Exponential Distribution, Mean, and Lamda confusion

[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?

Homework Equations

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?

 

[Charles Link]

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?

Homework Equations

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 $P(X \leq x)=1-e^{- \lambda x}$.

 

[Ray Vickson]

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?

Homework Equations

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.

 

[Charles Link]

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.)

 

[Of Mike and Men]

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.

 

[Charles Link]

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.

 

[Charles Link]

@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 $\nu=Np$).