First, note that the variables are discrete, taking on integer values such that \(X_i \in [1,k]\) .
The question is asking for the distribution of the order statistic \(Y = min(X_i) = X_{(1)}\) .
Because each of the \(X_i\) is distributed uniformly on \([1,k]\), the probability that any individual \(X_i\) equals 1 is \(\frac{1}{k}\), and thus the probability that any individual \(X_i\) is greater than 1 is \(\frac{k-1}{k}\) .
\[Pr(X_i=1)=\frac{1}{k},\quad for \ each \ i, 1 \le i \le n\] \[Pr(X_i>1)=\frac{k-1}{k},\quad for \ each \ i, 1 \le i \le n\]
In order for the minimum of all \(X_i\) to be greater than 1, this requires that all \(X_i\) are greater than 1, which would happen with probability \(\left( \frac{k-1}{k} \right)^n\) .
\[Pr(Y>1)=Pr(min(X_i)>1)=Pr(X_1>1\ ; \ X_2>1 \ ; \ ... \ ; \ X_n>1)=\left( \frac{k-1}{k} \right)^n\]
Therefore, \[Pr(Y=1)=1-Pr(Y>1)=1-\left( \frac{k-1}{k} \right)^n\] .
Note that this can be written as \[\left( \frac{k-0}{k} \right)^n - \left( \frac{k-1}{k} \right)^n\] .
The probability that any individual \(X_i\) is greater than 2 is \(\frac{k-2}{k}\) , so the probability that \(X_i > 2, \forall i\) is \(\left( \frac{k-2}{k} \right)^n = Pr(min(X_i)>2)=Pr(Y>2)\).
So, the probability that the minimum is equal to 2 is \[\begin{aligned} Pr(Y=2) &= 1 - Pr(Y>2) - Pr(Y=1) \\ &= 1 - \left( \frac{k-2}{k} \right)^n - \left[ 1-\left( \frac{k-1}{k} \right)^n \right] \\ &= \left( \frac{k-1}{k} \right)^n - \left( \frac{k-2}{k} \right)^n \end{aligned}\]
\[ \begin{aligned} Pr(Y=min(X_i)=3) &= 1- Pr(Y>3) - Pr(Y=2) - Pr(Y=1)\\ &= 1 - \left( \frac{k-3}{k} \right)^n - \left[ \left( \frac{k-1}{k} \right)^n - \left( \frac{k-2}{k} \right)^n \right] - \left[ 1 - \left( \frac{k-1}{k} \right)^n\right] \\ &= 1 - \left( \frac{k-3}{k} \right)^n - \left( \frac{k-1}{k} \right)^n + \left( \frac{k-2}{k} \right)^n - 1 + \left( \frac{k-1}{k} \right)^n \\ &= \left( \frac{k-2}{k} \right)^n - \left( \frac{k-3}{k} \right)^n \end{aligned} \]
\[Pr(Y=min(X_i)=y) = \left( \frac{k-y+1}{k} \right)^n - \left( \frac{k-y}{k} \right)^n\]
Probability of 1 failure in 10 years –> Probability of failure in 1 year is 1/10 = 10% = 0.1 .
Annual = 10
pAnnual = 1/Annual # 0.1
qAnnual = 1-pAnnual # 0.9
YearsToFail = 8
pgeom(YearsToFail-1,pAnnual) # 0.56953279## [1] 0.56953279## [1] 0.56953279## [1] 0.43046721# annual table
cbind(year=0:10,
Prob_Fail=pgeom(0:10,pAnnual),
Prob_Not_fail=pgeom(0:10,pAnnual,lower.tail=F)) %>%
kable() %>% kable_styling(c("striped", "bordered"))| year | Prob_Fail | Prob_Not_fail |
|---|---|---|
| 0 | 0.10000000000 | 0.90000000000 |
| 1 | 0.19000000000 | 0.81000000000 |
| 2 | 0.27100000000 | 0.72900000000 |
| 3 | 0.34390000000 | 0.65610000000 |
| 4 | 0.40951000000 | 0.59049000000 |
| 5 | 0.46855900000 | 0.53144100000 |
| 6 | 0.52170310000 | 0.47829690000 |
| 7 | 0.56953279000 | 0.43046721000 |
| 8 | 0.61257951100 | 0.38742048900 |
| 9 | 0.65132155990 | 0.34867844010 |
| 10 | 0.68618940391 | 0.31381059609 |
\(p=0.1\) ; \(q=1-p = 0.9\) ; \(Pr(X=n)=p \cdot q^{(n-1)}\)
Probability of failing within the first 8 years (where the first year is enumerated by 0, the second year by 1, … the eighth year by 7): \[Pr(X<8) = \sum \limits _{i=0}^7 {p \cdot q^i} =p \cdot\sum \limits _{i=0}^7 { q^i} =p \left[\frac{1-q^8}{1-q}\right] =p \left[\frac{1-q^8}{p}\right] =1-q^8 =1-(0.9)^8 =1-.43046721 =0.56953279 \]
This is pgeom(7,1/10):
\(Pr(X \ge 8) = 1-Pr(X<8) = 1-(1-q^8)=q^8=(0.9)^8=0.43046721\) .
