The exponential probability density function is built upon the general exponential function where the variable is an exponent: \(f(x)=a(b)^x\). This equation can be converted to a natural system of logarithms with a base e that has an approximate value of 2.71828 . The exponential probability density function is valuable with a number of practical applications. The exponential distribution is often concerned with the amount of time until some specific event occurs. For example, in epidemiology, the amount of time (beginning now) until the next confirmed case of a specific communicable disease is reported in a community often follows an exponential distribution. Other examples include the length of time, in minutes, between incoming emergency calls to a 911 dispatch center or the amount of time, in years, until a critical piece of diagnostic imaging equipment, such as an MRI machine, requires a major component replacement. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution.
Values for an exponential random variable occur in a specific pattern: there are fewer large values and many more small values. For example, community health studies have shown that the number of miles a resident travels to reach a local health clinic often follows an exponential distribution. There are many more people who live within a short distance of a clinic (small values) and far fewer people who must travel very long distances to access care (large values).
Exponential distributions are also commonly used in clinical reliability, such as calculating the lifespan of medical implants or prosthetic devices. A manufacturer or surgical team might use exponential distribution analysis to determine the recommended replacement interval for a pacemaker battery based on the probability of the time until the device fails.
The random variable for the exponential distribution is continuous and most often measures the passage of time, though it can be applied to other continuous metrics. In a clinical setting, typical questions might be:
- "What is the probability that a patient will experience a recurring symptom within the next x days?"
- "What is the probability that a rehabilitative session will last between x1 and x2 minutes?"
- "What is the probability that it will take more than x1 hours for a post-surgical patient to achieve a specific mobility milestone?"
In short, the random variable X equals (a) the time between health-related events or (b) the passage of time to complete a clinical action (e.g., waiting for a physician in triage).
It is crucial to differentiate the exponential distribution from the Poisson distribution. The Poisson uses a discrete random variable to give us the probability that a certain number of occurrences (like the number of patients arriving at an ER) will happen in a fixed time period. Conversely, the continuous exponential distribution provides the probability of the length of time (X) that will pass between now and the very next occurrence. We will explore the mathematical link between these two distributions further.
The exponent probability density function is given by:
\[f(x)=\dfrac{1}{\mu} e^{\dfrac{x}{\mu}}\]
where \(\mu\) is the historical average waiting time.
The exponential distribution has a mean and standard deviation both equal to \(\mu\).
To find a probability with a continuous random variable, the cumulative distribution is integrated between the relevant values of \(x\). The definite integral of a function provides the area under the function between two values of \(x\).
For the exponential distribution, there are three cases that depend upon the question asked. Case I would answer the question "What is the probability that the time to completion is less than \(X\) ?" The probability question asked would be written as \(P(X \leq x)\).
For example, to calculate the probability that a patiet will be served within the next hour given that the historical waiting time was 30 minutes, we integrate the exponential probability distribution function \(\dfrac{1}{\mu} e^{\dfrac{x}{\mu}}\) from \(x=0\) to \(x=60\).
\[F(x)=P(X \leq x)=\int_0^x \dfrac{1}{\mu} e^{\dfrac{x}{\mu}} d x=1-e^{\dfrac{x}{\mu}}\]
Case I graphs the exponential probability density function for this example. The area under the function from zero to 60 minutes is the probability for the question asked. Note that the exponential probability distribution at zero is at the value of \(1 / \mu\). Solving for the probability density function equation above where \(\mu=30\) is the historical waiting time, and \(x=60\), which is the time we are interested in.
\[P(x \leq 60)=1-e^{\dfrac{x}{\mu}}=1-e^{\dfrac{60}{30}}=0.865\]
With a historical waiting time of 30 minutes, certainly we would expect the high probability of 0.865 of being waited on in an hour's time. (A probability of 0.865 is almost a certainty.)
Case II would answer questions as "What is the probability more than \(X\) ?" What is the probability that a person will have to wait MORE THAN \(x\) minutes? The relevant probability density function would be
\[P\left(X \geq x_\alpha\right)=1-\left(1-e^{\dfrac{x}{\mu}}\right)=e^{\dfrac{x}{\mu}}\]
and is presented in Figure \(\PageIndex{1}\).
