Exponential Distribution -‘waiting time density’ - the time until the next event for a Poisson...

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Exponential Distribution-‘waiting time density’

- the time until the next event for a Poisson distribution.

-The mean number of events per unit time is represented by λ.

-X ~ Exp(λ)

• In Poisson – the variable is the number of events in an interval (discrete)

• In the Exp dist – the variable is the waiting time until the next event (time is continuous).

• The pdf: • x ≥ 0 as time cannot be negative.• Exp dist is consider ‘memoryless’ – the mean

waiting time can start at any moment. If you have waited 30 mins without the next event occuring, the mean waiting time is still 10 mins.

xexf )(

-The exp dist is always a decreasing function- The mode of the exp dist is always 0- λ is a parameter affecting the decay rate.

If the mean number of events per hour is 5 then λ = 5 and the mean waiting time will be 1/5 so 12 minutes.

• The mean =

0 0

1))((

dxexdxex xx

• Variance =

- The std dev = σ =

0

212)()2(

dxxfx

1

)var(x

• The probability that the waiting time is a minutes or less when X ~ Exp(λ) is

• P(X ≤ a) =

- Thus the probability of waiting at least a minutes is

P(X ≥ a) = 1 – P(X ≤ a) = 1 – ( ) =

a ax edxe0

1

ae1 ae

• Median waiting time

• So

mt a dxe02

1

2ln

mt

An online statistics forum gets 3 postings per randomly distributed per hour.

• A) If a posting was just made, find the mean waiting time to the next posting.

• Soln: the mean psoting time is 1/λ which is 1/3. this has to be converted to minutes. It is 1/3rd of an hour which is 20 mins.

• B) If a posting was made 10 minutes ago, find the mean waiting time to the next posting.

• Soln: ‘memoryless’, so 20 mins.

• C) Find the standrad deviation of the waiting time to the next posting.

• Soln: μ = σ so also 20 mins.• D) Find the probability that the waiting time will

be 30 minutes or less.• Soln: P(X≤ 30) =

• One can also use P(X≤a) = 1 –

which is = 0.777

30

0

20

1

7769.020

1dxex

ae

20

30

1

e

• E) Find the median waiting time to the next posting

• Soln: this can be solved using:

m mt ttm

xxeedxe

0

20

1

020

1

20

1

1][20

1

2

1

86.132log2012

1 20

1

em

tte

m

• We could also use the formula

• Here

2ln

mt

9.132ln20

2012ln

mt