Problem 1 - Pg 363 Ex. 11

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. \(\ge 100\)
  2. \(\ge 110\)
  3. \(\ge 120\)



Solution 1
Since \(X_n\) is an independent random variable, then its sum \(S_n\) is normally distributed.

std_dev = sqrt(365 * (1/4))
mean_value = 0


Solution 1(a); \(Y_{365} \geq\) 100
Since \(S_n\) is normally distributed with mean \(0\) and normal distribution is symmetric, exactly half of values will be greater than mean. Alternatively:

prob_greater_100 <- pnorm(100 - 100, mean = mean_value, sd = std_dev, lower.tail = FALSE)
prob_greater_100
## [1] 0.5


Solution 1(b); \(Y_{365} \geq\) 110

prob_greater_110 <- round(pnorm(110 - 100, mean = mean_value, sd = std_dev, lower.tail = FALSE), 4)
prob_greater_110
## [1] 0.1476


Solution 1(c); \(Y_{365} \geq\) 120

prob_greater_120 <- round(pnorm(120 - 100, mean = mean_value, sd = std_dev, lower.tail = FALSE), 4)
prob_greater_120
## [1] 0.0181

Problem 2 - Binomial Distribution

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


Solution 2

For the binomial distribution, the probability mass function (PMF) is given by: \(P(X=k) = {n \choose k} p^k q^{n-k}\) where \(q=1-p\).

The moment generating function is given as: \(M_X(t)=(q+pe^t)^n\).

The first moment is \(M'_X(t) = n(q+pe^t)^{n-1}pe^t\) which is the first derivative of \(M_X(t)\) above.

The expected value is the first moment evaluated at \(t=0\), therefore:

\[ \begin{split} E(X)=M'_X(0) &= n(q+pe^0)^{n-1}pe^0\\ &= n(q+p)^{n-1}p\\ &= np(1-p+p)^{n-1}\\ &= np(1)^{n-1}\\ &=np \end{split} \] Therefore, the expected value of the Bionomial distribution is \(E(X) = np\).

The variance is given by: \(Var(X)=E(X^2)-E(X)^2\); where \(E(X^2)\) is the second derivative of the moment generating function and \(E(X)^2\) is the square of the first derivative of the moment generating function which is also the square of the expected value when evaluated at \(t=0\).

The second moment is \(M''_X(t) = n(n-1)(q+pe^t)^{n-2}p^2 e^{2t}+n(q+pe^t)^{n-1}pe^t\). second order derivative of \(M_X(t)\)

Evaluating the second moment at \(t=0\):

\[ \begin{split} E(X^2)=M''_X(0) &= n(n-1)(q+pe^0)^{n-2}p^2 e^0+n(q+pe^0)^{n-1}pe^0\\ &= n(n-1)(1-p+p)^{n-2}p^2+n(1-p+p)^{n-1}p\\ &= n(n-1)p^2+np \end{split} \]

Since, the variance is \(Var(X)=E(X^2)-E(X)^2\), substituting the values gives:

\[ \begin{split} Var(X) &= n(n-1)p^2+np-n^2p^2 \\ &= np((n-1)p+1-np) \\ &= np(np-p+1-np) \\ &= np(1-p) \\ &= npq \end{split} \] Therefore, we see that \(Var(X) = npq\) for a binomial probability distribution.



Problem 3 - Exponential Distribution

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


Solution 3

The probability of the exponential distribution is given by, \(f(x)=\lambda e^{-\lambda x}\).
Also, the moment generating function is \(M_X(t)=\frac{\lambda}{\lambda-t}, t<\lambda\).

The first moment is, \(M'_X(t) = \frac{\lambda}{(\lambda-t)^2}\) and the second moment is given by: \(M''_X(t) = \frac{2\lambda}{(\lambda-t)^3}\).

To get the expected value, we evaluate the first moment at \(t=0\):

\[ \begin{split} E(X)=M'_X(0) &= \frac{\lambda}{(\lambda-0)^2} \\ &= \frac{\lambda}{\lambda^2}\\ &= \frac{1}{\lambda} \end{split} \]

The variance is given by: \(Var(X)=E(X^2)-E(X)^2\); where \(E(X^2)\) is the second derivative of the moment generating function and \(E(X)^2\) is the square of the first derivative of the moment generating function which is also the square of the expected value when evaluated at \(t=0\). Evaluating the variance function for where \(t=0\):

\[ \begin{split} Var(X) = E(X^2)-E(X)^2 &= M''_X(0)-M'_X(0)^2 \\ &=\frac{2\lambda}{(\lambda-0)^3} - \frac{1}{\lambda^2}\\ &=\frac{2\lambda}{\lambda^3} - \frac{1}{\lambda^2}\\ &=\frac{2}{\lambda^2} - \frac{1}{\lambda^2}\\ &=\frac{1}{\lambda^2} \end{split} \]

Hence, we get the expected value and variance for exponential probability distribution as \(E(X)=1/\lambda\) and \(Var(X)=1/\lambda^2\) respectively.