Case II
Imagine a medical device company produces a specific type of cardiac pacemaker lead that is guaranteed for two years (730 days). If the lead fails within this two-year guarantee, the hospital is reimbursed for the cost of the replacement part. Through clinical monitoring, the manufacturer knows the historical mean life of this lead is 750 days (approximately two and a half years).
Is the manufacturer pursuing a financially risky policy by offering a two-year guarantee on a device with such a mean life?
We can calculate the probability that a lead will last longer than 730 days, meaning no reimbursement is required. Using the exponential probability density function, where the historical mean (μ) is 750 days (making the decay parameter m=1/750 or 0.00133), we find the probability shown in Figure \(\PageIndex{2}\).
\[
P(X \geq x)=e^{\dfrac{x}{\mu}}=e^{\dfrac{730}{750}}=0.378
\]
Because the functional life of each pacemaker lead is an independent random variable, we can conclude that only approximately 38 percent of these devices will survive the two-year mark. This means 62 percent of the healthcare facilities using this device will likely be eligible for a full reimbursement under the guarantee. In short, the device’s reliability rate is too low to support such a long guarantee period without causing significant financial loss to the manufacturer. From a health administration perspective, we could expect that this two-year guarantee will either be shortened or the device will be recalled for a quality redesign.
Case III allows us to calculate the probability that a health-related event will occur within a specific window of time. For example, when estimating the recovery time for a patient in a post-anesthesia care unit (PACU), a clinician might be given a range of time, such as "plus or minus from the historical expected recovery time" (measured in hours from the end of surgery).
In this scenario, the mean is the historical average time a patient takes to regain full consciousness and stable vitals. Assume that the historical average recovery time for this specific procedure is four hours. Case III requires us to calculate the probability between two points in time—for example, the probability that a patient will be ready for discharge to a general ward between 3.5 and 4.5 hours after surgery.
Figure \(\PageIndex{3}\): Case III shows the two points that are one hour longer and one hour sooner than the recovery time, which is midway between these points.
If the historical mean recovery time is four hours, we may want to find the probability that a patient’s recovery finishes within a window of x0=3 hours and x1=5 hours. The mathmatical expression is:
\[\begin{array}{c}
P\left(x_0 \leq x \leq x_1\right)=P\left(X \geq x_0\right)-P\left(X \geq x_1\right)=e^{\dfrac{x_0}{\mu}}-e^{\dfrac{x_1}{\mu}} \\
P(3 \leq x \leq 5)=e^{\dfrac{3}{4}}-e^{\dfrac{5}{4}}=0.186
\end{array}\]
The probability that a patient’s recovery time will fall within the window of plus or minus one hour of the four-hour historical average (between 3 and 5 hours) is only 0.186. Intuitively, this seems like a very small probability to hit such a broad two-hour window. At its core, this tells us that the variability in patient recovery times is quite high, meaning the standard deviation is large. In clinical practice, this high variance makes it difficult to predict exact bed turnover times based on the mean alone.
Exponential Distribution and Decay Factor
The core of the exponential distribution is e, the natural logarithm in an exponential function with the variable \(-\dfrac{x}{\mu}\) in the exponent. The exponent is negative, and thus it describes an exponentially declining function as we have seen in Figure \(\PageIndex{1}\), Figure \(\PageIndex{2}\), and Figure \(\PageIndex{3}\) above.
This gives rise to an alternative formula for the exponential probability distribution:
\[\begin{array}{c}
f(x)=m e^{-m} \\
F(x)=\int_0^x m e^{-m t} d t=1-e^{-m x}
\end{array}\]
Where \(m=\dfrac{1}{\mu}\) is given the name decay factor and measures the speed of decay. The decay factor simply measures how rapidly the probability of an event declines as the random variable \(X\) increases.
