# Exponential Distribution, Mean, and Lamda confusion

• Of Mike and Men
In summary, for a Poisson distribution, the mean is ## \mu ## (in days) and the probability of at least 10 days between accidents is ## P(X>=10) ##.
Of Mike and Men

## Homework Statement

Accidents at a busy intersection follow a Poisson distribution with three accidents expected in a week.
What is the probability that at least 10 days pass between accidents?

F(X) = 1- e-λx
μ = 1/λ

## The Attempt at a Solution

Let x = amount of time between accidents in days
My r.v. is continuous so x~Exponential(λ=?)

E(x) = 3/7 (in days)
Since E(x) = μ = 1/λ = 3/7
λ = 7/3

Thus x~Exponential(λ=7/3)
P(X≥10) = 1-P(X≤10) = 1- F(10) = 1- e-70/3

Answer in the back of the book says:
λ = 3/7
P(X≥10) = 1-P(X≤10) = 1- F(10) = 1- e-30/7

I'm confused why λ = 3/7 and not 7/3 if my expected value is 3/7. Shouldn't lambda, by definition, be its reciprocal?

Of Mike and Men said:

## Homework Statement

Accidents at a busy intersection follow a Poisson distribution with three accidents expected in a week.
What is the probability that at least 10 days pass between accidents?

F(X) = 1- e-λx
μ = 1/λ

## The Attempt at a Solution

Let x = amount of time between accidents in days
My r.v. is continuous so x~Exponential(λ=?)

E(x) = 3/7 (in days)
Since E(x) = μ = 1/λ = 3/7
λ = 7/3

Thus x~Exponential(λ=7/3)
P(X≥10) = 1-P(X≤10) = 1- F(10) = 1- e-70/3

Answer in the back of the book says:
λ = 3/7
P(X≥10) = 1-P(X≤10) = 1- F(10) = 1- e-30/7

I'm confused why λ = 3/7 and not 7/3 if my expected value is 3/7. Shouldn't lambda, by definition, be its reciprocal?
I agree that ## \lambda=3/7 ##. ## \lambda ## is the rate at which things occur. I think your very last line is in error though. As I understand it, the exponential distribution comes from the differential equation ## \frac{dN}{dt}=-\lambda N ##. This would give ## P(X \geq x)=e^{- \lambda x} ## and ## P(X \leq x)=1-e^{- \lambda x} ##.

Of Mike and Men said:

## Homework Statement

Accidents at a busy intersection follow a Poisson distribution with three accidents expected in a week.
What is the probability that at least 10 days pass between accidents?

F(X) = 1- e-λx
μ = 1/λ

## The Attempt at a Solution

Let x = amount of time between accidents in days
My r.v. is continuous so x~Exponential(λ=?)

E(x) = 3/7 (in days)
Since E(x) = μ = 1/λ = 3/7
λ = 7/3

Thus x~Exponential(λ=7/3)
P(X≥10) = 1-P(X≤10) = 1- F(10) = 1- e-70/3

Answer in the back of the book says:
λ = 3/7
P(X≥10) = 1-P(X≤10) = 1- F(10) = 1- e-30/7

I'm confused why λ = 3/7 and not 7/3 if my expected value is 3/7. Shouldn't lambda, by definition, be its reciprocal?

For a Poisson random variable ##N## with mean ##m## the probability mass function is
$$P(N = k) = \frac{m^k}{k!} e^{-m}, \; k = 0,1,2,\ldots$$
In your case if you measure arrivals per day your ##m = 3/7##.

On the other hand, the times between successive arrivals are iid exponentials with mean ##\mu = 1/m##, so ##\mu = 7/3## (days) in your case.

That makes perfect sense: on average, fewer than one accident per day occurs, so it is reasonable that the mean time between accidents is greater than 1 day.

I have used different names ##m## and ##\mu##, but you can relate them to some ##\lambda## if you want.

Ray Vickson said:
For a Poisson random variable ##N## with mean ##m## the probability mass function is
$$P(N = k) = \frac{m^k}{k!} e^{-m}, \; k = 0,1,2,\ldots$$
In your case if you measure arrivals per day your ##m = 3/7##.

On the other hand, the times between successive arrivals are iid exponentials with mean ##\mu = 1/m##, so ##\mu = 7/3## (days) in your case.

That makes perfect sense: on average, fewer than one accident per day occurs, so it is reasonable that the mean time between accidents is greater than 1 day.

