This question comes from page 363, question #11 of “Introduction to Probability” by Charles M. Grinstead.
The price of one share of stock in the Pilsdorff Beer Company (see Exercise 8.2.12) 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
We know that \(Y_{365} = Y_1 + X_1 + X_2 + ... + X_{364} = Y_1 + \sum_{i=1}^{364}X_i\). Given that \(\sum_{i=1}^{364}X_i\) is a sum of independent random variables, we can assume it follows a normal distribution via the Central Limit Theorem. Furthermore, we can define the mean \(\mu_{364}\) and standard deviation \(\sigma_{364}\) of this normal distribution as:
\[ \begin{align} \mu_{364} &= 364\mu = 0 \\ \sigma_{364} &= \sqrt{\sigma^2_{X_1}+ \sigma^2_{X_1} + ... + \sigma^2_{364}} = \sqrt{\frac{364}{4}} \end{align} \]
We can use this mean and standard deviation to sample from a normal distribution and simulate 100,000 different values of \(Y_{365}\):
sd = sqrt(364/4)
sims = 100000
Y_365s = 100 + rnorm(sims, mean=0, sd=sd)We can now use these simulated \(Y_{365}\) values to estimate the probabilities:
P_a <- length(which(Y_365s > 100)) / sims
P_b <- length(which(Y_365s > 110)) / sims
P_c <- length(which(Y_365s > 120)) / sims
str1 <- paste('Estimated probability of Y_365 >= 100:', P_a)
str2 <- paste('Estimated probability of Y_365 >= 110:', P_b)
str3 <- paste('Estimated probability of Y_365 >= 120:', P_c)
str_final <- paste(str1, str2, str3, sep = '\n')
writeLines(str_final)## Estimated probability of Y_365 >= 100: 0.49974
## Estimated probability of Y_365 >= 110: 0.14841
## Estimated probability of Y_365 >= 120: 0.01865# the below can be used to check the above values
# using z-score and pnorm distribution
1-pnorm((100-100)/(0.5*sqrt(364)))
1-pnorm((110-100)/(0.5*sqrt(364)))
1-pnorm((120-100)/(0.5*sqrt(364)))Calculate the expected value and variance of the binomial distribution using the moment generating function.
Given a discrete random variable \(X\) with probability mass function \(f_X(x)\) and dummy parameter \(t\), the moment generating function of \(X\) is defined as the expectation value of \(e^{tX}\):
\[ \begin{equation} M_X(t) = E[e^{tX}] \end{equation} \]
Using the fact that, for a discrete random variable, \(E[X] = \sum_{x \in X} x \cdot p(x)\), we have:
\[ \begin{equation} M_X(t) = \sum_{x\in X}e^{tx} \cdot f_X(x) \end{equation} \]
Since \(X\) follows a binomial distribution in which \(0 \leq x \leq n\), we can write:
\[ \begin{align} M_X(t) &= \sum_{x=0}^ne^{tx} \cdot {n \choose x} p^x q^{x-n} \\ M_X(t) &= \sum_{x=0}^n {n \choose x}(pe^t)^xq^{x-n} \end{align} \]
The binomial theorem tells us that \((u+v)^n = \sum_{x=0}^n {n \choose x}u^xv^{n-x}\), and as such we can simplify the above sum:
\[ \begin{equation} M_X(t) = (pe^t + q)^n \end{equation} \]
Once we have the moment generating function of \(X\), we can use it to determine \(E[X]\) and \(E[X^2]\):
\[ \begin{align} E[X] &= M'_X(0) \\ E[X^2] &= M''_X(0) \end{align} \]
Thus, we first find \(M'_X(t)\):
\[ \begin{align} M'_X(t) &= \frac{d}{dt}((pe^t + q)^n) \\ M'_X(t) &= npe^t(pe^t+q)^{n-1} \\ \end{align} \]
Using the above in conjunction with the fact that \(p+q=1\), we can determine \(E[X]\):
\[ \begin{align} E[X] &= M'_X(0) \\ E[X] &= npe^0(pe^0+q)^{n-1} \\ E[X] &= np(p+q)^{n-1} \\ E[X] &= np(1)^{n-1} \\ E[X] &= np \end{align} \]
Next, we determine \(M''_X(t)\):
\[ \begin{align} M''_X(t) &= \frac{d}{dt}(npe^t(pe^t+q)^{n-1}) \\ M''_X(t) &= n(n-1)(pe^t)^2+npe^t(pe^t+q)^{n-2} \\ \end{align} \]
Once again we evaluate at 0 to find \(E[X^2]\):
\[ \begin{align} E[X^2] &= M''_X(0) \\ E[X^2] &= n(n-1)(pe^0)^2+npe^0(pe^0+q)^{n-2} \\ E[X^2] &= n(n-1)(p^2)+np(p+q)^{n-2} \\ E[X^2] &= n^2p^2-np^2+ np\\ \end{align} \]
To find the variance \(\sigma^2\), we use the fact that \(\sigma^2 = E[X^2] - E[X]^2\):
\[ \begin{align} \sigma^2 &= n^2p^2-np^2+ np -(np)^2 \\ \sigma^2 &= np -np^2 \\ \sigma^2 &= np(1-p) \\ \sigma^2 &= npq \end{align} \]
Thus, to summarize, we have confirmed using the binomial distribution’s moment generating function that: \(E[X] = \mu = np\) and \(\sigma^2 = npq\).