Figure \(\PageIndex{4}\) shows the exponential distribution for three different decay factors. As before, the intercept on the vertical axis, the probability density, is the decay factor. A decay factor of 1.5 , as in the first exponential function, is three time the speed of the third exponential distribution with a decay factor of 0.5 . Looking at the probability for each of the three functions at the same value of \(x_1\), it is clear that the first function has a significantly greater probability of occurrence at \(x_1\) than the other two (i.e., \(P\left(X_\alpha \leq x_1\right)>P\left(X_b \leq x_1\right)>P\left(X_c \leq x_1\right)\) comparing the relevant probabilities for each of the functions \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) in order). Finally, the exponential distribution shows with drama the true meaning of "exponential." In each unit, a change in \(x\) in an exponential distribution has a significant change in probability. For example, if the simple exponential distribution function \(Y=2^x\) changes from \(x=4\) to \(x=5\), the \(y\)-value increases from 16 to 32 , but a one-unit change from \(x=10\) to \(x=11\) causes the \(y\)-value to increase from 1,024 to 2,048 . Likewise, in an exponential function with a negative exponent such as the exponential probability distribution, the changes are startling for each unit change at higher levels of \(X\). At \(X=-4\) to \(x=-5\), the \(y\)-value changes from 0.063 to 0.031 , while a one-unit change from \(X=-10\) to \(X=-11\) results in a change in the \(y\)-value from 0.00097 to 0.00048 .
The conclusion is that probabilities in an exponential probability distribution at large \(x\)-values are scarce almost to the point of nonexistence.
Let X= the amount of time (in minutes) a triage nurse spends performing an initial assessment on a patient. The time is known from historical hospital data to have an average duration equal to four minutes.
It is given that μ=4 minutes; that is, the average time the nurse spends with a patient is 4 minutes. Because we are performing probability analysis, we must work with known population parameters like the mean. Knowing this historical mean allows for the calculation of the decay parameter, m.
\[m=\dfrac{1}{\mu} . \text { Therefore, } m=\dfrac{1}{4}=0.25\]
When the notation used the decay parameter, \(m\), the probability density function is presented as \(f(x)=m e^{-m x}\), which is simply the original formula with \(m\) substituted for \(\dfrac{1}{\mu}\), or \(f(x)=\dfrac{1}{\mu} e^{-\dfrac{1}{\mu} x}\).
To calculate probabilities for an exponential probability density function, we need to use the cumulative density function. As shown below, the curve for the cumulative density function is \(f(x)=0.25 e^{-0.25 x}\) where \(x\) is at least zero and \(m=0.25\).
For example, \(f(5)=0.25 e^{(-0.25)(5)}=0.072\). In other words, the function has a value of .072 when \(x=5\).
The graph is as follows:
Notice the graph is a declining curve. When \(x=0\), \(f(x)=0.25 e^{(-0.25)(0)}=(0.25)(1)=0.25=m\). The maximum value on the \(y\)-axis is always \(m\), one divided by the mean.
The amount of time a physical therapist spends performing a preliminary gait analysis screening can be modeled by an exponential distribution with the average amount of time equal to eight minutes. Write the distribution, state the probability density function, and graph the distribution.
Using the information in Exercise, find the probability that a triage nurse spends performing an initial assessment on a patient.
- Answer
-
Find \(P(4<x<5)\).
The cumulative distribution function (CDF) gives the area to the left.
\[\begin{array}{l}
P(x<x)=1-e^{-m x} \\
P(x<5)=1-e^{(-0.25)(5)}=0.7135 \text { and } P(x<4)=1-e^{(-0.25)(4)}=0.6321 \\
P(4<x<5)=0.7135-0.6321=0.0814
\end{array}\]Figure \(\PageIndex{6}\):
On average, a specific type of titanium hip replacement component lasts ten years. The length of time the component lasts is exponentially distributed.
What is the probability that a hip replacement component lasts more than 7 years?
- Answer
-
Figure \(\PageIndex{7}\):
A motorized wheelchair uses a specific type of rechargeable battery pack. On average, one battery pack lasts ten years before it can no longer hold a charge. The lifespan of these packs is exponentially distributed.
If a patient has a supply of five battery packs to be used one after another (replacing each only when the previous one fails), on average, how long will the total supply last?
Answer:
Because the expected value (mean) of the sum of independent random variables is the sum of their individual means:
- On average, one battery pack lasts ten years.
- Therefore, five battery packs used sequentially would last, on average:
(5)(10)=50 years
What is the probability that a wheelchair battery lasts between nine and 11 years?
- Answer
-
d. Find \(P(9<x<11)\). Draw the graph.
