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