605: hw9

Jie Zou

2021-10-24

ex11 pg363

The price of one share of stock in the Pilsdorff Beer Company is given by \(Y_n\) on the n\(th\) 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 \(\mu = 0\) and variance \(\sigma^2 = 1/4\). If \(Y_1=100\), estimate the probability that \(Y_{365}\) is

  1. \(\geq 100\)
  2. \(\geq 110\)
  3. \(\geq 120\)

from the text above, we have

  • \(X_n = Y_{n+1} -Y_n\)
  • \(\mu_X = 0\)
  • \(\sigma_X^2 = 1/4\)
  • \(Y_1 = 100\)

we also know that,

\(S_n = X_1 + X_2 + ...+X_n\)

so,

\(S_n = (Y_2-Y_1)+(Y_3-Y_2)+...+(Y_n-Y_{n-1})+(Y_{n+1}-Y_n)\\ = Y_{n+1} - Y_1 \\= Y_{n+1} - 100\)

when \(Y_{n+1} = Y_{365}\), n = 365-1 = 364,

so, \(S_{364} = Y_{365} - 100\)

then, \(Y_{365} = S_{364} + 100\)

As a result,

  • \(\mu_S = np = n\mu_X = 364*0 = 0\)
  • \(\sigma_S^2 = n\sigma_X^2 = 364 * 1/4 = 91\)
  • \(\sigma_S = \sqrt{91}\)

then,

\(P(Y_{365} \geq100) = P(S_{364}+100 \geq100) = P(S_{364} \geq0)\)

# >=100
1-pnorm(0,mean = 0, sd = sqrt(91))
## [1] 0.5

\(P(Y_{365} \geq110) = P(S_{364}+100 \geq110) = P(S_{364} \geq10)\)

# >=110
1-pnorm(10,mean = 0, sd = sqrt(91))
## [1] 0.1472537

\(P(Y_{365} \geq120) = P(S_{364}+100 \geq120) = P(S_{364} \geq20)\)

# >=120
1-pnorm(20,mean = 0, sd = sqrt(91))
## [1] 0.01801584

Moment generation

binominal distribution

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

binomial distribution function: \(n\choose j\)\(p^jq^{n-j}\)

moment generating function:

\(M(t) = \sum_{j = 0}^{n} e^{tj}\)\(n\choose j\)\(p^jq^{n-j}\)

\(=\sum_{j=0}^n\)\(n\choose j\)\((pe^t)^jq^{n-j}\)

\(=(pe^t + q)^n\)

because \(\mu = \mu_1, \sigma^2 = \mu_2-\mu_1^2\), we need to calculate \(\mu_1\) and \(\mu_2\)

with derivative calculator, we get

  • \(M'(t) = npe^t(pe^t+q)^{n-1}\)
  • \(M''(t) = (n-1)np^2(pe^t+q)^{n-2}e^{2t}+npe^t(pe^t+q)^{n-1}\)

then,

  • \(\mu_1 = M'(t=0)\\=npe^t(pe^t+q)^{n-1} |_{t=0}\\=npe^0(pe^0+q)^{n-1}\\=np\)

  • \(\mu_2 = M''(t=0)\\=(n-1)np^2(pe^t+q)^{n-2}e^{2t}+npe^t(pe^t+q)^{n-1}|_{t=0}\\=(n-1)np^2(pe^0+q)^{n-2}e^{2*0}+npe^0(pe^0+q)^{n-1}\\=(n-1)np^2+np\\=np(np+q)\)

As a result,

  • \(\mu = \mu_1 = np\)
  • \(\sigma^2 = \mu_2 - \mu_1^2 = npq\) where \(q = 1-p\)

exponential distribution

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

exponential distribution function: \(\lambda e^{-\lambda x}\)

moment generating function(with integral calculator):

\(M(t) = \int e^{tx} \lambda e^{-\lambda x}\\=\int \lambda e^{(-\lambda+t)x} \\=-\frac{\lambda e^{(t-\lambda)x}}{\lambda-t}|_{0}^{\infty}\\=\frac{\lambda}{\lambda-t}\)

using derivative calculator, we have:

  • \(M'(t) = \frac{\lambda}{(\lambda-t)^2}\)
  • \(M''(t) = \frac{2\lambda}{(\lambda-t)^3}\)

therefore,

  • \(\mu_1 = M'(t=0)\\=\frac{\lambda}{(\lambda-0)^2}\\=\frac{1}{\lambda}\)
  • \(\mu_2 = M''(t=0)\\=\frac{2\lambda}{(\lambda-0)^3}\\=\frac{2}{\lambda^2}\)

finally,

  • \(\mu = \mu_1 = \frac{1}{\lambda}\)
  • \(\sigma^2 = \mu_2-\mu_1^2 = \frac{2}{\lambda^2}-(\frac{1}{\lambda})^2=\frac{1}{\lambda^2}\)