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Essentials of Mathematical Probability and Statistics

John Travis

Section 8.3 Exponential Distribution

Once again, consider a Poisson Process where you start with an interval of variable length X so that X measures the interval needed in order to obtain a first success with \(R = (0,\infty)\text{.}\) The resulting distribution of X will be called an Exponential distribution.
To derive the probability function for this distribution, consider finding f(x) by first considering F(x).
We found that if \(\lambda\) is the parameter for the Poisson process, then the Poisson distribution had mean \(\lambda T\) or if one presumed that T=1, then for that distribution, the mean is just \(\lambda\text{.}\) We will find out below that the mean for the exponential distribution will be \(\frac{1}{\lambda}\text{.}\) Therefore, we will eventually present this formula using the exponental mean \(\mu = \frac{1}{\lambda}\) rather than using \(\lambda\) in the actual formula.
You will also often find that exercises in other textbooks and in online WeBWorK will just provide \(\lambda\) for the underlying Poisson process and that the time interval T will be presumed to be T=1. For those problems, if the exercise asks for a Poisson probability, use \(\mu = \lambda\) while if the exercise asks for an Exponential probability, use \(\mu = \frac{1}{\lambda}\text{.}\)
\begin{align*} & \int_0^{\infty} \lambda e^{-\lambda x} dx\\ & = \int_0^{\infty} e^{-u} du\\ & = -e^{-u} \big |_0^{\infty} = 1 \end{align*}
where we used the substitution \(u = \lambda x\text{.}\)
For the mean, use integration by parts with \(u = x\) and \(dv = \lambda e^{-\lambda \cdot x}\) to get (eventually)
\begin{align*} \text{Mean} = \mu & = \int_0^{\infty} x \cdot \lambda e^{-\lambda x} dx\\ & = \left [ -(x+\frac{1}{\lambda}) e^{-\lambda \cdot x} \right ] \big |_0^{\infty} \\ & = \frac{1}{\lambda}. \end{align*}
and so the use of \(\lambda = \frac{1}{\mu}\) in f(x) is warranted.
The remaining statistics are derived similarly using repeated integration by parts. The interactive Sage cell below calculates those for you automatically.
Simply replace the mean as derived above in the original function.
Using \(f(x) = \frac{1}{\mu} e^{-\frac{x}{\mu}}\text{,}\) note
\begin{align*} F(x) & = \int_0^x \frac{1}{\mu} e^{-\frac{u}{\mu}} du\\ & = - e^{-\frac{u}{\mu}} \big |_0^x\\ & = 1 - e^{-\frac{x}{\mu}} \end{align*}
Once again, let’s consider a router which, over time, has been shown to receive on average 1000 such requests in any given 10 minute period during regular working hours. This would mean that, on average, it would take \(\mu = \frac{10}{1000} = \frac{1}{100} = 0.01\) minutes (i.e., less than a second) to receive the first request. If X were to measure the time interval until the first actual request comes in, then the Exponential distribution would be a good model using
\begin{equation*} f(x) = \frac{1}{0.01} e^{-\frac{x}{0.01}}. \end{equation*}
Let’s determine the probability that a first request arrives in the next two seconds. First, note that since X is a continuous variable that f(x) is NOT the probability of exactly X minutes but you must integrate to compute all probabilities. Also, the next 2 seconds is actually the next \(\frac{2}{60} = \frac{1}{30}\) of a minute. Therefore, F(x) is what you need in general and you find
\begin{equation*} P(X \le \frac{1}{30}) = F(\frac{1}{30}) = 1 - e^{-\frac{\frac{1}{30}}{0.01}} \approx 0.96433. \end{equation*}
Suppose that the time (in hours) required to repair a machine is an exponentially distributed random variable with parameter \(\lambda = 0.8\text{.}\) What is
(a) the probability that a repair time exceeds \(10\) hours?
(b) the conditional probability that a repair takes at least \(9\) hours, given that it takes more than \(6\) hours?
Answer 1.
\(0.000335462627902512\)
Answer 2.
\(0.0907179532894125\)
Using the definition of conditional probability,
\begin{align*} P( X > a + b | X > b ) & = P( X > a + b \cap X > b ) \ P( X > b)\\ & = P( X > a + b ) / P( X > b)\\ & = e^{-(a+b)/ \mu} / e^{-b / \mu}\\ & = e^{-a/ \mu}\\ & = P(X > a) \end{align*}
It can be presumed that the life span in years of a certain brand of lightbulbs follows the Poisson process assumptions. Suppose that the mean life span till failure is known to be 0.7 years. A tester makes random observations of the life times of this particular brand of lightbulbs and records them one by one as either a success if the life time exceeds 1 year, or as a failure otherwise.
Determine the probability that a randomly tested bulb lasts more than 1 year:
Determine probability that the first success occurs in the fifth observation:
Determine the probability that the second success occurs in the 8th observation given that the first success occurred in the 3rd observation:
Determine the probability that the first success occurs in an odd-numbered observation:
Hint.
Use the exponential distribution to get the first answer. Use the geometric to get the remainder.
Answer 1.
\(0.2397\)
Answer 2.
\(0.0801\)
Answer 3.
\(0.0801\)
Answer 4.
\(0.5681\)
Solution.
Using the exponential distribution,
\begin{equation*} p = P(X \gt 1) = 1 - F(1) = 1 - (1-e^{-\frac{1}{0.7}}) \approx 0.2397. \end{equation*}
Now, use the geometric distribution with \(p = 0.2397\text{...}\)
\begin{equation*} P(X = 5) = f(5) = (1-0.2397)^4 \cdot 0.2397 \approx 0.0801. \end{equation*}
\begin{equation*} P(\text{2nd success on 8} | \text{1st success on 3}) = P(X = 5) \approx 0.0801 \end{equation*}
\begin{equation*} \begin{aligned} P(\text{1st success on an odd trial}) \amp = f(1) + f(3) + f(5) + ...\\ \amp = p + (1-p)^2 \cdot p + (1-p)^4 \cdot p + (1-p)^6 \cdot p + ... \\ \amp = p \sum_{x=0}^{\infty} \left ( (1-p)^2 \right)^k \\ \amp = p \frac{1}{1-(1-p)^2} \approx 0.5681 \end{aligned} \end{equation*}
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