Figure \(\PageIndex{8}\): \[P(9<x<11)=P(x<11)-P(x<9)=\left(1-e^{(-0.1)(11)}\right)-\left(1-e^{(-0.1)(9)}\right)=0.6671-0.5934=0.0737 . \]
The probability that a wheelchair battery lasts between nine and 11 years is 0.0737 .
Suppose that the length of time a patient spends using a self-service health screening kiosk, in minutes, is an exponential random variable with a decay parameter \(\dfrac{1}{12}\). The decay parameter is another way to view \(1 / \lambda\). If you arrive at the kiosk just as another person begins their screening, find the probability that you will have to wait more than five minutes. Let X= the length of a screening session in minutes.
What is \(m, \mu\), and \(\sigma\) ? The probability that you must wait more than five minutes is \(\qquad\).
- Answer
-
- \(m=\dfrac{1}{12}\)
- \(\mu=12\)
- \(\sigma=12\)
\[P(x>5)=0.6592\]
The time spent waiting between events is often modeled using the exponential distribution. For example, suppose that an average of 30 patients per hour arrive at a clinic and the time between arrivals is exponentially distributed.
Problem
- On average, how many minutes elapse between two successive arrivals?
- When the clinic first opens, how long on average does it take for three patients to arrive?
- After a patient arrives, find the probability that it takes less than one minute for the next patient to arrive.
- After a patient arrives, find the probability that it takes more than five minutes for the next patient to arrive.
- Is an exponential distribution reasonable for this situation?
- Answer
-
a. Since we expect 30 patients to arrive per hour ( 60 minutes), we expect on average one patient to arrive every two minutes on average.
b. Since one patient arrives every two minutes on average, it will take six minutes on average for three patients to arrive.
c. Let \(X=\) the time between arrivals, in minutes. By part a, \(\mu=2\), so \(m=\dfrac{1}{2}=0.5\).
The cumulative distribution function is \(P(X<x)=1-e^{(-0.5)(x)}\)
Therefore \(P(X<1)=1-\mathrm{e}^{(-0.5)(1)}=0.3935\).Figure \(\PageIndex{9}\): d. \(P(X>5)=1-P(X<5)=1-\left(1-e^{(-0.5)(5)}\right)=e^{-2.5} \approx 0.0821\).
Figure \(\PageIndex{10}\): e. This model assumes that a single customer arrives at a time, which may not be reasonable since people might shop in groups, leading to several patients arriving at the same time. It also assumes that the flow of patients does not change throughout the day, which is not valid if some times of the day are busier than others.
Memorylessness of the Exponential Distribution
Recall that the amount of time between patients for the clinic discussed earlier is exponentially distributed with a mean of two minutes. Suppose that five minutes have elapsed since the last patient arrived. Since an unusually long amount of time has now elapsed, it would seem to be more likely for a patient to arrive within the next minute. With the exponential distribution, this is not the case-the additional time spent waiting for the next patient does not depend on how much time has already elapsed since the last patient. This is referred to as the memoryless property. The exponential and geometric probability density functions are the only probability functions that have the memoryless property. Specifically, the memoryless property says that
\(P(X>r+t \mid X>r)=P(X>t)\) for all \(r \geq 0\) and \(t \geq 0\)
For example, if five minutes have elapsed since the last patient arrived, then the probability that more than one minute will elapse before the next patient arrives is computed by using \(r=5\) and \(t=1\) in the foregoing equation.
\[P(X>5+1 \mid X>5)=P(X>1)=e^{(-0.5)(1)}=0.6065 .\]
This is the same probability as that of waiting more than one minute for a patient to arrive after the previous arrival.
The exponential distribution is often used to model the longevity of an electrical or mechanical medical device. In our previous example, the functional lifetime of a cardiac pacemaker lead followed an exponential distribution with a mean of ten years. The memoryless property states that knowledge of what has occurred in the past has no effect on future probabilities. In the context of this pacemaker, it implies that a lead that has already been functioning for several years is "statistically" no more likely to fail in the next hour than a brand-new lead. In other words, the device does not "wear out" gradually in a way the distribution can track; it stays functionally "as good as new" until the moment it suddenly fails.
