pg. 363 Ex. 11 The price of one share of stock in the Pilsdorff Beer Company (see Exercise 8.2.12) is given by Y(n) on the nthnth 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=1/4. If Y(1)=100, estimate the probability that Y(365) is

Since X(n)=Y(n+1)−Y(n), then X(n)+Y(n)=Y(n+1). So say n=1, then X(1)+Y(1)=Y(2), and now look at n=2 X(2)+Y(2)=Y(3) which is also X(2)+X(1)+Y(1)=Y(3), so for Y(365), we can say Y(365)=X(1)+…+X(364)+Y(1). We know Y(1)=100 and that μ=0 for X(n), so the expected value of Y(365)= 0+0+…+0+100=100 and the variance is V(365)=365*1/4=91.25, so the standard deviation is (91.25)^.5

  1. ≥ 100
pnorm(100 - 100, mean = 0, sd = sqrt(91.25), lower.tail = FALSE)
## [1] 0.5
  1. ≥ 110
pnorm(110 - 100, mean = 0, sd = sqrt(91.25), lower.tail = FALSE)
## [1] 0.1475849
  1. ≥ 120
pnorm(120 - 100, mean = 0, sd = sqrt(91.25), lower.tail = FALSE)
## [1] 0.01814355
  1. Calculate the expected value and variance of the binomial distribution using the moment generating function.

The binomial distribution function is \(\binom{n}{k}p^{k}(1−p)^{n-k}\).

So moment generating function is g(t)=\(\sum_{j=1}^{n} \binom{n}{k}p^{k}(1−p)^{n-k}e^{tj}\)= \((1−p+pe^{t})n\)

so \(g'(t)=npe^{t}(1−p+pe^{t})^{n-1}\)

so \(\mu_{1}=g'(0)=npe^{0}(1−p+pe^{0})^{n-1}=np\).

and \(g''(t)=npe^{t}(1−p+pe^{t})^{n-1}(1−p+npe^{t})\) so \(\mu_{2}=g''(0)=npe^{0}(1−p+pe^{0})^{n-1}(1−p+npe^{0})=np(1-p+np)\)

so we have the expected value \(\mu=\mu_{1}=np\) and the variance is \(\mu_{2}-\mu_{1}^{2}=np(1-p+np)-np^{2}\)

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

The exponential distribution function is \(\lambda e^{-\lambda x}\).

So moment generating function for \(t< \lambda\) is g(t)=\(\frac{\lambda}{\lambda-t}\)

so \(g'(t)=\frac{\lambda}{(\lambda-t)^{2}}\)

so \(\mu_{1}=g'(0)=\frac{\lambda}{(\lambda-0)^{2}}=\frac{1}{\lambda}\).

and \(g''(t)=\frac{2\lambda}{(\lambda-t)^{3}}\) so \(\mu_{2}=g''(0)=\frac{2\lambda}{(\lambda-0)^{3}}=\frac{2}{\lambda^{2}}\)

so we have the expected value \(\mu=\mu_{1}=\frac{1}{\lambda}\) and the variance is \(\mu_{2}-\mu_{1}^{2}=\frac{2}{\lambda^{2}}-(\frac{1}{\lambda})^{2}=\frac{1}{\lambda^{2}}\).