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 \(\sigma^2=\frac{1}{4}\) If \(Y_1 = 100\), estimate the probability that \(Y_{365}\) is

  1. \(\geq 100\)

Given that each observation is an independent random variable, we know that the distribution of financial results is roughly normal centered about the mean of 0. This indicates a gain/loss of 0 dollars, leaving the original value of 100 intact. The normal distribution is evenly split on this point and thus there is a \(0.5\) or \(50%\) chance of this occuring. No calculations required.

  1. \(\geq 110\)

In order to end up with at least $110, this indicates a gain of $10 or more. Finding the likelyhood of this using the fact that according to the Central Limit Theorem, the independent random variables can be modeled with a normal distribution.

pnorm(10/sqrt(365), mean=0, sd=sqrt(1/4), lower.tail = FALSE)
## [1] 0.1475849

There is about a \(15\%\) chance of this happening.

  1. \(\geq 120\)

Following a similar pattern as in part (b) but finding a gain of $20 or more.

pnorm(20/sqrt(365), mean=0, sd=sqrt(1/4), lower.tail = FALSE)
## [1] 0.01814355

There is about a \(2\%\) chance of this happening.

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

The binomial moment generating function is defined as: \(M(t)=[(1-p)+pe^t]^n\)

The mean is calculated by solving the equation \(M'(0)\).

\(M'(t)=n(pe^t)[(1-p)+pe^t]^{n-1}\) \(M'(0)=n(pe^0)[(1-p)+pe^0]^{n-1}\) \(M'(0)=n(p)[1-p+p]^{n-1}\) \(M'(0)=n(p)1^{n-1}\) \(M'(0)=n(p)\)

The variance is calculated by solving the equation \(M''(0)-M'(0)^2\).

\(M''(t)=n(n-1)(pe^t)^2[(1-p)+pe^t]^{n-2}+n(pe^t)[(1-p)+pe^t]^{n-1}\) \(M''(0)=n(n-1)(pe^0)^2[(1-p)+pe^0]^{n-2}+n(pe^0)[(1-p)+pe^0]^{n-1}\) \(M''(0)=n(n-1)(p^2)+np\)

Substituting and reducing:

\(M''(0)-M'(0)^2=n(n-1)(p^2)+np-(np)^2\) \(=n(p)(1-p)\)

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

The exponential moment generating function is defined as: \(M(t)=\frac{\lambda}{\lambda-t}\)

The mean is calculated by solving the equation \(M'(0)\)

\(M'(t)=\frac{0(\lambda-t) - \lambda(-1)}{(\lambda-t)^2}=\frac{\lambda}{(\lambda-t)^2}\) \(M'(0)=\frac{1}{\lambda}\)

The variance is calculated by solving the equation \(M''(0)-M'(0)^2\)

\(M''(t)=\frac{2\lambda}{(\lambda-t)^3}\) \(M''(0)=\frac{2\lambda}{(\lambda-0)^3}=\frac{2}{\lambda^2}\)

Subsituting and reducing:

\(M''(0)-M'(0)^2=\frac{2}{\lambda^2}-\frac{1}{\lambda^2}=\frac{1}{\lambda^2}\)