Calculate the expected value and variance of the exponential distribution using the moment generating function.
Given a continuous random variable \(X\) with probability density function \(f_X(x)\) and dummy parameter \(t\), the moment generating function of \(X\) is defined as the expectation value of \(e^{tX}\):
\[ \begin{equation} M_X(t) = E[e^{tX}] \end{equation} \]
Using the fact that, for a continuous random variable, \(E[X] = \int_{-\infty}^{\infty} x \cdot p(x)\), we have:
\[ \begin{equation} M_X(t) = \int_{-\infty}^{\infty}e^{tx} \cdot f_X(x)\,dx \end{equation} \]
Since \(X\) follows an exponential distribution in which:
\[ \begin{equation} f_X(x) = \begin{cases} \lambda e^{-\lambda x} & \text{if } 0 \leq x < \infty\\ 0 & \text{otherwise } \end{cases} \end{equation} \] we can write:
\[ \begin{align} M_X(t) &= \int_{0}^{\infty}e^{tx} \lambda e^{-\lambda x} \, dx \\ M_X(t) &= \lambda \int_{0}^{\infty}e^{x(t-\lambda)} \, dx \\ M_X(t) &= \frac{\lambda}{t-\lambda}e^{x(t-\lambda)}|_0^{\infty} \end{align} \]
For \(t<\lambda\), this integral evaluates to:
\[ \begin{align} M_X(t) &= \frac{\lambda}{t-\lambda}(0-1) \\ M_X(t) &= \frac{\lambda}{\lambda - t} \end{align} \]
Now that we have the moment generating function of \(X\), we can use it to determine \(E[X]\) and \(E[X^2]\):
\[ \begin{align} E[X] &= M'_X(0) \\ E[X^2] &= M''_X(0) \end{align} \]
We first find \(M'_X(t)\) below:
\[ \begin{align} M'_X(t) &= \frac{d}{dt}\left( \frac{\lambda}{\lambda - t} \right) \\ M'_X(t) &= \frac{\lambda}{(\lambda-t)^2} \\ \end{align} \]
Using the above, we can determine \(E[X]\):
\[ \begin{align} E[X] &= M'_X(0) \\ E[X] &= \frac{\lambda}{(\lambda-0)^2} \\ E[X] &= \frac{\lambda}{\lambda^2} \\ E[X] &= \frac{1}{\lambda} \\ \end{align} \]
Next, we determine \(M''_X(t)\):
\[ \begin{align} M''_X(t) &= \frac{d}{dt}(\frac{\lambda}{(\lambda-t)^2}) \\ M''_X(t) &= \frac{2\lambda}{(\lambda-t)^3} \\ \end{align} \]
Once again we evaluate at 0 to find \(E[X^2]\):
\[ \begin{align} E[X^2] &= M''_X(0) \\ E[X^2] &= \frac{2\lambda}{(\lambda-0)^3} \\ E[X^2] &= \frac{2}{\lambda^2} \end{align} \]
To find the variance \(\sigma^2\), we use the fact that \(\sigma^2 = E[X^2] - E[X]^2\):
\[ \begin{align} \sigma^2 &= \frac{2}{\lambda^2} - \left(\frac{1}{\lambda} \right)^2 \\ \sigma^2 &= \frac{1}{\lambda^2} \end{align} \]
Thus, to summarize, we have confirmed using the exponential distribution’s moment generating function that: \(E[X] = \mu = \frac{1}{\lambda}\) and \(\sigma^2 = \frac{1}{\lambda^2}\).