Let X1, X2, . . . , Xn be n mutually independent random variables, each of which is uniformly distributed on the integers from 1 to k. Let Y denote the minimum of the Xi’s. Find the distribution of Y.
Drawing from: https://www2.stat.duke.edu/courses/Spring12/sta104.1/Lectures/Lec15.pdf
Slide 7 gives the main argument to this problem.
and
The beginning part of this page takes the calculation done on slide 7 above and shows how the products of the CDF come about more clearly.
For an individual \(X_i\) the CDF, \(F(x)\) where \(x \in (1,k)\) is:
\[ F(x) = \int_1^x \frac{dx}{k-1} = \frac{x-1}{k-1} \]
Note also that Probability, \(P(X_i<x) = F(x)\). We will substitute this expression in the final stages of the calculation below. Which \(X_i\) is the minimum \(minX_i\) is arbitrary, and can be defined by it’s compliment:
\[ P(min X_i < x) = 1 - P(X_i \geq x) \\ P(min X_i < x) = 1 - \prod_{i=1}^n[1 -P(X_i \leq x)] \\ P(min X_i < x) = 1-[1 - P(X_i \leq x)]^n \\ P(min X_i < x) = 1- [1 - \frac{x-1}{k-1}]^n \\ \bf{F(Y) = 1-[1- \frac{x-1}{k-1}]^n} \]
Where Y is the distribution of the minimum \(X_i\), as defined by the problem text.
Your organization owns a copier (future lawyers, etc.) or MRI (future doctors). This machine has a manufacturer’s expected lifetime of 10 years. This means that we expect one failure every ten years. (Include the probability statements and R Code for each part.).
If we can expect 1 failure every 10 years this means that the probability (p) for failure in a given year is 0.10 since \(E(x) = \sum xm(x) = 10(p) = 1\). We expect 1 failure, there are 10 years and each year has an equal chance of failure.
What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a geometric. (Hint: the probability is equivalent to not failing during the first 8 years..)
From above \(p = 0.1\), and \(q = 1- p =0.9\) According to the hint we assume the copier fails on or after the 9th year so n = 9.
\[ P(X = n) = q^{n-1}*p \\ P(X \leq 9) = \sum_{i=0}^{8}q^i*p = \\ 0.1+0.9*0.1+0.9^2*0.1+0.9^3*0.1+...+0.9^80.1=0.6125795 \] To get the probability that the copier fails after 8 years we take the compliment \(P(X \geq 9) = 1- 0.6125795 = 0.3874205\)
However we can refine our model by counting over seconds instead of years. We can verify using pgeom in r with x = 8*365.25*86400 and p = 0.1/(365.25*86400) where 365.25 is days in a year and 86400 is seconds in a day, and taking the compliment:
1-pgeom((8*365.25*86400),(0.1/(365.25*86400)))## [1] 0.449329According to the Geometric Distribution there is a 44.9329% probability of the copier failing after 8 years if you reassign the time step and probabilty to seconds.
For the Geometric Distribution the Expected Value is simply \(E[X] =\frac{1}{p} = 10\space years\) and the Standard Deviation is \(SD[X] = \sqrt{\frac{q}{p^2}} = 9.486833\space years\).
What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
For the exponential density we will let \(\lambda = 0.1\) since \(\mu = 10\) and \(\lambda = \frac{1}{\mu}\).
From the text, the probability that there is a failure within 8 years is:
\[ P(T \leq 8) = \int_0^8 0.1e^{-0.1t}dt = 1-e^{-0.1*8} = 0.550671 \]
The probability of failure after 8 years failure is the compliment \(e^{-0.1*8} = 0.449329\)
1-pexp(8,0.1)## [1] 0.449329According to Exponential Density there is a 44.93% probability of the copier failing after 8 years.
The expected value, taken on the interval \(0 to \infty\) is defined as:
\[ E[X] = \int_0^{\infty}xf(x)dx = \int_0^{\infty}\lambda xe^{-\lambda x}dx \\ \int_0^{\infty}\lambda xe^{-\lambda x}dx = \frac{\lambda}{\lambda^2} = \frac{1}{\lambda}\\ E[X] = \mu = 10\space years \]
I used wolframalpha.com to evaluate \(\int_0^{\infty}\lambda xe^{-\lambda x}dx\)
Now for Standard Deviation: \[ Var[X] = \int_0^{\infty}(x - \mu)^2f(x)dx = \int_0^{\infty}\lambda x^2e^{-\lambda x}dx - \mu^2 \\ \int_0^{\infty}\lambda x^2e^{-\lambda x}dx = \frac{2\lambda}{\lambda^3} = \frac{2}{\lambda^2}\\ Var[X] = 2\mu^2 - \mu^2 = 100\space years^2 \\ SD[X] = \sqrt{Var[X]} = 10\space years \]
I used wolframalpha.com to evaluate \(\int_0^{\infty}\lambda x^2e^{-\lambda x}dx\)
What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a binomial. (Hint: 0 success in 8 years)
We can take \(p = 0.1\) and \(q =0.9\)
The probability that there are 0 successes and 8 failures is:
\[ P(X\leq n) = p^0q^n\\ P(X \leq 8) = p^0q^8 = 0.9^8 = 0.4304672 \]
We can verify in r using pbinom with x=0, size = 8, p = 0.1
pbinom(0,8,0.1)## [1] 0.4304672According to the Binomial Distribution there is a 43.05% chance of failure after 8 years. As there are no failures within 8 years, the copy machine will fail some time after that.
The expected value for a binomial distribution is:
\(E[X] = np = 8*0.1 = 0.8\space failures\)
And Standard Deviation is:
\(SD[X] = \sqrt{Var[X]} = \sqrt{npq} =\sqrt{8*0.1*0.9} = 0.8485281\space failures\)
What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
\[ P(X=x) = \frac{\lambda^x}{x!}e^{-\lambda} \] In this case we let x = 0 (0 copy machine failures) such that:
\[ \lambda^0 = 1 \\ 0! =1 \\ P(X=0) = e^{-\lambda} \]
Where \[ \lambda = \frac{np}{t} = 0.8\\ n = 8 \\ p =0.1 \\ t =1 \\ \]
t is the time interval of interest, 1 year, n is the total number of time periods of interest, 8 years, and p is the probability of the event happening within a time period, 0.1 or 10%.
Finally,
\(P(X=0) = e^{-0.8} = 0.449329\)
We can check using ppois with q =0 and lambda =0.8
ppois(0,0.8)## [1] 0.449329According to the Poisson Distribution there is a 44.93% chance of failure after 8 years. As there are no failures within 8 years, the copy machine will fail some time after that.
For the Poisson Distribution the Expected Value is simply \(E[X] =\lambda = 0.8\space failures\) and the Standard Deviation is \(SD[X] = \sqrt{\lambda} = 0.8944272\space failures\)