# Exponential/Continuous Distributions

• whitehorsey
In summary: At first you might think that this would be the probability that 't' goes by and nothing happens---i.e. that the time between successive events is at least 't'. But you'd be wrong. The reason is that this is a continuous distribution, not a discrete one; so we have to ask for something else: what is the probability that at least time 't' goes by *without* an event having occurred? This is the probability that the first event occurs after 't'. This is called the survival function, and is denoted as S(t) (or you might also see Fbar(t) or 'P(T>t)').So, for the discrete case, if you have a
whitehorsey
1. A particular nuclear plant releases a detectable amount of radioactive gases twice a month on the average. Find the probability that at least 3 months will elapse before the release of the first detectable emission. What is the average time that one must wait to observe the first emission?

2. P[X ≤ x] = 1 - e-αx

3. Find the probability that at least 3 months will elapse before the release of the first detectable emission.
P[T ≥ 3] = 1 - P[T ≤ 2]
= 1 - (1 - e-.5(2)) - (1 - e-.5(1)) - (1 - e-.5(0))
= -0.0256
I ended up getting a negative number which shouldn't happen but I'm not sure what's wrong.

What is the average time that one must wait to observe the first emission?
P[T = 1] = 1 - e-.5(1)
= 0.3935?

The negative number appeared since you are considering the distribution to be discrete when it's not. P(T<=3) = 1 - P(T<3) = 1-P(T<=3) since the distribution is continuous.

In 2. I think you are not calculating what you think you are... First of all P(T=1) = P(X<=1)-P(X<1) = 0.
You really are computing P(T<=1). Makes sense?

As for the average time before the first emission can't say I am entirely sure. I'd go for 0.5 months since the "nuclear plant releases a detectable amount of radioactive gases twice a month, on the average".
Don't trust me 100% on this one though, it's been quite a while since I've taken nuclear physics.

whitehorsey said:
1. A particular nuclear plant releases a detectable amount of radioactive gases twice a month on the average. Find the probability that at least 3 months will elapse before the release of the first detectable emission. What is the average time that one must wait to observe the first emission?

2. P[X ≤ x] = 1 - e-αx

3. Find the probability that at least 3 months will elapse before the release of the first detectable emission.
P[T ≥ 3] = 1 - P[T ≤ 2]
= 1 - (1 - e-.5(2)) - (1 - e-.5(1)) - (1 - e-.5(0))
= -0.0256
I ended up getting a negative number which shouldn't happen but I'm not sure what's wrong.

What is the average time that one must wait to observe the first emission?
P[T = 1] = 1 - e-.5(1)
= 0.3935?

What is the course context of this problem? Are you taking (or have taken) material about Poisson processes, exponential distributions and the link between exponential and Poisson? If so, this is a simple application of standard formulas; if not, you have a lot of reading to do. Your solution gets almost everything wrong, and it is hard to know where to begin without having you answer my questions above.

To answer your questions: The course context is continuous distributions. I have learned about Poisson for discrete and this problem comes from the exponential distribution section. As for the link between Poisson and exponential, I'm not really sure what ties them together besides the β in exponential density formula equals 1/λ and λ is Poisson parameter.

Rikardus said:
The negative number appeared since you are considering the distribution to be discrete when it's not. P(T<=3) = 1 - P(T<3) = 1-P(T<=3) since the distribution is continuous.

In 2. I think you are not calculating what you think you are... First of all P(T=1) = P(X<=1)-P(X<1) = 0.
You really are computing P(T<=1). Makes sense?

As for the average time before the first emission can't say I am entirely sure. I'd go for 0.5 months since the "nuclear plant releases a detectable amount of radioactive gases twice a month, on the average".
Don't trust me 100% on this one though, it's been quite a while since I've taken nuclear physics.

Oh! So looking at the formula again I got this:
P[T ≤ 3] = 1 - (1 - e-.5(3))
= 0.7769

Part 2:
I got 2 because its asking for average so that means the mean (E[X])?

Last edited:
whitehorsey said:
To answer your questions: The course context is continuous distributions. I have learned about Poisson for discrete and this problem comes from the exponential distribution section. As for the link between Poisson and exponential, I'm not really sure what ties them together besides the β in exponential density formula equals 1/λ and λ is Poisson parameter.