I have used different names ##m## and ##\mu##, but you can relate them to some ##\lambda## if you want.
I can add a hint to the OP to do the calculation in this manner. For a Poisson distribution, let's use the letter ## \nu ## for the mean, i.e. ## \nu=Np ##. They give you data for a 7 day period, and ## \nu=3 ##. You need to compute ## \nu ## (which is ## \lambda ## in my post #2 and ## m ## in post #3 above) for a 10 day period . Once you have that, the formula given by @Ray Vickson should be able to lead you to the answer. Additional comment=I believe formula as given by @Ray Vickson is the result for the probability of ## k ## accidents on a single day. (Note: In some ways this method by @Ray Vickson is a better solution than the route I presented in post #2, because this way also allows you to compute the probability that there is 1 collision in the 10 day period,the probability that there are 2 collisions, the probability that there are 3 collisions, the probability that there are 4 collisions in the 10 day period, etc.)

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Ray Vickson said:
For a Poisson random variable ##N## with mean ##m## the probability mass function is
$$P(N = k) = \frac{m^k}{k!} e^{-m}, \; k = 0,1,2,\ldots$$
In your case if you measure arrivals per day your ##m = 3/7##.

On the other hand, the times between successive arrivals are iid exponentials with mean ##\mu = 1/m##, so ##\mu = 7/3## (days) in your case.

That makes perfect sense: on average, fewer than one accident per day occurs, so it is reasonable that the mean time between accidents is greater than 1 day.

I have used different names ##m## and ##\mu##, but you can relate them to some ##\lambda## if you want.

Ahhh, I think I found out where I was confused. I was confusing the continuous exponential distribution with a discrete Poisson distribution (where mean = μ = m = lamda). Rather than viewing them as their own independent definitions, I got used to viewing them as synonyms.

Thanks.

Of Mike and Men said:
Ahhh, I think I found out where I was confused. I was confusing the continuous exponential distribution with a discrete Poisson distribution (where mean = μ = m = lamda). Rather than viewing them as their own independent definitions, I got used to viewing them as synonyms.

Thanks.
It actually is advantageous to learn both methods. See my edited post #4. For the calculation at hand, using the @Ray Vickson approach, you would compute the probability that there are 0 collisions in the 10 day period.

@Of Mike and Men Notice the Poisson distribution has ## e^{- \nu} ## built into it. This is certainly necessary because ## \sum\limits_{k=0}^{+\infty} \frac{\nu^k}{k!}=e^{\nu} ##. (The result follows because of the Taylor expansion for ## e^{\nu} ##.) Thereby the Poisson distribution is correctly normalized to 1... The Poisson distribution is a limiting case of the binomial distribution, which is in fact properly normalized, so that the Poisson distribution that is computed from it would of course be properly normalized. ## \\ ## Additional comment: This problem of the intersection with accidents can be treated similar to the problem of the life of a light bulb, where the probability that a given bulb survives without an event for a time ## X \geq x ## is ## P(X \geq x)=e^{- \lambda x} ## (Comes from ## \frac{dN}{dt}=-\lambda N ##) . ## \\ ## Alternatively, it can be solved by using the Poisson distribution which gives the probability of ## k ## events in a given time span as a function of integer ## k ##, and in this case setting ## k=0 ## for the ten day time span. ## \\ ## In setting up the Poisson distribution for the case at hand, you don't know ## N ## (although you know it is a large number ,which will be larger for a longer time span in whatever you are analyzing), and you don't know ## p ## (## p ## is small), but you do know the product ## \nu=Np ## (the letters ## \lambda ## and ## m ## are also used in place of ## \nu ##), which is the mean number of events that occurs with the scenario you are analyzing over the time span that is selected. ## \\ ## (In some cases you will know both ## N ## and ## p ##, but the only requirement is that ## N ## is large and ## p ## is small and that you know the product ## \nu=Np ##).

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## 1. What is the exponential distribution?

The exponential distribution is a probability distribution that is commonly used to model the time between events that occur at a constant rate. It is a continuous distribution that takes into account the probability of an event occurring at any given point in time.

## 2. How is the mean calculated for the exponential distribution?

The mean, or average, for the exponential distribution is calculated by taking the reciprocal of the lambda parameter. This can also be written as 1/lambda or λ^-1. The lambda parameter represents the rate at which events occur, so the mean is a measure of the average time between events.

## 3. What is the relationship between lambda and the shape of the exponential distribution?

The value of lambda directly affects the shape of the exponential distribution. A larger lambda value results in a steeper curve, while a smaller lambda value results in a flatter curve. This means that a higher rate of events leads to a shorter average time between events, while a lower rate leads to a longer average time between events.

## 4. Why is there confusion between mean and lambda in the exponential distribution?

The confusion between mean and lambda often stems from the fact that both are used to represent the average time between events in the exponential distribution. However, the two are not interchangeable. While the mean is a measure of the average time between events, lambda represents the rate at which events occur.

## 5. How is the exponential distribution used in real-world applications?

The exponential distribution is commonly used in various fields, including engineering, economics, and biology. It is used to model the time between equipment failures, the time between customer arrivals, and the time between radioactive decays, among others. It can also be used to calculate probabilities and make predictions about future events based on past data.

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