Transcript Exponential Distribution

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:
f ( x )  e  x
• 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.
-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.
• The mean =


 x e dx    ( x)(e
0
x
0
x
)dx 
1

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.
• Variance =

1
2
2
 ( x ) f ( x)dx    2

0
- The std dev = σ =
var( x) 
1


• The probability that the waiting time is a minutes or
less when X ~ Exp(λ) is
• P(X ≤ a) =

a
0
ex dx  1  ea
- Thus the probability of waiting at least a minutes is
 a
 a
e
P(X ≥ a) = 1 – P(X ≤ a) = 1 – ( 1  e ) =
• Median waiting time
• So
tm 
tm
1
 a
  e dx
2 0
ln 2

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.
1
 x
30 1
e 20 dx  0.7769
• Soln: P(X≤ 30) = 0
20
• One can also use P(X≤a) = 1 – e
30

which is 1  e 20 = 0.777
 a
• E) Find the median waiting time to the next
posting
• Soln: this can be solved using:
1
1

e
2 0 20
tm
1

1
x
20
dx  [e

1
x
20 tm
0
]  1 e

1
tm
20
 tm
1
 1  e 20  tm  20 log e 2  13.86
2
• We could also use the formula
tm 
• Here
ln 2

ln 2
tm 
 20 ln 2  13.9
1
20