For example, if a patient’s pacemaker lead has already lasted ten years, then the probability that it lasts another seven years is \(P(X>17 \mid X>10)=P(X>7)=0.4966\), where the vertical line is read as "given".
Refer back to our previous analysis of clinic efficiency, where the time a physician spends with a patient follows an exponential distribution with a mean of four minutes.
Suppose a patient has already spent four minutes in consultation with the physician. What is the probability that the consultation will last for at least an additional three minutes?
The decay parameter of \(X\) is \(m=\dfrac{1}{4}=0.25\), so \(X \sim \operatorname{Exp}(0.25)\).
The cumulative distribution function is \(P(X<x)=1-e^{-0.25 x}\).
We want to find \(P(X>7 \mid X>4)\). The memoryless property says that \(P(X>7 \mid X>4)=P(X>3)\), so we just need to find the probability that a physican spends more than three minutes with a patient.
This is \(P(X>3)=1-P(X<3)=1-\left(1-e^{-0.25 \cdot 3}\right)=e^{-0.75} \approx 0.4724\).
There is an interesting relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of \(\mu\) units of time. Also assume that these times are independent, meaning that the time between events is not affected by the times between previous events. If these assumptions hold, then the number of events per unit time follows a Poisson distribution with mean \(\mu\). Recall that if \(X\) has the Poisson distribution with mean \(\mu\), then \(P(X=x)=\dfrac{\mu^x e^{-\mu}}{x!}\).
The formula for the exponential distribution: \(P(X=x)=m e^{-m x}=\dfrac{1}{\mu} e^{-\dfrac{1}{\mu} x}\) Where \(m=\) the rate parameter, or \(\mu\) = average time between occurrences.
We see that the exponential is the cousin of the Poisson distribution and they are linked through this formula. There are important differences that make each distribution relevant for different types of probability problems.
First, the Poisson has a discrete random variable, \(x\), where time; a continuous variable is artificially broken into discrete pieces. We saw that the number of occurrences of an event in a given time interval, \(x\), follows the Poisson distribution.
For example, the Poisson distribution is used to count discrete events over a fixed period. For example, a hospital administrator might use it to track the number of patients admitted to the Emergency Room per hour. By contrast, the exponential distribution focuses on the time between those occurrences. Instead of counting how many people arrive, we ask: "A patient just checked in; how long will it be until the next patient arrives?" Here, we are measuring the length of the interval, which is a continuous random variable.
The Exponential Distribution v. the Poisson Distribution
A visual way to show both the similarities and differences between these two distributions is with a time line.
The random variable for the Poisson distribution is discrete and thus counts events during a given time period, \(t_1\) to \(t_2\) on Figure 5.24, and calculates the probability of that number occurring. The number of events, four in the graph, is measured in counting numbers; therefore, the random variable of the Poisson is a discrete random variable.
The exponential probability distribution calculates probabilities of the passage of time, a continuous random variable. In Figure 5.24 this is shown as the bracket from \(t_1\) to the next occurrence of the event marked with a triangle.
Classic Poisson distribution questions are "how many patients will arrive at my clinic in the next hour?".
Classic exponential distribution questions are "how long it will be until the next patient arrives," or a variant, "how long will the patient remain here once they have arrived?".
Again, the formula for the exponential distribution is:
\[f(x)=m e^{-m x} \operatorname{or} f(x)=\dfrac{1}{\mu} e^{-\dfrac{1}{\mu} x}\]
We see immediately the similarity between the exponential formula and the Poisson formula.
\[P(x)=\dfrac{\mu^x e^{-\mu}}{x!}\]
Both probability density functions are based upon the relationship between time and exponential growth or decay. The " \(e\) " in the formula is a constant with the approximate value of 2.71828 and is the base of the natural logarithmic exponential growth formula. When people say that something has grown exponentially this is what they are talking about.
An example of the exponential and the Poisson will make clear the differences been the two. It will also show the interesting applications they have.
Poisson Distribution
Suppose that historically 10 patients arrive at the clinic each hour. Remember that this is still probability so we have to be told these historical values. We see this is a Poisson probability problem.
We can put this information into the Poisson probability density function and get a general formula that will calculate the probability of any specific number of patients arriving in the next hour.
