Q 11 (Page 363)

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

  1. \(\geq\) 100
pnorm(100 - 100, mean = 0, sd = sqrt(364/4), lower.tail = FALSE)
## [1] 0.5
  1. \(\geq\) 110
pnorm(110 - 100, mean = 0, sd = sqrt(364/4), lower.tail = FALSE)
## [1] 0.1472537
  1. \(\geq\) 120
pnorm(120 - 100, mean = 0, sd = sqrt(364/4), lower.tail = FALSE)
## [1] 0.01801584

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

Moment generating function for a random variable X: \(M(t)=[(1-p) +pe^t]^n\)

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

for t = 0: \(M'(0) = n [ 1 - p + p ]^{n-1} p(1) = np\)

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

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

for t = 0: \(M''(0) = np + n^2p^2 -np^2\)

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

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

Moment generating function for exponential distribution: \(f(x)= \lambda e^{−\lambda x}\)

\[g(t)=∫_{ 0 }^{ \infty }e^{ tx }λe^{ -λe }dx\]

\[g(t)=\frac { λe^{ (t-λ)x } }{ t-λ } |_{ 0 }^{ ∞ }\]

\[g(t)=\frac { λ }{ λ-t }\]

\[g(t)=\frac { λ }{ (λ-t)^{ 2 } }\]

\[g(t)=\frac { λ }{ λ^{ 2 } } =\frac { 1 }{ λ }\]

\[g''(t)=\frac { 2λ }{ (λ-t)^{ 3 } }\]

\[g''(0)=\frac { 2λ }{ λ^{ 3 } } =\frac { 2 }{ λ^{ 2 } }\]

Expected Value:

\[\mu =g'(0)=λ^{ -1 }\]

Variance:

\[σ^{ 2 }=g''(0)-g'(0)^{ 2 }=\frac { 2 }{ λ^{ 2 } } =\frac { 1 }{ λ^{ 2 } } =λ^{ -2 }\]