#Understanding Poisson distribution and exponential distribution

58 messages · Page 1 of 1 (latest)

rain bramble
#

I have difficulties to understand Poisson distribution and exponential distribution, mostly finding the λ parameter, and unit of X for each.

Here is an example: by average, a taxi passes every 5 min. Graph and find parameter for both pois(λ) , and exp(λ). Also find probability of getting a taxi within 30 min.
So can we say, if it's one taxi per 5-min, so it's 12 per 1-hour? Or it's wrong?
So parameter for pois(λ) is λ=12 ? and for exp(λ) is λ=1/12? And unit for X is hour?

And about "find probability of getting a taxi within 30 min."
Is it (exponential) P(X >= 30)≈0.91 or P(X >= 0.5)≈0.04 (4%! make sense?!) ? I mean if it's ? per hour, so the X must be in hour too ig?
How do I find λ, and unit of X to calculate cdf? Thanks.

worthy needleBOT
#
  1. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
  2. Wait patiently for a helper to come along.
  3. Once someone helps you, say thank you and close the thread with:
    +close
  4. Feel free to nominate the person for helper of the week in #helper-nominations
  5. Do not ping the mods, unless someone is breaking the rules. If there is a conflict amongst multiple helpers feel free to ping “Helper Mod”
  6. If you're happy with the help you got here, and the server overall, you can contribute financially as well:
rain bramble
#

Here is another question.
by average, there are 2 calls per hour. So the λ for pois(λ) is 2? and for exp(λ) is 1/2? If yes, then cdf(1) of exponential returns 0.39 (around 40% chance to get a call within first hour. 40%? is it correct?)

hard cradle
#

are you counting taxis or hours?

#

poisson is discrete so it makes sense X ~ Po(lambda) is the number of taxis

#

with the frequency of lambda taxis per 1 unit of time

#

exponential is continuous so it makes sense to say Z ~ Exp(lambda) shows the time until next taxi with the frequency of 1/12 units of time per taxi

#

is this about Markov chains? @rain bramble

rain bramble
# hard cradle

Hi
It's "probability of getting a taxi within 30 min.", so it's exponential.
So far, here is my knowledge(still cannot bet my life on it).
• Let X ~ Pois(3) (where λ=3), which means "1 taxi per unit of time, the unit=5min, 1 hour, etc..."
• Let Z ~ exp(3) (where λ=3, and NOT 1/3), which means "the probability of getting one taxi in an interval"

It seems, in a problem, once we found λ for Poisson(λ), we use the same λ(not the 1/λ) for the exp(λ)?

rain bramble
hard cradle
#

Here's how I read it. On average 12 taxis pass in 1 hour. If X is a continuous variable it makes sense to declare "X measures the time in between arrivals"

so probability of getting a taxi in the next half hour would be P(X< 1/2)

rain bramble
hard cradle
#

the probability that next arrival takes at most 1/2 hour

#

you are describing a memoriless process

rain bramble
hard cradle
#

, w 1-\exp(-1/24)

plush hornetBOT
hard cradle
#

units are messed up then

#

try 1 unit of time = 5 min

rain bramble
plush hornetBOT
hard cradle
#

that's more believable

rain bramble
hard cradle
#

see you are describing exactly interarrival times in a markov process :d

#

wdym it's not markov

#

ok i looked up some theory so you don't invert or anything

rain bramble
#

Let say, by average, the rate taxi is 2 per 5 min. SO we assume 5min the unit of time, so λ=2 for X ~ Pois(λ)
Now how to find the related Z ~ exp(λ)?
The λ for Z should be same 2? or 1/2?
If we assume Z ~ exp(2), then P(Z<=1) (1 means 5 min here) returns 86%
For W ~ exp(1/2), the P(W<=1) returns 40%
Where Z makes more sense
So it seems once we find λ of Pois(λ), we should use the same λ for exp(λ) ?

hard cradle
#

the poisson process has intensity lambda, therefore interarrival times are exponentially distributed with parameter lambda

rain bramble
hard cradle
#

12 taxis per hour is the intensity

#

the the time in between taxis is exponentially distributed

#

if the taxi counting follows Po(2), then interarrival times are Exp(2)

#

, w 1-exp(-2\cdot 6)

plush hornetBOT
hard cradle
#

probability of getting a taxi within 30 minutes time provided on average 2 taxis per 5 minutes

#

or

#

,w exp(-2)

plush hornetBOT
hard cradle
#

probability of having to wait for a taxi more than 5 minutes

rain bramble
hard cradle
#

yes, same paramter

#

but pay attention to units

rain bramble
hard cradle
#

if it says 12 taxis / h on average, then you can do 1 taxi per 5 min or also 12 per hour

#

e.g we can do hours as well

#

,w 1-exp(-12/2)

plush hornetBOT
hard cradle
#

fml

#

, w 1 - exp(-6)

plush hornetBOT
rain bramble
hard cradle
#

it's part of markov process theory

#

more specifically, this an example of a poisson process

rain bramble
mental jungleBOT
#

@rain bramble has given 1 rep to @hard cradle

hard cradle
#

🍺