The price of one share of stock in the Pilsdorff Beer Company is given by \(Y_n\) on the nth 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 \(\sigma ^2 = \frac{1}{4}\). If \(Y_1\) = 100, estimate the probability that \(Y_{365}\) is
Number of trials (n) = \(365 - 1 = 364\)
V(X) = \(n \times \sigma ^2 = \frac{364}{4} = 91\)
\(\geq 100\)
\(P(Y_{365} \geq 100) =\)
\(\mu = 0\)
\(\bar x = Y_{365} - Y_1\)
\(z = \frac{\bar x - \mu}{\sqrt{Var(x)}} = \frac{0 - 0}{9.54} = 0\)
p <- pnorm(0, lower.tail = F)
print(paste0("Probabilty is ", round(p*100,2), "%"))## [1] "Probabilty is 50%"\(\geq 110\)
\(P(Y_{365} \geq 110) =\)
\(\mu = 0\)
\(\bar x = Y_{365} - Y_1\)
\(z = \frac{\bar x - \mu}{\sqrt{Var(x)}} = \frac{10 - 0}{9.54} = 1.048\)
p <- pnorm(10/sqrt(91), lower.tail = F)
print(paste0("Probabilty is ", round(p*100,2), "%"))## [1] "Probabilty is 14.73%"\(\geq 120\)
\(P(Y_{365} \geq 110) =\)
\(\mu = 0\)
\(\bar x = Y_{365} - Y_1\)
\(z = \frac{\bar x - \mu}{\sqrt{Var(x)}} = \frac{20 - 0}{9.54} = 2.10\)
p <- pnorm(20/sqrt(91), lower.tail = F)
print(paste0("Probabilty is ", round(p*100,2), "%"))## [1] "Probabilty is 1.8%"Calculate the expected value and variance of the binomial distribution using the moment generating function.
MGF for \(x \rightarrow\) \(M(t) = E(e^{tx}) = \sum^\infty_{x=0} e^{tx}f(x)\)
Binomial Distribution function:
\[f(x) = \left( \begin{array}{c} n \\ x \end{array} \right) p^x(1-p)^{n-x}\]
Binomial Expansion Theorem (Thereom 3.7 page 103):
\[(a + b)^n = \sum^n_{x=0} \left( \begin{array}{c} n \\ x \end{array} \right) a^xb^{n-x} \]
Substitute \(f(x)\) into the MGF:
\[M(t) = \sum^\infty_{x=0} e^{tx}\left( \begin{array}{c} n \\ x \end{array} \right) p^x(1-p)^{n-x}\]
Simplify
I had to look up the rules of exponents to figure out that \(e^{tx}p^x\) = \((pe^t)^x\).
\[ M(t) = \sum^\infty_{x=0} \left( \begin{array}{c} n \\ x \end{array} \right) (pe^t)^x(1-p)^{n-x} \] Substitute the Binomial Expansion function
\[(a + b)^n = \sum^n_{x=0} \left( \begin{array}{c} n \\ x \end{array} \right) a^xb^{n-x} \]
\(a^x = (pe^t)^x\)
\(a = pe^t\)
\(b^{n-x} = (1-p)^{n-x}\)
\(b = (1-p)\)
Therefore, The MGF is:
\((a + b)^n = [(pe^t) + (1-p)]^n\)
Find the E(x) by solving the first derivative of the MGF at \(t=0\) in respect to \(t\)
\(M'(0)\) = \([1-p+pe^0]^n = n[1-p+pe^0]^{n-1}pe^0\)
= \(n[1-p+p]^{n-1}p\)
= \(n[1]^{n-1}p\)
= \(np\)
Find the second derivative
\(E(X) = M'(0) = np = n[1-p+pe^t]^{n-1}pe^t\)
Factor out the constants
\(M''(t) = M'(t) + np[1-p+pe^t]^{n-1}e^t\)
= \(np(n-1)[1-p+pe^t]^{n-2}pe^te^t\)
t = 0
\(M''(0) = M'(t) + np(n-1)[1-p+pe^0]^{n-2}pe^0e^0\)
= \(M'(t) + np^2(n-1) = np + np^2(n-1)\)
Find the Variance = \(E(X^2) - [E(X)]^2 = M''(0) - (M'(0))^2\)
\(np + np^2(n-1) - (np)^2 = np + n^2p^2 - np^2 - n^2p^2\)
= \(np - np^2\)
= \(np(1-p)\)
Calculate the expected value and variance of the exponential distribution using the moment generating function.
MGF for \(x \rightarrow\) \(M(t) = E(e^{tx}) = \sum^\infty_{x=0} e^{tx}f(x)\)
Exponential density function:
\(f(x) = (1-e^{-\lambda x})\)
Substitute \(f(x)\) into the MGF:
\(\sum^\infty_{x=0} e^{tx}(\lambda e^{-\lambda x})\)
Since exponential function is continuous, we need to integrate.
\(\int^\infty_0 e^{tx}\lambda e^{-\lambda x}\)
Simplify
\(\int^\infty_0 \lambda e^{(\lambda-t)x}\)
Integrate
\(a = (t - \lambda)\)
\(\lambda \int^\infty_0 \frac{1}{a} e^{(a)x} = \frac{\lambda}{t - \lambda} e^{(t - \lambda)x}\)
If x \(\rightarrow \infty\), then \(e^{(t-\lambda)x}\) goes to infinity. But if \(t\) is less than \(\lambda\), then \({(t-\lambda)x}\) which \(e^{(t-\lambda)x}\) goes towards zero. So this only works when \(t \leq \lambda\).
\(\frac{\lambda}{t - \lambda} \large[ e^{(t - \lambda)x} \large]^\infty_0\)
= \(\frac{\lambda}{t - \lambda} \large[0 - 1 \large] = \frac{\lambda}{\lambda - t}\)
MGF
\(\frac{\lambda}{\lambda - t}\)
Find the E(x) by solving the first derivative of the MGF at \(t=0\) in respect to \(t\)
\(\frac{\lambda}{\lambda - t} = (1 - \frac{t}{\lambda})^{-1}\)
\(M'(0) = \frac{\lambda}{(\lambda - t)^2} = \frac{1}{\lambda}\)
Find the second derivative.
\(M'(0) = \frac{1}{\lambda}\)
\(M''(0) = 0?\)
I’m not sure if I’m on the right track here. There is no \(t\) to differentiate.
Find the Variance = \(E(X^2) - [E(X)]^2 = M''(0) - (M'(0))^2\)
\(M''(0) - (m'(0))^2 = 0 - \frac{1}{\lambda ^2} = - \frac{1}{\lambda ^2}\)