The price of one share of stock in the Pilsdorff Berr 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 \(\mu = 0\) and variance \(\sigma ^2 = \frac{1}{4}\). If \(Y1 = 100\) estimate the probability that \(Y365\) is

These probabilities will be solved using the Central Limit Theorem and the answers will be calculated using the pnorm function in R. Different from earlier submissions, the answer will be subtracted from 1.

\[(A)\ge 100\]

For exercise A we have the mean where 100 is subtracted from \(Y365=100\) and then divided by the standard deviation.

cat("Probability of Greater Than or Equal to 100","\n")
## Probability of Greater Than or Equal to 100
(A<- 1 - pnorm((100-100)/(0.5*sqrt(365-1))))
## [1] 0.5

\[(B)\ge 110\] For exercise A we have the mean where 100 is subtracted from \(Y365=110\) and then divided by the standard deviation.

cat("Probability of Greater Than or Equal to 110","\n")
## Probability of Greater Than or Equal to 110
(B<- round(1 - pnorm((110-100)/(0.5*sqrt(365-1))),digits=3))
## [1] 0.147

\[(C)\ge 120\]

cat("Probability of Greater Than or Equal to 110","\n")
## Probability of Greater Than or Equal to 110
(C<- round(1 - pnorm((120-100)/(0.5*sqrt(365-1))),digits=3))
## [1] 0.018

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

According to the following sites https://www.thoughtco.com/moment-generating-function-binomial-distribution-3126454

https://www.youtube.com/watch?v=gcpSImAQjlk&feature=youtu.be

https://www.youtube.com/watch?v=OKdjid9FeKk&feature=youtu.be

Moment Generating Function \[M(t)=[(1-p)+pe^t]^n\] 1st Derivative \[M'(t)=n(pe^t)[(1-p)+pe^t]^{n-1}\] \(\mu=E(X)\) \[\sum x.p(x)=np\] \(\sigma^2\) \[M^n(t)=n(n-1)(pe^t)[(1-p)+pe^t]^{n-2}+n(pe^t)[(1-p)+pe^t]^{n-1}\] Calculate the expected value and variance of the exponential distribution using the moment generating function.

Moment Generating Function \[Mx(t)=\lambda/(\lambda-t)\] Expected Value \[E(X)=\frac{1}{\lambda}\] \(\sigma^2\) \[E(X)-[E(X)]^2=\frac{1}{\lambda^2}\]