Problem 1

\(X_1\), \(X_2\) are mutually indeendent, random variables. They are uniformly distributed from 1 to \(k\). Let \(Y\) denote the minimum of the \(X\). Find the distribution of \(Y\).

Since \(Y\) is the minimum value of \(X\) over all the \(X_s\), we have enumberate the ways in which each \(X\) can be assinged different values between 1 and \(k\).

First, suppose that each \(X\) has \(k\) possibilities. Then, the total possible number of assignments for all random variables is \(k^n\). This will form the denominator later.

Next, the number of ways of getting Y = 1 is \(k^n-(k-1)^n\), since the first term represents the number of options and the second term represents the options were none of the \(X\) are equal to 1.

By proceeding in this manner for higher values of Y, we see that for $ 1j k$, and the \[ pdf(Y = j) = \frac{(k-j+1)^n - (k-j)^n}{k^n} \]

Problem 2

An organization has a machine that is expected to work for 10 years (i.e. its failure rate is .1 failures/year). For each of the below distributions I have shown the probability statement in latex and computed the result in R to verify.

Geometric

The geometric mass function can be given as \[ P(X=k) = (1-p)^{k}p = \frac{.9^8}{10} = .043 \]

dgeom(x = 8, prob = .1)
## [1] 0.04304672

The geometric distribution has a mean given by \(\frac{1}{p} = \frac{1}{.1} = 10\) with a variance of \[ \frac{1-p}{p^2} = \frac{.9}{.01} = 90 \]

Exponential

The exponential mass function can be given as \[ P(X=x) \lambda e^{-\lambda x} = .1e^.1*8 = .1e^.8 = .045 \]

dexp(x = 8, rate = .1)
## [1] 0.0449329

The exponential distribution has mean given by \(\frac{1}{\lambda}\), or 10, with a variance of $()^{-2} = 100 $. ## Binomial The binomial mass function can be given as \[ P(X = k) = {{n}\choose{k}}p^k(1-p)^{n-k} = {{8}\choose{0}}.1^0(.9)^8 = .43 \]

dbinom(0,8,.1)
## [1] 0.4304672

The mean can be modeled by \(np\), or .8. The variance can be modelled as \(np(1-p)\), or \(.8*.9 = .72\)

Poisson

The poisson mass function can be given as \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} = \frac{10^1 e^-10}{1!} = .112 \]

dpois(x = 8, lambda = 10)
## [1] 0.112599

It’s mean and variance are equal to its \(\lambda\), 10.