This is 1-pgeom(YearsToFail-1,pAnnual) = 1-pgeom(YearsToFail-1,pAnnual,lower.tail=FALSE) :
## [1] 0.43046721## [1] 0.43046721The formula for the expected value of a geometric distribution where there are k failures is \(E[x] = \mu = \frac{(1-p)}{p} =\frac{q}{p}\)
## [1] 9The formula for the variance of a geometric distribution where there are k failures is \(E[x] = \mu = \frac{(1-p)}{p^2} =\frac{q}{p^2}\)
## [1] 90## [1] 9.48683298051This result indicates that the expected time to failure is 9 years – however, we have made the assumption that failure could only occur at the beginning of each year. (i.e., when performing the averaging for the expected value, the probability of failure in the first year is multiplied by zero, which implied that such failure occurs IMMEDIATELY, rather than at some random time during the year.)
To be more realistic, it would be better to assume that the failure could occur any time within each year, which averages out to be at the middle of the year, making the expected time to fail closer to 9.5.
However, we expect to obtain a value of 10 because that was the initial time-to-failure included in the problem.
To get this, we need to take smaller timesteps:
Consider daily rather than annual. Then:
DaysInYear = 365.25
Daily = Annual * DaysInYear # 3652.5
pDaily = 1/Daily # 0.1 / 365.25 = 0.000273785078713
qDaily = 1-pDaily # 0.9 / 365.25 = 0.999726214921
DaysToFail = YearsToFail * DaysInYear # 8 * 365.25 = 2922
pgeom(DaysToFail-1,pDaily) # 0.550720249997 for daily, rather than 0.56953279 for annual## [1] 0.550720249997## [1] 0.550720249997## [1] 0.449279750003## [1] 0.5507202499971-pgeom(DaysToFail-1,pDailyAnnual) = 1-pgeom(DaysToFail-1,pDaily,lower.tail=FALSE) :
## [1] 0.449279750003## [1] 0.449279750003The formula for the expected value of a geometric distribution where there are k failures is \(E[x] = \mu = \frac{(1-p)}{p} =\frac{q}{p}\)
## [1] 3651.5# (This is one day less than 10 years.)
# Expected Years until failure:
expectedval_Daily_in_Years = expectedval_Daily / DaysInYear
expectedval_Daily_in_Years## [1] 9.99726214921This is much closer to the 10-year result that we were expecting.
The formula for the variance of a geometric distribution where there are k failures is \(E[x] = \mu = \frac{(1-p)}{p^2} =\frac{q}{p^2}\)
## [1] 13337103.75## [1] 3651.99996577# Stdev Daily (units expressed in years)
stdev_Daily_in_Years = stdev_Daily / DaysInYear
stdev_Daily_in_Years## [1] 9.9986309809The exponential distribution is quite similar to the geometric distribution, except it is continuous, while the geometric distribution is discrete.
The PDF is
\({\displaystyle f(x;\lambda )={\begin{cases}\lambda e^{-\lambda x}&x\geq 0,\\0&x<0.\end{cases}}}\)
and the CDF is \({\displaystyle F(x;\lambda )={\begin{cases}1-e^{-\lambda x}&x\geq 0,\\0&x<0.\end{cases}}}\)
Because the expected life of the device is 10 years, \(\lambda = \frac{1}{10}=0.1\) .
The probability that the device does not fail in the first 8 years is
## [1] 0.550671035883so the probability that it fails AFTER 8 years is
## [1] 0.449328964117For the exponential distribution, the expected value is \(E[X]=\frac{1}{\lambda}=10\) and the variance is \(VAR[X]=\frac{1}{\lambda^2}=100\) .
