9.3. CONTINUOUS INDEPENDENT TRIALS

11.The price of one share of stock in the Pilsdorff Beer Company (see Exercise 8.2.12) 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 \(\sigma^2 = 1/4\). If \(Y_1 = 100\), estimate the probability that \(Y_{365}\) is

\((a) \geq 100.\)

\((b) \geq 110.\)

\((c) \geq 120.\)

solution:

\((a) \geq 100.\)

\(Y_{365} - Y_1 = X_1 + X_2+...+X_{364}\)

\(E[X]=0\)

\(\mu = 0\)

transform to standardized form \(P(Y_{365} - Y_1 \geq 0)=P(\frac{Y_{365} - Y_1}{\sqrt{364}} \geq \frac{0}{\sqrt{364}})\)

a <- 0/sqrt(364)
sigma <- sqrt(1/4)
1 - pnorm(q = a, mean = 0, sd = sigma, lower.tail = TRUE)
## [1] 0.5

\((b) \geq 110.\)

transform to standardized form \(P(Y_{365} - Y_1 \geq 10)=P(\frac{Y_{365} - Y_1}{\sqrt{364}} \geq \frac{10}{\sqrt{364}})\)

a <- 10/sqrt(364)
sigma <- sqrt(1/4)
1 - pnorm(q = a, mean = 0, sd = sigma, lower.tail = TRUE)
## [1] 0.1472537

Aternatively, using one-sided Chebyshev’s inequality.

In one-sided Chebyshev’s inequality, if X has mean 0 and variance \(\sigma^2\) , then for any a>0;

\(P(X \geq a)\leq \frac{\sigma^2}{\sigma^2+a^2})\)

So for \(\sigma^2\) = var(Y_{365} - Y_1) = 364, a = 10

\(P(Y_{365} - Y_1 \geq 10)\leq \frac{364}{364+100}=0.8387\)

\((c) \geq 120.\)

transform to standardized form \(P(Y_{365} - Y_1 \geq 20)=P(\frac{Y_{365} - Y_1}{\sqrt{364}} \geq \frac{20}{\sqrt{364}})\)

a <- 20/sqrt(364)
sigma <- sqrt(1/4)
1 - pnorm(q = a, mean = 0, sd = sigma, lower.tail = TRUE)
## [1] 0.01801584

Aternatively, using one-sided Chebyshev’s inequality.

In one-sided Chebyshev’s inequality, if X has mean 0 and variance \(\sigma^2\) , then for any a>0;

\(P(X \geq a)\leq \frac{\sigma^2}{\sigma^2+a^2})\)

So for \(\sigma^2\) = var(Y_{365} - Y_1) = 364, a = 20

\(P(Y_{365} - Y_1 \geq 10)\leq \frac{364}{364+400}=0.4764\)

Aternatively, using two-sided Chebyshev’s inequality. \[\begin{equation} \begin{split} P(Y_{365} - Y_1 \geq 10) \\ & =P(|Y_{365} - Y_1| \geq 10) \\ & = P(|X| \geq 10)\\ & = \frac{Var(\displaystyle\sum_{i=1}^{10} X_i)}{400}\\ & = 364/400\\ & = 0.91 \end{split} \end{equation}\]

So using Chebyshev’s inequality is not accurate.