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 µ =
0 and variance \(σ^2\) = 1/4. If \(Y_1\) = 100, estimate the probability that
\(Y_{365}\) is:
(a) ≥ 100
(b) ≥ 110
(c) ≥ 120
The differences \(X_n\) are independent random variables, and the mean \(\mu\) and variance \(\alpha^2\) are well defined. \(X_n = Y_{n+1} − Y_n\), which we can rewrite as:
\[ Y_{n+1} = Y_1 + \sum_{i=1}^{n} X_i \]
Meaning that the price at day \(n+1\) can be found by adding the initial price at day \(Y_1\) to the differences in price up to day \(n\), better known as the accumulation of price changes.
We now have an expression that contains a sum. We are given that \(Y_1=100\), so that becomes:
\[ Y_{n+1} = 100 + \sum_{i=1}^{n} X_i \] We can easily rewrite this to model the price on current day \(n\):
\[ Y_{n} = 100 + \sum_{i=1}^{n-1} X_i \] Because of the aforementioned properties (independent with known mean and variance), the sum of \(X_i\) will follow a normal distribution, meaning \(Y_n\) will as well. We will need to find the mean \(\mu\) and the variance \(\alpha^2\) at n=365:
\[\mu_{Y_{365}} = 100+364*0 = 100\] and \[\alpha^2_{Y_{365}} = 364 * \frac{1}{4} = 91 \\ \alpha = \sqrt{91}\]
Using the properties of normal distribution we can find the Z scores such that:
\[ Z >= \frac{x-\mu}{\alpha} \]
Plugging in values:
\[ Z >= \frac{100-100}{\sqrt{}91} \\ Z = 0\]
A z-score of 0 means no deviation from the mean of the distribution which is 100. That means the probability of getting a raw value of 100 or more is 50%.
\[ Z >= \frac{110-100}{\sqrt{}91} \\ Z = 1.048\] The raw value 110 is 1.048 standard deviations above the mean, which means getting a raw value >= that is 14.725% likely.
\[ Z >= \frac{120-100}{\sqrt{}91} \\ Z = 2.097\]
120 is 2.097 standard deviations above the meanmm and so chances of
getting a value >= that is 1.08%
a)\(50\%\)
b)\(14.724\%\)
c)\(1.08\%\)