Finding the Standard Deviation

\(X_n = Y_{n+1} - Y_n\),

mean \(\mu = 0\)

variance \(\sigma^2 = 1/4\).

\(Y_1 = 100\), $ Y_{365} = Y_1 + X_1 + X_2 + + X_{364} $

\(E(X_n) = 0\)

\(Var(X_n) = \sigma^2 = 1/4\).\[ E(Y_{365}) = E(Y_1 + X_1 + X_2 + \ldots + X_{364}) = E(Y_1) + E(X_1) + E(X_2) + \ldots + E(X_{364}) = 100 + 0 + 0 + \ldots + 0 = 100 \]

\[ Var(Y_{365}) = Var(Y_1 + X_1 + X_2 + \ldots + X_{364}) = Var(Y_1) + Var(X_1) + Var(X_2) + \ldots + Var(X_{364}) = 0 + \frac{1}{4} + \frac{1}{4} + \ldots + \frac{1}{4} = 91 \]

\(E(Y_{365}) = 100\)

SD = \(\sqrt{Var(Y_{365})} = \sqrt{91}\).

(a) ≥ 100.

mean_Y365 <- 100
sd_Y365 <- sqrt(91)
value <- 100


Z <- (value - mean_Y365) / sd_Y365

probability <- 1 - pnorm(Z)
probability
## [1] 0.5

(b) ≥ 110.

mean_Y365 <- 100
sd_Y365 <- sqrt(91)
value <- 110

Z <- (value - mean_Y365) / sd_Y365

probability <- 1 - pnorm(Z)
probability
## [1] 0.1472537

(c) ≥ 120.

mean_Y1 <- 100
var_X <- 1/4
stddev_Y365 <- sqrt(365 * var_X)

value_c <- 120

Z_c <- (value_c - mean_Y1) / stddev_Y365

prob_c <- 1 - pnorm(Z_c)

print(prob_c)
## [1] 0.01814355

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

\[ M_X(t) = E(e^{tX}) \]

\[ E(X^r) = \frac{d^r M_X(t)}{dt^r} \Bigg|_{t=0} \]

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

\[ M_X(t) = E(e^{tX}) = \sum_{k=0}^{n} e^{tk} \binom{n}{k} p^k (1-p)^{n-k} \]

\[ M_X(t) = \sum_{k=0}^{n} \binom{n}{k} (pe^t)^k (1-p)^{n-k} \]

\[ M_X(t) = \sum_{k=0}^{n} \binom{n}{k} (pe^t)^k (1-p)^{n-k} \]

\[ M_X(t) = (1-p + pe^t)^n \]

  1. Expected value:

\[ E(X) = \frac{d M_X(t)}{dt} \Bigg|_{t=0} \]

\[ E(X) = n(1-p + pe^t)^{n-1} \cdot p \Bigg|_{t=0} \]

\[ E(X) = np \]

  1. Variance:

\[ E(X^2) = \frac{d^2 M_X(t)}{dt^2} \Bigg|_{t=0} \]

\[ E(X^2) = n(n-1)(1-p + pe^t)^{n-2} \cdot p^2 \Bigg|_{t=0} \]

\[ E(X^2) = n(n-1)p^2 \]

\[ \text{Var}(X) = E(X^2) - (E(X))^2 \]

\[ \text{Var}(X) = n(n-1)p^2 - (np)^2 \]

\[ \text{Var}(X) = np(1-p) \]

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

\[ f(x;\lambda) = \lambda e^{-\lambda x} \text{ for } x \geq 0 \]

\[ M_X(t) = E(e^{tX}) \]

\[ M_X(t) = E(e^{tX}) = \int_{0}^{\infty} e^{tx} \lambda e^{-\lambda x} \, dx \]

\[ M_X(t) = \lambda \int_{0}^{\infty} e^{-(\lambda - t)x} \, dx \]

\[ M_X(t) = \lambda \left[ \frac{-1}{\lambda - t} e^{-(\lambda - t)x} \right]_{0}^{\infty} \]

\[ M_X(t) = \frac{\lambda}{\lambda - t} \left[ e^{-(\lambda - t)x} \right]_{0}^{\infty} \]

\[ M_X(t) = \frac{\lambda}{\lambda - t} (0 - 1) \]

\[ M_X(t) = \frac{\lambda}{\lambda - t} \]

  1. Expected value:

\[ E(X) = \frac{d M_X(t)}{dt} \Bigg|_{t=0} \]

\[ E(X) = \frac{d}{dt} \left( \frac{\lambda}{\lambda - t} \right) \Bigg|_{t=0} \]

\[ E(X) = \frac{\lambda}{(\lambda - t)^2} \Bigg|_{t=0} \]

\[ E(X) = \frac{\lambda}{\lambda^2} \]

\[ E(X) = \frac{1}{\lambda} \]

  1. Variance:

\[ E(X^2) = \frac{d^2 M_X(t)}{dt^2} \Bigg|_{t=0} \]

\[ E(X^2) = \frac{d}{dt} \left( \frac{\lambda}{(\lambda - t)^2} \right) \Bigg|_{t=0} \]

\[ E(X^2) = \frac{\lambda \cdot 2(\lambda - t)}{(\lambda - t)^3} \Bigg|_{t=0} \]

\[ E(X^2) = \frac{2\lambda}{\lambda^3} \]

\[ E(X^2) = \frac{2}{\lambda^2} \]

\[ \text{Var}(X) = E(X^2) - (E(X))^2 \]

\[ \text{Var}(X) = \frac{2}{\lambda^2} - \left( \frac{1}{\lambda} \right)^2 \]

\[ \text{Var}(X) = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} \]

\[ \text{Var}(X) = \frac{1}{\lambda^2} \]