** DATA_605_Assignment_9_Thonn - Probability Central Limit **
# install libraries if needed
#install.packages("permutations")
#library(permutations)
#install.packages('gtools')
#library(gtools)Assignment-9: 1. #11 page 363 2. Calculate the expected value and variance of the binomial distribution using the moment generating function. 3. Calculate the expected value and variance of the exponential distribution using the moment generating function.
** 1. - Problem-Ch.9 - #11, pg. 363 **
The price of one share of stock in the Pilsdorff Beer Company (see Exercise 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 sigma^2 = 1/4. If Y_1 = 100, estimate the probability that Y_365 is
= 100.
p(Y_365 - Y_1 >= 0) = P( (Y_365-Y_1)/(364^(1/2)) >= 0/(364^(1/2))
xa <- 0/sqrt(364)
sigma1 <- sqrt(1/4)
pa <- pnorm(q = xa, mean = 0, sd = sigma1, lower.tail = FALSE)
pa## [1] 0.5# [1] 0.5 = 110.
p(Y_365 - Y_1 >= 10) = P( (Y_365-Y_1)/(364^(1/2)) >= 10/(364^(1/2))
xb <- 10/sqrt(364)
sigma1 <- sqrt(1/4)
pb <- pnorm(q = xb, mean = 0, sd = sigma1, lower.tail = FALSE)
round(pb,4)## [1] 0.1473# [1] 0.1473= 120.
p(Y_365 - Y_1 >= 20) = P( (Y_365-Y_1)/(364^(1/2)) >= 20/(364^(1/2))
xc <- 20/sqrt(364)
sigma1 <- sqrt(1/4)
pc <- pnorm(q = xc, mean = 0, sd = sigma1, lower.tail = FALSE)
round(pc,4)## [1] 0.018# [1] 0.018** 2. - Problem-Ch.9 - moment generating function **
Problem: Calculate the expected value and variance of the binomial distribution using the moment generating function.
Binomial: \[M_x(t) = E(e^{tx})\]
\[ \sum_{x=0}^n e^{tx} (\begin{array}{cc} n \\ k \end{array})p^x(1-p)^{n-x}\] \[ = ((1-p) + pe^{t})^{n}\]
Expected Value:
\[M_x(t) = \sum_{x=0}^{n} e^{xt} C(n,x)p^x(1-p)^{n-x}\] \[M_x(t) = \sum_{x=0}^{n} (pe^t)^x(1-p)^{n-x}\] \[M_x(t) = ((1-p) + pe^t)^n\]
Differentiate with respect to t: \[M'(t) = n((1-p) + pe^t)^{n-1}pe^t\]
substitute \(t = 0\)
Then obtain the Expected Value \[E(x) = M'(0) = np((1-p) + p)^{n-1} = np\]
Variance:
differentiate Mx(0)
\[\frac{d^2M_x(t)}{dt^2} = npe^t((n-1)((1-p) + pe^t)^{n-2}pe^t) + ((1-p) + pe^t)^{n-1}(npe^t)\] \[E(X^2) = M''(0)\]
with t = 0: \[E(X^2) = \frac{d^2M_x(0)}{dt^2} = np[1 * (n-1)* 1^{n-2}p*1 + [p*1 + (1-p)]^{n-1} *1 \] \[ = n(n-1)p^2 + np \]
Variance = \(V(x) = E(X^2) - (E(x))^2\)
\[V(x) = E(X^2) - (E(x))^2 = n(n-1)p^2 + np - n^2p^2 = npq\] Variane, V(x): \[E(x) = np , and , V(x) = npq\]
** 3. - Problem-Ch.9 - moment generating function **
exponential distribution: \[f(x) = \lambda e^{-\lambda x}\]
obtain the moment generating function with probability density function \(X\): \[M_x(t) = \int_{0}^{\infty} e^{tx}\lambda e^{-\lambda x} dx\]
\[M_x(t) = \lambda \int_{0}^{\infty}e^{(t-\lambda) x} dx\]
\[M_x(t) = \lambda / \lambda - t\]
Expected Value:
\[M'(t) = - \lambda (\lambda - t)^{-2}\]
\[E(X) = M'(0) = \lambda(\lambda)^{-2} = \lambda^{-1}\]
Variance: \[M''(t) = - \lambda (-2)(\lambda - t)^{-3}(-1)\]
\[ = - \lambda (-2)(\lambda - t)^{-3}(-1) = 2 \lambda(\lambda - t)^{-3} = 2 \lambda^{-2}\] \[Var(X) = E(X^2) = M''(0) = 2 \lambda^{-2} - \lambda^2 = \lambda^-2 \]
** END **