The formula is for any value of the random variable we chose, and so the \(x\) is put into the formula. In this example, \(\mu=10\) because we expect 10 patients to arrive each hour. This is the formula:
\[f(x)=\dfrac{10^x e^{-10}}{x!}\]
As an example, the probability of 15 patients arriving in the next hour would be
\[P(x=15)=\dfrac{10^{15} e^{-10}}{15!}=0.0347\]
Here we have inserted \(x=15\) and calculated the probability that in the next hour 15 people will arrive is 0.035.
Exponential Distribution
If we keep the same historical facts that 10 patients arrive each hour, but we now are interested in the time a physician spends with each patient, then we would use the exponential distribution. The exponential probability function for any value of \(x\), the random variable, for this particular clinic historical data is:
\[f(x)=\dfrac{1}{.1} e^{-x / 1}=10 e^{-10 x}\]
To calculate \(\mu\), the historical average patient time we simply divide the number of patients that arrive per hour, 10 , into the time period, one hour, and have \(\mu=0.1\). Historically, people spend 0.1 of an hour at the clinic, or 6 minutes. This explains the .1 in the formula.
There is a natural confusion with \(\mu\) in both the Poisson and exponential formulas. They have different meanings, although they have the same symbol. The mean of the exponential is one divided by the mean of the Poisson. If you are given the historical number of arrivals you have the mean of the Poisson. If you are given an historical length of time between events you have the mean of an exponential.
Continuing with our example at the clinic; if we wanted to know the probability that a physician would spend 9 minutes or less with a patient, then we use this formula.
First, we convert to the same time units which are parts of one hour. Nine minutes is 0.15 of one hour. Next we note that we are asking for a range of values. This is always the case for a continuous random variable. We write the probability question as:
\[p(x \leq 9)=1-10 e^{-10 x}\]
We can now put the numbers into the formula and we have our result.
\[p(x=.15)=1-10 e^{-10(.15)}=0.7769\]
The probability that a physician will spend 9 minutes or less with a patient is 0.7769 .
We see that we have a high probability of being seen in less than nine minutes and a tiny probability of having 15 patients arriving in the next hour.
At a regional EMS Dispatch Center, emergency medical calls arrive at an average rate of four calls per minute. We are concerned only with the rate of arrival and assume the time spent between calls is independent (the "memoryless" property). This allows us to use the Poisson distribution for the number of calls and the exponential distribution for the time between them.
Problem
- Find the average time between two successive emergency calls.
- Find the probability that after a call is received, the next call occurs in less than ten seconds.
- Find the probability that exactly five calls occur within a minute.
- Find the probability that fewer than five calls occur within a minute.
- Find the probability that more than 40 calls occur in an eight-minute period.
- Answer
-
a. On average, four calls occur per minute, so one call occurs every 15 seconds, or \(=0.25\) minutes occur between successive calls on average.
b. Let \(T=\) time elapsed between calls. From part a, \(\mu=0.25\), so \(m=4\). Thus, \(T \sim \operatorname{Exp}(4)\). The cumulative distribution function is \(P(T<t)=1-e-4 t)\).
The probability that the next call occurs in less than ten seconds ( ten seconds \(=\dfrac{1}{6}\) minute) is \(P\left(T<\dfrac{1}{6}\right)=1-e^{-4\left(\dfrac{1}{6}\right)}=0.4866\).
Figure \(\PageIndex{13}\): c. Let \(X=\) the number of calls per minute. As previously stated, the number of calls per minute has a Poisson distribution, with a mean of four calls per minute.
Therefore, \(X \sim\) Poisson(4) and so
\[P(X=5)=\dfrac{4^5 e^{-4}}{5!} \approx 0.1563 . \quad 5!=(5)(4)(3)(2)(1)\]
d. Keep in mind that \(X\) must be a whole number, so \(P(X<5)=P(X \leq 4)\).
To compute this, we could take
\[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)\]
Using technology, we see that \(P(X \leq 4)=0.6288\).
e. Let \(\mathrm{Y}=\) the number of calls that occur during an eight-minute period.
Since there is an average of four calls per minute, there is an average of (8)(4) = 32 calls during each eight-minute period.
Hence, \(Y \sim\) Poisson (32). Therefore,
\[P(Y>40)=1-P(Y \leq 40)=1-0.9294=0.0707\]