## 1. Exercise 11 - Page 363

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 $$μ=0$$ and variance $$σ^2$$=$$\frac{1}{4}$$. If $$Y_1=100$$, estimate the probability that $$Y_365$$ is

Since $$X_n=Y_{n+1}−Y_n$$, also we can say $$Y_{n+1}=X_n−Y_n$$ In year 1 and 2 $$Y_2 = X_1 + Y_1$$ And then $$Y_{365} = Y_1 + X_1 + X_2 + X_3 + ... + X_{364}$$ Because we know that $$Y_1 = 100$$ and $$X_1 , X_2 , X_3 ...X_{364}$$ have $$μ=0$$, We know the expected value for $$Y_1$$ is 100+0+0+0+…+0=100. As variance $$σ^2$$=$$\frac{1}{4}$$, standard deviation is can be calculated as below, $$Y_{365}$$ = 365 * $$\frac{1}{4}$$ = 91.25 Therefor, SD = $$\sqrt{91.25} = 9.552$$

Using mean, $$μ=0$$ and SD, $$σ = \sqrt{91.25}$$

### (a)$$≥ 100$$

pnorm(100 - 100, mean = 0, sd = sqrt(91.25), lower.tail = FALSE)
##  0.5

### (a)$$≥ 110$$

pnorm(110 - 100, mean = 0, sd = sqrt(91.25), lower.tail = FALSE)
##  0.1475849

### (a)$$≥ 120$$

pnorm(120 - 100, mean = 0, sd = sqrt(91.25), lower.tail = FALSE)
##  0.01814355

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

The probability mass function for the binomial distribution is:
$$\binom{n}{k} (p)^k(1-p)^{n-k}$$
The moment generating function is given as $$g(t) = \sum_{j=0}^{n}(p)^k(1-p)^{n-k}e^{jt}$$ = $$(1-p+pe^t)^n$$

First derivative, $$g'(t) = npe^t(pe^t-p+1)^{n-1}$$

Evaluating for $$t=0$$ to be determine the mean $$μ_1$$ $$g'(0) = n*p*e^0(pe^0-p+1)^{n-1} = np$$

Second derivative, $$g''(t) = npe^t(pe^t-p+1)^{n-2}(npe^t-p+1)$$

Evaluating for $$t=0$$ to be determine the mean $$μ_2$$ $$g''(0) = npe^0(pe^0-p+1)^{n-2}(npe^0-p+1) = np(np+1-p)$$

The variance is $$μ_2-μ_{1}^{2} = np(np+1-p)np = n^2p^2(np+1-p)$$

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

The probability density function for the exponential distribution is: $$\lambda e^{-\lambda x}$$

The moment generating function is givenn as $$g(t) = \frac{\lambda}{\lambda-t}$$ for $$t < \lambda$$

First derivative, $$g'(t) = \frac{\lambda}{(\lambda-t)^2}$$

Evaluating for $$t=0$$ to be determine the mean $$μ_1$$ $$g'(0) = \frac{\lambda}{(\lambda-0)^2} = \frac{1}{\lambda}$$

Second derivative, $$g''(t) = \frac{2}{(\lambda-t)^3}$$

Evaluating for $$t=0$$ to be determine the mean $$μ_2$$ $$g''(0) = \frac{2}{(\lambda-0)^3} = \frac{2}{\lambda^2}$$

The variance is $$μ_2-μ_{1}^{2} =\frac{2}{\lambda^2} - (\frac{1}{\lambda})^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}$$