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 \(\mu^2 = 0\) and variance \(\sigma^2 = 1/4\). If \(Y_{1} = 100\), estimate the probability that \(Y_{365}\) is
.\[ P(y_{365} \geqslant 100) \]
std = sqrt(364/4)
pnorm(100-100, mean = 0, sd = std, lower.tail = FALSE)## [1] 0.5.\[ P(y_{365} \geqslant 110) \]
pnorm(110-100, mean = 0, sd = std, lower.tail = FALSE)## [1] 0.1472537.\[ P(y_{365} \geqslant 120) \]
pnorm(120-100, mean = 0, sd = std, lower.tail = FALSE)## [1] 0.01801584Calculate the expected value and variance of the binomial distribution using the moment generating function.
The probability mass function for the binomial distribution is:
\(g(t) = E(e^{tx}) = \sum^n_{x=0} e^{tx}f(x)\)
Binomial Distribution function:
\(f(x) = \left( \begin{array}{c} n \\ x \end{array} \right) p^x(1-p)^{n-x}\)
Substitute f(x)in g(t):
\(g(t) = \sum^n_{j=0} e^{tx}\left( \begin{array}{c} n \\ x \end{array} \right) p^x(1-p)^{n-x}\)
\(g(t) = \sum^n_{j=0} \left( \begin{array}{c} n \\ x \end{array} \right) (pe^t)^x(1-p)^{n-x}\)
\(g(t) = ((1-p) + pe^t)^n\)
First derivative:
\(g'(t) = \frac{dg(t)}{dt} = n((1-p) + pe^t)^{n-1}pe^t \)
Find expected value E(x) by 1st derivative of MGF at t=0:
\(E(x) = \frac{dg(0)}{dt} = np((1-p) + p)^{n-1} = np\)
Find variance V(x) by 2nd derivative of MGF at t=0:
\(g'(t) = npe^t((1-p) + pe^t)^{n-1} \)
\(g''(t) = \frac{d^2M_x(t)}{dt^2} = npe^t((n-1)((1-p) + pe^t)^{n-2}pe^t) + g'(t)\)
\(g''(t) = npe^t((n-1)((1-p) + pe^t)^{n-2}pe^t) + g'(t)\)
when t=0:
\(g''(0) = np(n-1)[1-p+pe^0]^{n-2}pe^0e^0 + g'(t)\)
\(g''(0) = np^2(n-1) + g'(t) = np^2(n-1) + np\)
Variance: \(V(x) = E(x^2) - (E(x))^2\)
\(V(x) = np^2(n-1) + np - n^2p^2 = n^2p^2 -np^2 +np - n^2p^2 = np - np^2 = np(1-p)\)
Calculate the expected value and variance of the exponential distribution using the moment generating function.
Moment generating function:
\(g(t) = E(e^{tx}) = \sum^\infty_{x=0} e^{tx}f(x))\)
Exponential distribution function:
\(f(x) = \lambda e^{-\lambda x}\)
Substitute f(x)in g(t):
\(g_x(t) = \int_{0}^{\infty} e^{tx}\lambda e^{-\lambda x} dx\)
\(g_x(t) = \lambda \int_{0}^{\infty}e^{(t-\lambda) x} dx)\)
\(g_x(t) = \frac{\lambda}{t-\lambda}\)
First derivative:
\(g_x^{'}(t) = \frac{\lambda}{(\lambda - t)^2}\)
Find expected value E(x) by 1st derivative of MGF at t=0:
\(E(x) = {g_x}^{'}(0) = \frac{\lambda}{(\lambda)^2} = \frac{1}{\lambda}\)
Find variance V(x) by 2nd derivative of MGF at t=0:
\(g_x^{''}(t) = \frac{2\lambda}{(\lambda - t)^3}\)
\(E(X^2) = {g_x}^{''}(0) = \frac{2\lambda}{\lambda^3} = \frac{2}{\lambda^2}\)
Variance: \(V(x) = E(x^2) - (E(x))^2\)
\(V(X) = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}\)