The price of one share of stock in the Pilsdorff Beer Company (see Exercise 8.2.12) is given by Yn on the nth day of the year. Finn observes that the differences Xn = Yn+1 − Yn appear to be independent random variables

with a common distribution having mean µ = 0 and variance σ 2 = 1/4. If Y1 = 100, estimate the probability that Y365 is

  1. ≥ 100.
  2. ≥ 110.
  3. ≥ 120.

Answer:

using Central limit theorem Daily price Yn of a common stock is independent variable. The sigma of daily price Xn =Yn+1 - Yn is a random variable with mean 0 and sd 0.5.
use confidence interval Yn−Y1ϵ(nE(X)−z∗δn−−√,nE(X)+z∗δn−−√) to find the density. Given μ= 0 and δ2= 1/4, z is a critical value for normal distribution of the price change Yn=Y1+{ i=1 }^{ 365-1 }{ { x }{ i } } E(x)= =0 { i=1 }^{ 365-1 }{ { x }{ i } } z= = =

### In (a) Yn>= 100, n=365-1 
1-pnorm((100-100)/(0.5*sqrt(365-1)))
## [1] 0.5
### In (b) Yn>= 110, n=365-1
1-pnorm((110-100)/(0.5*sqrt(365-1)))
## [1] 0.1472537
### In (c) Yn>= 120, n=365-1
1-pnorm((120-100)/(0.5*sqrt(365-1)))
## [1] 0.01801584

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

Answer: Let p be a probability of success and 1-p is probability of failure in binomial distribution, thus

forjn,Xin{ 0,1,2,3,…,n}

Monment generating function: g(t)=E({ e }^{ tX })={ j=1 }^{ }{ { e }^{ t{ x }{ j } }p({ x }_{ j }) }

So we get g(t)=_{ j=0 }^{ n }{ { e }^{ tj }( \[\begin{matrix} n \\ j \end{matrix}\] ){ p }^{ j }{ q }^{ n-j } } =_{ j=0 }^{ n }{ ( \[\begin{matrix} n \\ j \end{matrix}\]

){ (p{ e }^{ t }) }^{ j }{ q }^{ n-j } } ={ { (p{ e }^{ t }+q) }^{ n } }

expected value is: { }_{ 1 }={ g }^{ ’ }(0)=n{ { (p{ e }^{ t }+q) }^{ n-1 } }p{ e }^{ t }=np(t=0)

variance is: { }_{ 2 }={ g }^{ " }(0)=n(n-1){ { { p }^{ 2 }+np } }=np

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

Answer:

Exponential function: { f }{ X }(x)=^{ -x }ifx(0,);otherwise{ X }(x)=0.

Monment generating function :

{ M }{ X }(t)=E({ e }^{ t }X)={ -}^{ }{ { e }^{ tx }{ f }_{ X }(x)dx. }

${ M }{ X }(t)={ -}^{ }{ { e }^{ tx }{ e }^{ -x }dx=_{ 0 }^{ }{ { e }^{ (t-)x }dx= } } $

Expected value is:

(X)={ M }_{ X }^{ ’ }(0)= =

Variance is:

E({ X }^{ 2 })={ M }_{ X }^{ (2) }(0)= = { }^{ 2 }=E({ X }^{ 2 })-E(X)= - =