1. The price of one share of stock in the Pilsdorff Beer Company (see Exer- cise 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 σ2 = 1/4. If Y1 = 100, estimate the probability that Y365 is
  1. ≥ 100.

μ = 0 σ2 = 1/4

\[P(S_{365} > 100) = P(S_{365} > \frac{100 - 100}{\sqrt{.25 * (365-1)}})\]

1-pnorm((100-100)/(sqrt(.25*(365-1))))
## [1] 0.5
  1. ≥ 110.

\[P(S_{365} > 100) = P(S_{365} > \frac{110 - 100}{\sqrt{.25 * (365-1)}})\]

1-pnorm((110-100)/(sqrt(.25*(365-1))))
## [1] 0.1472537
  1. ≥ 120.

\[P(S_{365} > 100) = P(S_{365} > \frac{120 - 100}{\sqrt{.25 * (365-1)}})\]

1-pnorm((120-100)/(sqrt(.25*(365-1))))
## [1] 0.01801584
  1. Calculate the expected value and variance of the binomial distribution using the moment generating function.

Suppose that X has range {0,1,2,3,…,n} and

\[p_X(j) = {n \choose j}p^jq^{n−j}\] for 0 ≤ j ≤ n (binomial distribution). Then

\[g(t) = \sum_{j=0}^{\infty}e^{tj}{n \choose j}p^jq^{n−j}\]

\[= \sum_{j=0}^{\infty}{n \choose j}(pe^t)^jq^{n−j}\] \[= (pe^t+q)^{n}\] So that..

\[u_1 = g'(0) = n(pe^t+q)^{n-1}pe^t|_{t=0} = np\] \[u_2 = g''(0) = n(n-1)p^{2}+np\]

μ = np \[σ^2 =μ2−μ1^2 =np(1−p)\]

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

Exponential Distribution is as follows:

\[f(x; \lambda ) = \lambda e^{ -\lambda x } \] for x > or = 0 and 0 for x < 0. Then

\[g(t) = \int_{-∞}^∞e^{tj}\lambda e^{ -\lambda x }dx\] \[= \lambda \int_{0}^∞ e^{(t -\lambda )x} dx\]

\[= \lambda [\frac {1}{t - \lambda} e^{(t - \lambda)x}]_{0}^∞\] \[= \frac {\lambda}{\lambda - t}\] Expected Value \[ = \frac {1}{\lambda}\]

Variance \[= \frac {1}{\lambda^2}\]