Chapter 9.3 Exercise 11 page 363
The price of one share of stock in the Pilsdorff Beer Company is given by \(Y_{n}\) Finn notices that \(X_{n} = Y_{n+1} - Y_{n}\) appear to be independent random variables with \(\mu = 0\) and \(\sigma =\frac{1}{2}\) If \(Y_{1} = 100\), estimate the probability that: a) \(Y_{365}\geq100\) b) \(Y_{365}\geq110\) c) \(Y_{365}\geq120\)
Solution:
Calculate the expected value and variance for \(Y_{365}\)
\(E(X_{n}) = E(Y_{n+1}-Y_{n}) = E(Y_{n+1})-E(Y_{n})\)
\(E(X_{i}) = 0 \implies E(Y_{n+1})=E(Y_{n})\)
\(E(Y_{1}) = E(100) = 100 \implies E(Y_{365}) = 100\)
\(V(Y_{365}) = V(Y_{364} + X_{365}) = V(Y_{364}) + V(X_{365}) = V(Y_{364}) + \sigma\)
\(= V(Y_{363}) + 2\sigma = ... = 365\sigma = \frac{365}{4}\)
Since \(E(Y_{365}) = 100 \implies P(Y_{365} \geq 100) = 0.5\)
\(\frac{110 - 100}{\sqrt{\frac{365}{4}}} = 1.04685\)
pnorm(-1.04685)## [1] 0.1475844\(\implies P(Y_{365}\geq 100) = 0.1476\)
pnorm(-2.0937)## [1] 0.01814336\(\implies P(Y_{365}\geq 100) = 0.01814\)
Calculate the expected value and variance of the binomial distribution using the moment generating function.
Solution:
For binomial distribution, \(P(X=k) = {n \choose k} p^k q^{n-k}\), where \(q=1-p\).
The moment generating function is \(M_X(t)=(q+pe^t)^n\).
The first moment is \(M'_X(t) = n(q+pe^t)^{n-1}pe^t\).
The expected value is the first moment evaluated at \(t=0\):
\[ \begin{split} E(X)=M'_X(0) &= n(q+pe^0)^{n-1}pe^0\\ &= n(q+p)^{n-1}p\\ &= np(1-p+p)^{n-1}\\ &= np1^{n-1}\\ &=np \end{split} \] The second moment is \(M''_X(t) = n(n-1)(q+pe^t)^{n-2}p^2 e^{2t}+n(q+pe^t)^{n-1}pe^t\).
Evaluating the second moment at \(t=0\): \[ \begin{split} E(X^2)=M''_X(0) &= n(n-1)(q+pe^0)^{n-2}p^2 e^0+n(q+pe^0)^{n-1}pe^0\\ &= n(n-1)(1-p+p)^{n-2}p^2+n(1-p+p)^{n-1}p\\ &= n(n-1)p^2+np \end{split} \] The variance is \(V(X)=E(X^2)-E(X)^2\): \[ \begin{split} V(X) &= n(n-1)p^2+np-n^2p^2 \\ &= np((n-1)p+1-np) \\ &= np(np-p+1-np) \\ &= np(1-p) \\ &= npq \end{split} \] We get the known definitions for binomial distribution - \(E(X)=np\) and \(V(X)=npq\).
Calculate the expected value and variance of the exponential distribution using the moment generating function.
Solution:
For exponential distribution, \(f(x)=\lambda e^{-\lambda x}\).
The moment generating function is \(M_X(t)=\frac{\lambda}{\lambda-t}, t<\lambda\).
\(M'_X(t) = \frac{\lambda}{(\lambda-t)^2}\) and \(M''_X(t) = \frac{2\lambda}{(\lambda-t)^3}\).
https://www.youtube.com/watch?v=ROpBZalIt3Q
Expected value: \[ \begin{split} E(X)=M'_X(0) &= \frac{\lambda}{(\lambda-0)^2} \\ &= \frac{\lambda}{\lambda^2}\\ &= \frac{1}{\lambda} \end{split} \] Variance: \[ \begin{split} V(X) = E(X^2)-E(X)^2 &= M''_X(0)-M'_X(0)^2 \\ &=\frac{2\lambda}{(\lambda-0)^3} - \frac{1}{\lambda^2}\\ &=\frac{2\lambda}{\lambda^3} - \frac{1}{\lambda^2}\\ &=\frac{2}{\lambda^2} - \frac{1}{\lambda^2}\\ &=\frac{1}{\lambda^2} \end{split} \] We get the known definitions for exponential distribution - \(E(X)=1/\lambda\) and \(V(X)=1/\lambda^2\).