Oh! So looking at the formula again I got this:
P[T ≤ 3] = 1 - (1 - e-.5(3))
= 0.7769

Part 2:
I got 2 because its asking for average so that means the mean (E[X])?

Notation is just notation, so whether you have β or 1/λ does not matter; what matters is *meaning* and *use*. So, in this type of situation there are two types of random variables occurring: (i) N--- a count of the number of 'events' occurring in spans of time (for example, the number of releases in a month); and (ii) T---a time between successive events (for example, the time until the first event, or the time between event 1 and event 2). You need to keep them separate, although they are closely related.

So, if you have Poisson-distributed number of events, with the distribution of the number N(t) in time span (0,t) given by
$$p_k(t) \equiv P(N(t) = k) = \frac{(at)^k}{k!} e^{-at}, \; k = 0, 1, 2, \ldots$$
then the mean number is ##E N = at.## We call 'a' the *rate* parameter---you may have seen it called λ or μ or some other Greek letter, but that does not matter.

So, how does this relate to T (the time until an event occurs)? Look at ##p_0(t)##, which is the probability that NO event happens in the interval (0,t). That means that the time of the first event is later than t; that is, if T_1 is the time until the first event, we have
$$\{ N(t) = 0 \} = \{ \text{no events in }(0,t) \} = \{ T_1 \:> t \}.$$
Therefore, ##P(T_1 > t) = p_0(t) = e^{-at},## and that means that ##T_1## has an exponential distribution with mean ##1/a##, and probability density function
$$f_{T_1}(t) = a e^{-at} 1\{t > 0 \},$$
where ##1\{ t > 0 \}## is the indicator function of the set ##\{ t > 0 \} ##.

It turns out that the inter-event times are ##T_1, T_2, T_3, \ldots## are independent and have the same distribution as ##T_1##; that is, they are all exponentially distributed with mean 1/a.

In your problem you are told the mean of N(1), so you can figure out the value of 'a'. You are asked to get P(T1 > 3) and E(T1).

Ray Vickson said:
Notation is just notation, so whether you have β or 1/λ does not matter; what matters is *meaning* and *use*. So, in this type of situation there are two types of random variables occurring: (i) N--- a count of the number of 'events' occurring in spans of time (for example, the number of releases in a month); and (ii) T---a time between successive events (for example, the time until the first event, or the time between event 1 and event 2). You need to keep them separate, although they are closely related.
...

In your problem you are told the mean of N(1), so you can figure out the value of 'a'. You are asked to get P(T1 > 3) and E(T1).

I think I got it. Would 'a' be .5?

P(T1 > 3) = e-.5(3) = 0.22313016
E(T1) = 2

## What is an exponential distribution?

An exponential distribution is a probability distribution that describes the time between events in a Poisson process. It is used to model events that occur randomly over time, such as the arrival of customers at a store or the decay of a radioactive substance.

## What are the characteristics of an exponential distribution?

An exponential distribution is characterized by a single parameter, lambda (λ), which represents the average rate of events occurring per unit of time. The shape of the distribution is determined by this lambda value, with larger values resulting in a steeper curve. The distribution is also right-skewed, with a longer tail on the right side.

## What is the relationship between the exponential and Poisson distributions?

The exponential distribution is closely related to the Poisson distribution, as it is often used to model the time between events in a Poisson process. In fact, if the events in a Poisson process occur at a constant rate, then the time between events will follow an exponential distribution.

## How is an exponential distribution calculated?

The probability density function (PDF) for an exponential distribution is f(x) = λe^-λx, where x is the time between events and λ is the rate parameter. To calculate the probability of a specific value of x occurring, we can substitute that value into the PDF and solve. The cumulative distribution function (CDF) of an exponential distribution is F(x) = 1 – e^-λx, which gives the probability of x being less than or equal to a certain value.

## What are some real-world applications of exponential distributions?

Exponential distributions are commonly used in various fields, such as reliability engineering, queueing theory, and finance. They can be used to model the time between equipment failures, the waiting time in a queue, or the time until a loan is repaid. They are also useful in survival analysis, where they can estimate the probability of survival for a given time period.

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