1. 11 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 â Yn appear to be independent random variables with a common distribution having mean Î¼ = 0 and variance Ï2 = 1/4. If Y1 = 100, estimate the probability that Y365 is

365*0.25

## [1] 91.25

The probability distribution is given by N(100,91.25)

Answer:

pnorm(100-100,0,sqrt(91.25))

## [1] 0.5

Answer:

pnorm(110-100,0,sqrt(91.25), lower.tail = F)

## [1] 0.1475849

Answer:

pnorm(120-100,0,sqrt(91.25), lower.tail = F)

## [1] 0.01814355

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

Answer:

\(g(t) = \sum_{j=0}^n e^{tj} {{n}\choose{j}}p^jq^{n-j}\)

\(g(t) = \sum_{j=0}^n {{n}\choose{j}}(pe^t)^jq^{n-j}\)

\(g(t) = (pe^t+q)^n\)

\(g'(t) = n(pe^t+q)^{n-1}pe^t\)

\(g''(t) = n(n-1)(pe^t+q)(pe^t)^2 + n(pe^t+q)^npe^t\)

\(g'(0) = n(p+q)^{n-1}p =np\)

\(g''(0) = n(n-1)p^2 + np\)

\(\boxed{\mu = \mu_1 = g'(0) = np}\)

\(\sigma^2 = \mu_2-\mu_1^2 = g''(0) - g'(0)^2\)

\(\sigma^2 = n(n-1)p^2 + np - (np)^2\)

\(\sigma^2 = np[(np-p) +1 -np]\)

\(\boxed{\sigma^2 = np[1-p]}\)

since p + q = 1.

Answer:

\(g(t) = \int_0^\infty e^{tx}\lambda e^{-\lambda x} dx\)

=>\(g(t) = \frac{\lambda e^{(t-\lambda )x}}{t -\lambda}|_0^{\infty}\)

=>\(g(t) = \frac{\lambda}{\lambda-t}\)

Taking first derivative and equating to 0 we find

\(g'(t) = \frac{\lambda}{(\lambda-t)^2}\)

\(g'(0) = \frac{\lambda}{\lambda^2} = \frac{1}{\lambda}\)

Taking second derivative and equating to 0 we find

\(g''(t) = \frac{2\lambda}{(\lambda-t)^3}\)

\(g''(0) = \frac{2\lambda}{\lambda^3} = \frac{2}{\lambda^2}\)

Hence mean

\(\mu = g'(0) = \lambda^{-1}\)

variance : \(\mu_{2} - \mu_{1}^2\) =

\(\sigma^2 = g''(0) - g'(0)^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \lambda^{-2}\)