Week 9, CLT & Gen Fnct

page 363 Q 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

\(a)\quad \ge 100\)

set.seed(123)
y<-replicate(10000,{
    y_365<-sum(rnorm(364,0,0.5))+100})
sum(y>=100)/10000
## [1] 0.4927

\(b)\quad \ge 110\)

set.seed(123)
y<-replicate(10000,{
    y_365<-sum(rnorm(364,0,0.5))+100})
sum(y>=110)/10000
## [1] 0.1424

\(c)\quad \ge 120\)

set.seed(123)
y<-replicate(10000,{
    y_365<-sum(rnorm(364,0,0.5))+100})
sum(y>=120)/10000
## [1] 0.0191
  1. Calculate the expected value and variance of the binomial distribution using the moment generating function.

\({ p }_{ x }(k)\quad =\quad \left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }{ { \left( 1-p \right) } }^{ n-k }\quad \quad (0\le \quad k\quad \le n)\\ Discrete\quad MGF\quad :\quad \\ E[{ e }^{ tX }]\quad =\quad \sum _{k=0 }^{ n }{ { e }^{ tk }\left( \begin{matrix} n \\ k \end{matrix} \right) } \quad { p }^{ k }{ { \left( 1-p \right) } }^{ n-k }\\ =\quad \sum _{k=0 }^{ n }{ \left( \begin{matrix} n \\ k \end{matrix} \right) } { \left( p{ e }^{ t } \right) }^{ k }{ { \left( 1-p \right) } }^{ n-k }\\ =[p{ e }^{ t }+{ (1-p) }]^{ n }\\ \frac { { { dM }_{ X } }^{ ' }(t) }{ dt } =n*\quad [p{ e }^{ t }+{ (1-p) }]^{ n-1 }*p{ e }^{ t }\\ \frac { { { dM }_{ X } }^{ '' }(t) }{ dt^2 } =n*\quad (n-1)*[p{ e }^{ t }+{ (1-p) }]^{ n-2 }*{ p }^{ 2 }{ e }^{ 2t }+n[p{ e }^{ t }+{ (1-p) }]^{ n-1 }*{ p }{ e }^{ t }\\ E[X]\quad ={ { \quad M }_{ X } }^{ ' }(0)\quad =\quad np\\ E[{ X }^{ 2 }]\quad =\quad { { M }_{ X } }^{ '' }(0)\quad =\quad n(n-1)*{ p }^{ 2 }\quad +\quad np\\ Var[X]\quad =\quad E[{ X }^{ 2 }]\quad -\quad ({ E[X]) }^{ 2\quad }=\quad n(n-1)*{ p }^{ 2 }\quad +\quad np\quad -{ n }^{ 2 }{ p }^{ 2 }\\ \quad \quad \quad \quad \quad =np(1-p)\)

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

\({ p }_{ x }(x)\quad =\quad \lambda { e }^{ -\lambda x }\quad \quad (x\ge 0)\\ continous\quad MGF\quad :\quad \\ E[{ e }^{ tX }]\quad =\quad \int _{ 0 }^{ +\infty }{ { e }^{ tx } } \lambda { e }^{ -\lambda x }dx\\ =\quad \lambda \int _{ 0 }^{ +\infty }{ { e }^{ (t-\lambda )x } } dx\\ =\quad \frac { \lambda }{ t-\lambda } \\ \\ \frac { { { dM }_{ X } }^{ ' }(t) }{ dt } =\frac { \lambda }{ { (t-\lambda ) }^{ 2 } } \\ \frac { { { dM }_{ X } }^{ '' }(t) }{ d{ t }^{ 2 } } =\frac { 2\lambda }{ { (t-\lambda ) }^{ 3 } } \\ E[X]\quad ={ { \quad M }_{ X } }^{ ' }(0)\quad =\quad \frac { 1 }{ { \lambda }^{ 2 } } \\ E[{ X }^{ 2 }]\quad =\quad { { M }_{ X } }^{ '' }(0)\quad =\frac { 2 }{ { \lambda }^{ 2 } } \\ Var[X]\quad =\quad E[{ X }^{ 2 }]\quad -\quad ({ E[X]) }^{ 2\quad }=\quad \frac { 1 }{ { \lambda }^{ 2 } }\)