Thus, the standard deviation of the exponential is \(SD[X] = \sqrt{VAR[X]} =\sqrt {100}=10\) .
For the Binomial distribution, we again face the granularity problem that became apparent above when looking at the geometric distribution.
The probability mass function for the binomial is \({\displaystyle f(k,n,p)=\Pr(k;n,p)=\Pr(X=k)={\binom {n}{k}}p^{k}(1-p)^{n-k}}\)
If we are again considering annual results, we have zero successes in 8 years, where the probability is \(p=\frac{1}{10}=0.1\) .
This gives: \({\displaystyle f(0,8,0.10)=\Pr(0;8,0.10)=\Pr(X=0)={\binom {8}{0}}(0.10)^{0}(1-0.10)^{8-0}}=(0.9)^8=0.43046721\)
which is the same result as obtained from the Geometric above (under annual failures.)
## [1] 0.43046721When possible failures are only considered on an annual basis,
The expected value of the binomial distribution is \(E[X]=n\cdot p = 8 \cdot \frac{1}{10}=0.8\)
and the variance of the binomial is \(VAR[X] = n \cdot p \cdot (1-p) = n \cdot p \cdot q = 8 \cdot 0.1 \cdot 0.9 = 0.72\) .
Thus, the standard deviation is \(SD[X] = \sqrt{VAR[X]} =\sqrt {0.72} = 0.848528137424\) .
To obtain greater granularity, we could consider daily, rather than annual, opportunities for failure.
Again, we have
DaysInYear = 365.25
Daily = Annual * DaysInYear # 3652.5
pDaily = 1/Daily # 0.1 / 365.25 = 0.000273785078713
qDaily = 1-pDaily # 0.9 / 365.25 = 0.999726214921
DaysToFail = YearsToFail * DaysInYear # 8 * 365.25 = 2922
#### pbinom(k,n,p)
pbinom(0,DaysToFail,pDaily) # 0.449279750003 for daily, rather than 0.43046721 for annual## [1] 0.449279750003This is the same value as obtained above for the geometric distribution, when possible daily (rather than annual) failures are considered.
As the time interval becomes smaller, we eventually approach continuous time, as measured by the exponential model.
If we instead perform the above calculations on an hourly, minutely, or secondly basis, the results should converge.
The expected value of the binomial distribution is \(E[X]=n\cdot p\) , which here is 2922 * 0.000273785079 = 0.8 .
This is unchanged from the annual value.
## [1] 0.8The Daily variance of the binomial is
\[\begin{aligned} VAR[X] &= \left(n*365.25 \right) \cdot \left(\frac{p}{365.25}\right) \cdot \left(1-\frac{p}{365.25}\right) \\ &= 2922 \cdot 0.000273785079 \cdot 0.999726214921 \\ &= 0.799780971937 \end{aligned} \] .
## [1] 0.799780971937## [1] 0.894304742209Thus, the standard deviation is \(SD[X] = \sqrt{VAR[X]} =\sqrt {0.799780971937} = 0.894304742209\) .
The probability mass function for the Poisson distribution is \({\displaystyle \!f(k;\lambda )=\Pr(X=k)={\frac {\lambda ^{k}e^{-\lambda }}{k!}}}\)
We seek to know whether no failures have occurred during the first 8 years, or \(Pr(X=0| t=8 ; \lambda = \frac{1}{10}=0.10)\)
The probability mass function for the Poisson distribution across non-unit time is \({\displaystyle \!f(k;\lambda t )=\Pr(X=k)={\frac {(\lambda t) ^{k}e^{-\lambda t}}{k!}}}\)
Since here we are fixing \(k=0\), the above simply becomes \[{\displaystyle \!f(0;\lambda t )=\Pr(X=0)={\frac {(\lambda t)^{0}e^{-\lambda t}}{0!}}=e^{-\lambda t}=e^{-(0.1) 8}=e^{-0.8}=0.449328964117}\]
## [1] 8## [1] 0.1## [1] 0.8## [1] 0.449328964117Note that this result matches that from the exponential calculation above.
For the poisson distribution, the mean is \(E[X]=\mu = \lambda = 0.8\) and the variance is also \(VAR[X] = \lambda = 0.8\) .
Therefore, the standard deviation is \(SD[X] = \sqrt{VAR[X]} =\sqrt {0.8}= 0.894427191\) .