#HOMEWORK 9
This exercise asks the following:
The price of one share of stock in the Pilsdorff Beer Company (see Exercise 8.2.12) is given by Yn on the nth day of the year. Finn observes that the differences Xn = Yn+1 - Yn appear to be independent random variables with a common distribution having mean = 0 and variance = 1/4. If Y1 = 100, estimate the probability that Y365 is: \[ (a) ≥ 100, (b) ≥ 110, (c) ≥ 120 \]
Since the expected value of each Xn is = 0, the expected value of Y365 is = 100. We can then find the variance of Y365 like so: \[ σ²(Y365) = σ²(X1) + σ²(X2) + ... + σ²(X365) = 1/4 + 1/4 + ... + 1/4 = 365 * (1/4) = 91.25 \] We can calculate the Z-scores using the mean and the variance which we previously found: \[ (a) P(Y365 ≥ 100): \] \[ Z = (Y365 - μ) / √(σ²) = (100 - 100) / √(91.25) = 0 / √(91.25) = 0. P(Z ≥ 0) = 0.5 P(Y365 ≥ 100) = 0.5 \] We can use the same approach for b: \[ (b) P(Y365 ≥ 110): \] \[ Z = (110 - 100) / √(91.25) ≈ 1.04684784 \] \[ P(Z ≥ 1.05) = 0.8531 \] \[ P(Y365 ≥ 110) = 1- 0.8531 ≈ .147 \] And again for c: \[ P(Y365 ≥ 120): \] \[ Z = (120 - 100) / √(91.25) ≈ 2.09369569 \] \[ P(Z ≥ 2.09) ≈ .098169 \] \[ P(Y365 ≥ 120) = 1 - .098169 ≈ .018 \]
##2. Calculate the expected value and variance of the binomial distribution using the moment generating function.
I started with the probability mass function: \[ f(x) = C(n,x)p^x(1 – p)^n-^x \] where C(n,x) represents n value of numbers x (0 through n). 1-p is the probability of failure for each of the trials.
Using the probability mass function, we can also find the moment generating function like so: \[ M(t) = Σ(where x=0, through n)e^t^xC(n,x)>)p^x(1 – p)^n - ^x \] We can simplify: \[ M(t) = Σ(where x=0, through n) (pe^t)^xC(n,x)>)(1 – p)^n - ^x \] \[ M(t) = [(1 – p) + pe^t]^n \] We can calculate the expected value from here: \[ M(0) = n(pe^0)[(1 – p) + pe^0]^n - ^1 = np \] And the variance through differentiation: \[ M’’(t) = n(n - 1)(pe^t)^2[(1 – p) + pe^t]^n - ^2 + n(pe^t)[(1 – p) + pe^t]^n - ^1 \] \[ M’’(0) = n(n - 1)p^2 +np \] \[ σ2 = M’’(0) – [M’(0)]^2 = n(n - 1)p^2 +np - (np)^2 = np(1 - p) σ2 = np(1 - p) \]
This is the probability density we will need: \[ px(x) = 𝜆𝑒−^𝜆^x \]
I started with the moment generating function to calculate the expected value: \[ M_x = E[e^t^X] \] \[ E(X) = M'(0) \] \[ M'(t) = d/dt [E(e^tx)] \] \[ M'(t) = λ / (λ - t)^2 \] \[ M'(0) = λ / (λ - 0)^2 = λ / λ^2 = 1 / λ \] \[ M'(0) = 1 / λ \]
We can take the second derivative at t=0 to help find the variance:
\[ M''(t) = 2λ / (λ - 0)^3 = 2 / λ^2 \] \[ M''(0)-[M'(0)^2] \] \[ Variance = (2 / λ^2) - (1 / λ)^2 = 1/λ^2 \]