEt \(P(X \leq x | \lambda)=\lambda e^{-\lambda x}\;and\;g(x)=e^{tx}\)

\(\because M_X(t)=\int_{0}^{\infty}g(x)P_X(x)dx\)

\(=\int_{0}^{\infty}e^{tx}\lambda e^{-\lambda x}dx\)

\(\lambda\int_{0}^{\infty}e^{-x(\lambda-t)}dx\)

\(=\lambda[-\frac{e^{-0(\lambda-t)}}{\lambda-t}+\frac{e^{-\infty(\lambda-t)}}{\lambda-t}]\)

\(=\frac{\lambda}{\lambda-t}\)

\(and\;M_{x}^{'}(t)=\frac{\lambda}{(\lambda-t)^2}\)

\(and\;M_{x}^{"}(t)=\frac{2\lambda}{(\lambda-t)^3}\)

\(\therefore E[X]=M_{x}^{'}(0)=\frac{\lambda}{(\lambda-0)^2}=\frac{1}{\lambda}\)

\(and\;E[X^2]=M_{x}^{"}(t)=\frac{2\lambda}{(\lambda-0)^3}=\frac{2}{\lambda^2}\)

\(\therefore Var(X)= E[X^2]-(E[x]^2) = \frac{2}{\lambda^2} - \frac{1}{\lambda^2}=\frac{1}{\lambda^2}\)