11 page 363

11 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 σ2 = 1/4. If Y1 = 100, estimate the probability that Y 365 is:

\(X_n = Y_{n+1}-Y_n\)
\(E[X]=\mu=0\)
\(\sigma^2=\frac{1}{4}\)
\(\sigma=\sqrt(\frac{1}{4})=.5\)

(a) ≥ 100.

n=364
yn <- 100
yn_one <-100
xn <- (yn-yn_one)/sqrt(364)
mu <- 0
v <- .25
sigma <- round(sqrt(v),3)
a <- pnorm(yn-yn_one,mu,sigma,lower.tail = F)

The probability that \(Y_{365}\) equals 100 is 0.5.

(b) ≥ 110

yn <- 110
yn_one <-100
xn <- (yn-yn_one)/sqrt(364)

b <-pnorm(xn,mu,sigma,lower.tail = F)

The probability that \(Y_{365}\) equals 110 is 0.147.

(c) ≥ 120.

yn <- 120
yn_one <-100
xn <- (yn-yn_one)/sqrt(364)

c <-pnorm(xn,mu,sigma,lower.tail = F)

The probability that \(Y_{365}\) equals 120 is 0.018.

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

\(\mu = E[X]\)
\(\sigma^2=E[X^2]-\mu^2\)
\[Moment\hspace{.15cm} Generating\hspace{.15cm} function:\] \[M(t)=E(e^{tX})=\sum\limits_{x_\in S}e^{tx}f(x)dx \] \[\mu=E[X]=M'(t);\hspace{.5cm} o^2=E[X^2]-E[X]^2=M''(t)\] \[M(t)=[(1-p)+pe^t]^n\] \[M'(t) = n[1-p+pe^t]^{n-1}(pe^t)\] \[when\hspace{.15cm}t=0;\] \[\hspace{.3cm}\mu=np=E[X]\] \[M''(t)=n[1-p+pe^t]^{n-1}(pe^t)+(pe^t)n(n-1)[1-p+pe^t]^{n-2}(pe^t)\] \[when\hspace{.15cm}t=0;\] \[M''(0)=n(n-1)p^2+np\] \[\sigma^2=E[X^2]-E[X]^2=np(1-p)\]

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

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

\[when\hspace{.14cm}t=0;\] \[-\lambda \frac{e^{-x(\lambda-t)}}{\lambda-t}\Big|_{0}^{\infty}=\]

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

\[E[X]=\mu = M'(t);\hspace{.3cm}when\hspace{.15cm}t=0\] \[M'(0)=\frac{\lambda}{(\lambda-0)^2}=\frac{\lambda}{\lambda^2}=\] \[E[X]=\mu=M'(0)=\frac{1}{\lambda}\] \[M''(t)=\frac{2\lambda}{(\lambda-t)^3}\]

\[M''(0); \hspace{.3cm}when\hspace{.15cm} t=0 \]

\[M''(0) = \frac{2\lambda}{\lambda^3}=\frac{2}{\lambda^2}\]

\[\sigma^2=E(X^2)-E(X)^2=\frac{2}{\lambda^2}-\frac{1}{\lambda^2}=\frac{1}{\lambda^2}\]